The equation for u from the given equation.

Question

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Answer 1

Question

Chapter 39, Problem 39.74AP

(a)

To determine

The equation for $u$ from the given equation.

(a)

Expert Solution

Answer to Problem 39.74AP

The equation for

$u$ from the given equation is

$c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$ .

Explanation of Solution

Given info: The given equation is $K=(11−u2/c2−1)mc2$ .

The equation for the kinetic energy is given as,

$K=(11−u2/c2−1)mc2$

Here,

$k$ is the kinetic energy.

$m$ is the mass.

$u$ is the speed.

$c$ is the speed of light.

Rearrange the above equation for $u$ .

$Kmc2=(11−u2/c2−1)$

Let us assume $H=Kmc2$ .

$H=(11−u2/c2−1)11−u2/c2=1+H1−u2/c2=(11+H)2u2=c2(1−(11+H)2)$

Further solve the equation.

$u2=c2(1−(11+H)2)u2=c2(H2+2H+1−1(H2+2H+1))u=cH2+2H(H2+2H+1)$ (1)

Replace $H$ by $Kmc2$ in above equation.

$u=c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$ (2)

Conclusion:

Therefore, the equation for $u$ from the given equation is $c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$ .

(b)

To determine

The minimum possible value of speed and corresponding kinetic energy.

(b)

Expert Solution

Answer to Problem 39.74AP

The minimum possible value of speed can be zero and corresponding kinetic energy will also be zero.

Explanation of Solution

Given info: The given equation is $K=(11−u2/c2−1)mc2$ .

From equation (2), the expression for the speed is given as,

$u=c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$

From the above expression all the term is positive as well as the expression contains only positive sign so the minimum possible value that the speed can have according to the above expression is zero.

At zero speed the corresponding value of kinetic energy is also zero.

Conclusion:

Therefore, the minimum possible value of speed can be zero and corresponding kinetic energy will also be zero.

(c)

To determine

The maximum possible value of speed and corresponding kinetic energy.

(c)

Expert Solution

Answer to Problem 39.74AP

The maximum possible value of speed can be speed of light and corresponding kinetic energy will increases without any limit.

Explanation of Solution

Given info: The given equation is $K=(11−u2/c2−1)mc2$ .

From equation (2), the expression for the speed is given as,

$u=c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$

The maximum value of speed is equal to the speed of light according to relativistic concept if the speed becomes more than the speed of light then its energy become unstable that would not exist practically.

At this speed of light, the kinetic energy increases without any limit.

Conclusion:

Therefore, the maximum possible value of speed can be speed of light and corresponding kinetic energy will increases without any limit.

(d)

To determine

The equation for the acceleration of the particle as a function of kinetic energy and power input.

(d)

Expert Solution

Answer to Problem 39.74AP

The equation for the acceleration of the particle as a function of kinetic energy and power input is

$Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$ .

Explanation of Solution

Given info: The given equation is $K=(11−u2/c2−1)mc2$ .

From equation (1), the expression for the speed is given as,

$u=c(H)2+2(H)((H)2+2(H)+1)$

Write the expression for the acceleration of a particle.

$a=d2udt2$

Substitute $c(H)2+2(H)((H)2+2(H)+1)$ for $u$ to find $a$ .

$a=d2(c(H)2+2(H)((H)2+2(H)+1))dt2=c((H)2+2(H)((H)2+2(H)+1))−1/2(H+1(H+1)4)d(H)dt=cH1/2(H+2)1/2(H+1)2d(H)dt$

Replace $H$ by $Kmc2$ and $P$ by $dKdt$ .

$a=cH1/2(H+2)1/2(H+1)2d(Kmc2)dt=1mcH1/2(H+2)1/2(H+1)2dKdt=PmcH1/2(H+2)1/2(H+1)2$ (3)

Substitute $Kmc2$ for $H$ in above equation.

$a=Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$ (4)

Conclusion:

Therefore, the equation for the acceleration of the particle as a function of kinetic energy and power input is $Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$ .

(e)

To determine

The limiting form of the expression in part (d) at low energy and compare with the non-relativistic expression.

(e)

Expert Solution

Answer to Problem 39.74AP

The limiting form of the expression of acceleration at low energy is

$P(2mK)1/2$ and it is same as that of expression of acceleration of non-relativistic case.

Explanation of Solution

Given info: The non-relativistic expression for the acceleration is $a=P/(2mK)1/2$ .

From equation (4), the expression for the acceleration is given as,

$a=Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$

At low energy the value of $Kmc2$ is very small that $Kmc2<<<1$ . So equation becomes,

$a=Pmc(Kmc2)1/2(2)1/2(1)2=P(2mK)1/2$

Thus, the limiting form of the expression of acceleration at low energy is $P(2mK)1/2$ and it is same as that of expression of acceleration of non-relativistic case.

Conclusion:

Therefore, the limiting form of the expression of acceleration at low energy is $P(2mK)1/2$ and it is same as that of expression of acceleration of non-relativistic case.

(f)

To determine

The limiting form of the expression in part (d) at high energy and compare with the non-relativistic expression.

(f)

Expert Solution

Answer to Problem 39.74AP

The limiting form of the expression of acceleration at high energy is

$Pm2c5(K)3$ .

Explanation of Solution

Given info: The non-relativistic expression for the acceleration is $a=P/(2mK)1/2$ .

From equation (4), the expression for the acceleration is given as,

$a=Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$

At high energy the value of $Kmc2$ is very small that $Kmc2>>>1$ . So equation becomes,

$a=Pmc(Kmc2)1/2(Kmc2)1/2(Kmc2)2=Pmc(Kmc2)3=Pm2c5(K)3$

Thus, the limiting form of the expression of acceleration at low energy is $P(2mK)1/2$ and it is same as that of expression of acceleration of non-relativistic case.

Conclusion:

Therefore, the limiting form of the expression of acceleration at high energy is $Pm2c5(K)3$ .

(g)

To determine

The reason that answer to part (f) help account for the answer to part (c) at constant input power.

(g)

Expert Solution

Answer to Problem 39.74AP

The acceleration of the particle is very less at high energy that gives the velocity of the particle a constant value.

Explanation of Solution

Given info: The non-relativistic expression for the acceleration is $a=P/(2mK)1/2$ .

From the answer of part (f) the expression for the acceleration is,

$a=Pm2c5(K)3$

Here,

$p$ is the power.

$m$ is the mass of a particle.

$K$ is the kinetic energy of the particle.

In part (c), the speed at high energy approaches to the speed of light. But from the acceleration equation if the energy is imparted to the particle at constant input power the acceleration is steeply decreases because the acceleration is inversely proportional to the cube root of the kinetic energy. So at high energy acceleration is very less and the velocity of the particle approaches to a constant value as indicate in part (c).

Conclusion:

Therefore, the acceleration of the particle is very less at high energy that gives the velocity of the particle a constant value.

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Answer 2

Question

Chapter 39, Problem 39.74AP

(a)

To determine

The equation for $u$ from the given equation.

(a)

Expert Solution

Answer to Problem 39.74AP

The equation for

$u$ from the given equation is

$c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$ .

Explanation of Solution

Given info: The given equation is $K=(11−u2/c2−1)mc2$ .

The equation for the kinetic energy is given as,

$K=(11−u2/c2−1)mc2$

Here,

$k$ is the kinetic energy.

$m$ is the mass.

$u$ is the speed.

$c$ is the speed of light.

Rearrange the above equation for $u$ .

$Kmc2=(11−u2/c2−1)$

Let us assume $H=Kmc2$ .

$H=(11−u2/c2−1)11−u2/c2=1+H1−u2/c2=(11+H)2u2=c2(1−(11+H)2)$

Further solve the equation.

$u2=c2(1−(11+H)2)u2=c2(H2+2H+1−1(H2+2H+1))u=cH2+2H(H2+2H+1)$ (1)

Replace $H$ by $Kmc2$ in above equation.

$u=c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$ (2)

Conclusion:

Therefore, the equation for $u$ from the given equation is $c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$ .

(b)

To determine

The minimum possible value of speed and corresponding kinetic energy.

(b)

Expert Solution

Answer to Problem 39.74AP

The minimum possible value of speed can be zero and corresponding kinetic energy will also be zero.

Explanation of Solution

Given info: The given equation is $K=(11−u2/c2−1)mc2$ .

From equation (2), the expression for the speed is given as,

$u=c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$

From the above expression all the term is positive as well as the expression contains only positive sign so the minimum possible value that the speed can have according to the above expression is zero.

At zero speed the corresponding value of kinetic energy is also zero.

Conclusion:

Therefore, the minimum possible value of speed can be zero and corresponding kinetic energy will also be zero.

(c)

To determine

The maximum possible value of speed and corresponding kinetic energy.

(c)

Expert Solution

Answer to Problem 39.74AP

The maximum possible value of speed can be speed of light and corresponding kinetic energy will increases without any limit.

Explanation of Solution

Given info: The given equation is $K=(11−u2/c2−1)mc2$ .

From equation (2), the expression for the speed is given as,

$u=c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$

The maximum value of speed is equal to the speed of light according to relativistic concept if the speed becomes more than the speed of light then its energy become unstable that would not exist practically.

At this speed of light, the kinetic energy increases without any limit.

Conclusion:

Therefore, the maximum possible value of speed can be speed of light and corresponding kinetic energy will increases without any limit.

(d)

To determine

The equation for the acceleration of the particle as a function of kinetic energy and power input.

(d)

Expert Solution

Answer to Problem 39.74AP

The equation for the acceleration of the particle as a function of kinetic energy and power input is

$Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$ .

Explanation of Solution

Given info: The given equation is $K=(11−u2/c2−1)mc2$ .

From equation (1), the expression for the speed is given as,

$u=c(H)2+2(H)((H)2+2(H)+1)$

Write the expression for the acceleration of a particle.

$a=d2udt2$

Substitute $c(H)2+2(H)((H)2+2(H)+1)$ for $u$ to find $a$ .

$a=d2(c(H)2+2(H)((H)2+2(H)+1))dt2=c((H)2+2(H)((H)2+2(H)+1))−1/2(H+1(H+1)4)d(H)dt=cH1/2(H+2)1/2(H+1)2d(H)dt$

Replace $H$ by $Kmc2$ and $P$ by $dKdt$ .

$a=cH1/2(H+2)1/2(H+1)2d(Kmc2)dt=1mcH1/2(H+2)1/2(H+1)2dKdt=PmcH1/2(H+2)1/2(H+1)2$ (3)

Substitute $Kmc2$ for $H$ in above equation.

$a=Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$ (4)

Conclusion:

Therefore, the equation for the acceleration of the particle as a function of kinetic energy and power input is $Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$ .

(e)

To determine

The limiting form of the expression in part (d) at low energy and compare with the non-relativistic expression.

(e)

Expert Solution

Answer to Problem 39.74AP

The limiting form of the expression of acceleration at low energy is

$P(2mK)1/2$ and it is same as that of expression of acceleration of non-relativistic case.

Explanation of Solution

Given info: The non-relativistic expression for the acceleration is $a=P/(2mK)1/2$ .

From equation (4), the expression for the acceleration is given as,

$a=Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$

At low energy the value of $Kmc2$ is very small that $Kmc2<<<1$ . So equation becomes,

$a=Pmc(Kmc2)1/2(2)1/2(1)2=P(2mK)1/2$

Thus, the limiting form of the expression of acceleration at low energy is $P(2mK)1/2$ and it is same as that of expression of acceleration of non-relativistic case.

Conclusion:

Therefore, the limiting form of the expression of acceleration at low energy is $P(2mK)1/2$ and it is same as that of expression of acceleration of non-relativistic case.

(f)

To determine

The limiting form of the expression in part (d) at high energy and compare with the non-relativistic expression.

(f)

Expert Solution

Answer to Problem 39.74AP

The limiting form of the expression of acceleration at high energy is

$Pm2c5(K)3$ .

Explanation of Solution

Given info: The non-relativistic expression for the acceleration is $a=P/(2mK)1/2$ .

From equation (4), the expression for the acceleration is given as,

$a=Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$

At high energy the value of $Kmc2$ is very small that $Kmc2>>>1$ . So equation becomes,

$a=Pmc(Kmc2)1/2(Kmc2)1/2(Kmc2)2=Pmc(Kmc2)3=Pm2c5(K)3$

Thus, the limiting form of the expression of acceleration at low energy is $P(2mK)1/2$ and it is same as that of expression of acceleration of non-relativistic case.

Conclusion:

Therefore, the limiting form of the expression of acceleration at high energy is $Pm2c5(K)3$ .

(g)

To determine

The reason that answer to part (f) help account for the answer to part (c) at constant input power.

(g)

Expert Solution

Answer to Problem 39.74AP

The acceleration of the particle is very less at high energy that gives the velocity of the particle a constant value.

Explanation of Solution

Given info: The non-relativistic expression for the acceleration is $a=P/(2mK)1/2$ .

From the answer of part (f) the expression for the acceleration is,

$a=Pm2c5(K)3$

Here,

$p$ is the power.

$m$ is the mass of a particle.

$K$ is the kinetic energy of the particle.

In part (c), the speed at high energy approaches to the speed of light. But from the acceleration equation if the energy is imparted to the particle at constant input power the acceleration is steeply decreases because the acceleration is inversely proportional to the cube root of the kinetic energy. So at high energy acceleration is very less and the velocity of the particle approaches to a constant value as indicate in part (c).

Conclusion:

Therefore, the acceleration of the particle is very less at high energy that gives the velocity of the particle a constant value.

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Answer 3

Question

Chapter 39, Problem 39.74AP

(a)

To determine

The equation for $u$ from the given equation.

(a)

Expert Solution

Answer to Problem 39.74AP

The equation for

$u$ from the given equation is

$c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$ .

Explanation of Solution

Given info: The given equation is $K=(11−u2/c2−1)mc2$ .

The equation for the kinetic energy is given as,

$K=(11−u2/c2−1)mc2$

Here,

$k$ is the kinetic energy.

$m$ is the mass.

$u$ is the speed.

$c$ is the speed of light.

Rearrange the above equation for $u$ .

$Kmc2=(11−u2/c2−1)$

Let us assume $H=Kmc2$ .

$H=(11−u2/c2−1)11−u2/c2=1+H1−u2/c2=(11+H)2u2=c2(1−(11+H)2)$

Further solve the equation.

$u2=c2(1−(11+H)2)u2=c2(H2+2H+1−1(H2+2H+1))u=cH2+2H(H2+2H+1)$ (1)

Replace $H$ by $Kmc2$ in above equation.

$u=c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$ (2)

Conclusion:

Therefore, the equation for $u$ from the given equation is $c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$ .

(b)

To determine

The minimum possible value of speed and corresponding kinetic energy.

(b)

Expert Solution

Answer to Problem 39.74AP

The minimum possible value of speed can be zero and corresponding kinetic energy will also be zero.

Explanation of Solution

Given info: The given equation is $K=(11−u2/c2−1)mc2$ .

From equation (2), the expression for the speed is given as,

$u=c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$

From the above expression all the term is positive as well as the expression contains only positive sign so the minimum possible value that the speed can have according to the above expression is zero.

At zero speed the corresponding value of kinetic energy is also zero.

Conclusion:

Therefore, the minimum possible value of speed can be zero and corresponding kinetic energy will also be zero.

(c)

To determine

The maximum possible value of speed and corresponding kinetic energy.

(c)

Expert Solution

Answer to Problem 39.74AP

The maximum possible value of speed can be speed of light and corresponding kinetic energy will increases without any limit.

Explanation of Solution

Given info: The given equation is $K=(11−u2/c2−1)mc2$ .

From equation (2), the expression for the speed is given as,

$u=c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$

The maximum value of speed is equal to the speed of light according to relativistic concept if the speed becomes more than the speed of light then its energy become unstable that would not exist practically.

At this speed of light, the kinetic energy increases without any limit.

Conclusion:

Therefore, the maximum possible value of speed can be speed of light and corresponding kinetic energy will increases without any limit.

(d)

To determine

The equation for the acceleration of the particle as a function of kinetic energy and power input.

(d)

Expert Solution

Answer to Problem 39.74AP

The equation for the acceleration of the particle as a function of kinetic energy and power input is

$Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$ .

Explanation of Solution

Given info: The given equation is $K=(11−u2/c2−1)mc2$ .

From equation (1), the expression for the speed is given as,

$u=c(H)2+2(H)((H)2+2(H)+1)$

Write the expression for the acceleration of a particle.

$a=d2udt2$

Substitute $c(H)2+2(H)((H)2+2(H)+1)$ for $u$ to find $a$ .

$a=d2(c(H)2+2(H)((H)2+2(H)+1))dt2=c((H)2+2(H)((H)2+2(H)+1))−1/2(H+1(H+1)4)d(H)dt=cH1/2(H+2)1/2(H+1)2d(H)dt$

Replace $H$ by $Kmc2$ and $P$ by $dKdt$ .

$a=cH1/2(H+2)1/2(H+1)2d(Kmc2)dt=1mcH1/2(H+2)1/2(H+1)2dKdt=PmcH1/2(H+2)1/2(H+1)2$ (3)

Substitute $Kmc2$ for $H$ in above equation.

$a=Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$ (4)

Conclusion:

Therefore, the equation for the acceleration of the particle as a function of kinetic energy and power input is $Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$ .

(e)

To determine

The limiting form of the expression in part (d) at low energy and compare with the non-relativistic expression.

(e)

Expert Solution

Answer to Problem 39.74AP

The limiting form of the expression of acceleration at low energy is

$P(2mK)1/2$ and it is same as that of expression of acceleration of non-relativistic case.

Explanation of Solution

Given info: The non-relativistic expression for the acceleration is $a=P/(2mK)1/2$ .

From equation (4), the expression for the acceleration is given as,

$a=Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$

At low energy the value of $Kmc2$ is very small that $Kmc2<<<1$ . So equation becomes,

$a=Pmc(Kmc2)1/2(2)1/2(1)2=P(2mK)1/2$

Thus, the limiting form of the expression of acceleration at low energy is $P(2mK)1/2$ and it is same as that of expression of acceleration of non-relativistic case.

Conclusion:

Therefore, the limiting form of the expression of acceleration at low energy is $P(2mK)1/2$ and it is same as that of expression of acceleration of non-relativistic case.

(f)

To determine

The limiting form of the expression in part (d) at high energy and compare with the non-relativistic expression.

(f)

Expert Solution

Answer to Problem 39.74AP

The limiting form of the expression of acceleration at high energy is

$Pm2c5(K)3$ .

Explanation of Solution

Given info: The non-relativistic expression for the acceleration is $a=P/(2mK)1/2$ .

From equation (4), the expression for the acceleration is given as,

$a=Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$

At high energy the value of $Kmc2$ is very small that $Kmc2>>>1$ . So equation becomes,

$a=Pmc(Kmc2)1/2(Kmc2)1/2(Kmc2)2=Pmc(Kmc2)3=Pm2c5(K)3$

Thus, the limiting form of the expression of acceleration at low energy is $P(2mK)1/2$ and it is same as that of expression of acceleration of non-relativistic case.

Conclusion:

Therefore, the limiting form of the expression of acceleration at high energy is $Pm2c5(K)3$ .

(g)

To determine

The reason that answer to part (f) help account for the answer to part (c) at constant input power.

(g)

Expert Solution

Answer to Problem 39.74AP

The acceleration of the particle is very less at high energy that gives the velocity of the particle a constant value.

Explanation of Solution

Given info: The non-relativistic expression for the acceleration is $a=P/(2mK)1/2$ .

From the answer of part (f) the expression for the acceleration is,

$a=Pm2c5(K)3$

Here,

$p$ is the power.

$m$ is the mass of a particle.

$K$ is the kinetic energy of the particle.

In part (c), the speed at high energy approaches to the speed of light. But from the acceleration equation if the energy is imparted to the particle at constant input power the acceleration is steeply decreases because the acceleration is inversely proportional to the cube root of the kinetic energy. So at high energy acceleration is very less and the velocity of the particle approaches to a constant value as indicate in part (c).

Conclusion:

Therefore, the acceleration of the particle is very less at high energy that gives the velocity of the particle a constant value.

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Answer 4

Question

Chapter 39, Problem 39.74AP

(a)

To determine

The equation for $u$ from the given equation.

(a)

Expert Solution

Answer to Problem 39.74AP

The equation for

$u$ from the given equation is

$c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$ .

Explanation of Solution

Given info: The given equation is $K=(11−u2/c2−1)mc2$ .

The equation for the kinetic energy is given as,

$K=(11−u2/c2−1)mc2$

Here,

$k$ is the kinetic energy.

$m$ is the mass.

$u$ is the speed.

$c$ is the speed of light.

Rearrange the above equation for $u$ .

$Kmc2=(11−u2/c2−1)$

Let us assume $H=Kmc2$ .

$H=(11−u2/c2−1)11−u2/c2=1+H1−u2/c2=(11+H)2u2=c2(1−(11+H)2)$

Further solve the equation.

$u2=c2(1−(11+H)2)u2=c2(H2+2H+1−1(H2+2H+1))u=cH2+2H(H2+2H+1)$ (1)

Replace $H$ by $Kmc2$ in above equation.

$u=c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$ (2)

Conclusion:

Therefore, the equation for $u$ from the given equation is $c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$ .

(b)

To determine

The minimum possible value of speed and corresponding kinetic energy.

(b)

Expert Solution

Answer to Problem 39.74AP

The minimum possible value of speed can be zero and corresponding kinetic energy will also be zero.

Explanation of Solution

Given info: The given equation is $K=(11−u2/c2−1)mc2$ .

From equation (2), the expression for the speed is given as,

$u=c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$

From the above expression all the term is positive as well as the expression contains only positive sign so the minimum possible value that the speed can have according to the above expression is zero.

At zero speed the corresponding value of kinetic energy is also zero.

Conclusion:

Therefore, the minimum possible value of speed can be zero and corresponding kinetic energy will also be zero.

(c)

To determine

The maximum possible value of speed and corresponding kinetic energy.

(c)

Expert Solution

Answer to Problem 39.74AP

The maximum possible value of speed can be speed of light and corresponding kinetic energy will increases without any limit.

Explanation of Solution

Given info: The given equation is $K=(11−u2/c2−1)mc2$ .

From equation (2), the expression for the speed is given as,

$u=c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$

The maximum value of speed is equal to the speed of light according to relativistic concept if the speed becomes more than the speed of light then its energy become unstable that would not exist practically.

At this speed of light, the kinetic energy increases without any limit.

Conclusion:

Therefore, the maximum possible value of speed can be speed of light and corresponding kinetic energy will increases without any limit.

(d)

To determine

The equation for the acceleration of the particle as a function of kinetic energy and power input.

(d)

Expert Solution

Answer to Problem 39.74AP

The equation for the acceleration of the particle as a function of kinetic energy and power input is

$Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$ .

Explanation of Solution

Given info: The given equation is $K=(11−u2/c2−1)mc2$ .

From equation (1), the expression for the speed is given as,

$u=c(H)2+2(H)((H)2+2(H)+1)$

Write the expression for the acceleration of a particle.

$a=d2udt2$

Substitute $c(H)2+2(H)((H)2+2(H)+1)$ for $u$ to find $a$ .

$a=d2(c(H)2+2(H)((H)2+2(H)+1))dt2=c((H)2+2(H)((H)2+2(H)+1))−1/2(H+1(H+1)4)d(H)dt=cH1/2(H+2)1/2(H+1)2d(H)dt$

Replace $H$ by $Kmc2$ and $P$ by $dKdt$ .

$a=cH1/2(H+2)1/2(H+1)2d(Kmc2)dt=1mcH1/2(H+2)1/2(H+1)2dKdt=PmcH1/2(H+2)1/2(H+1)2$ (3)

Substitute $Kmc2$ for $H$ in above equation.

$a=Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$ (4)

Conclusion:

Therefore, the equation for the acceleration of the particle as a function of kinetic energy and power input is $Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$ .

(e)

To determine

The limiting form of the expression in part (d) at low energy and compare with the non-relativistic expression.

(e)

Expert Solution

Answer to Problem 39.74AP

The limiting form of the expression of acceleration at low energy is

$P(2mK)1/2$ and it is same as that of expression of acceleration of non-relativistic case.

Explanation of Solution

Given info: The non-relativistic expression for the acceleration is $a=P/(2mK)1/2$ .

From equation (4), the expression for the acceleration is given as,

$a=Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$

At low energy the value of $Kmc2$ is very small that $Kmc2<<<1$ . So equation becomes,

$a=Pmc(Kmc2)1/2(2)1/2(1)2=P(2mK)1/2$

Thus, the limiting form of the expression of acceleration at low energy is $P(2mK)1/2$ and it is same as that of expression of acceleration of non-relativistic case.

Conclusion:

Therefore, the limiting form of the expression of acceleration at low energy is $P(2mK)1/2$ and it is same as that of expression of acceleration of non-relativistic case.

(f)

To determine

The limiting form of the expression in part (d) at high energy and compare with the non-relativistic expression.

(f)

Expert Solution

Answer to Problem 39.74AP

The limiting form of the expression of acceleration at high energy is

$Pm2c5(K)3$ .

Explanation of Solution

Given info: The non-relativistic expression for the acceleration is $a=P/(2mK)1/2$ .

From equation (4), the expression for the acceleration is given as,

$a=Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$

At high energy the value of $Kmc2$ is very small that $Kmc2>>>1$ . So equation becomes,

$a=Pmc(Kmc2)1/2(Kmc2)1/2(Kmc2)2=Pmc(Kmc2)3=Pm2c5(K)3$

Thus, the limiting form of the expression of acceleration at low energy is $P(2mK)1/2$ and it is same as that of expression of acceleration of non-relativistic case.

Conclusion:

Therefore, the limiting form of the expression of acceleration at high energy is $Pm2c5(K)3$ .

(g)

To determine

The reason that answer to part (f) help account for the answer to part (c) at constant input power.

(g)

Expert Solution

Answer to Problem 39.74AP

The acceleration of the particle is very less at high energy that gives the velocity of the particle a constant value.

Explanation of Solution

Given info: The non-relativistic expression for the acceleration is $a=P/(2mK)1/2$ .

From the answer of part (f) the expression for the acceleration is,

$a=Pm2c5(K)3$

Here,

$p$ is the power.

$m$ is the mass of a particle.

$K$ is the kinetic energy of the particle.

In part (c), the speed at high energy approaches to the speed of light. But from the acceleration equation if the energy is imparted to the particle at constant input power the acceleration is steeply decreases because the acceleration is inversely proportional to the cube root of the kinetic energy. So at high energy acceleration is very less and the velocity of the particle approaches to a constant value as indicate in part (c).

Conclusion:

Therefore, the acceleration of the particle is very less at high energy that gives the velocity of the particle a constant value.

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Answer 5

Chapter 39, Problem 39.74AP

(a)

To determine

The equation for $u$ from the given equation.

(a)

Expert Solution

Answer to Problem 39.74AP

The equation for

$u$ from the given equation is

$c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$ .

Explanation of Solution

Given info: The given equation is $K=(11−u2/c2−1)mc2$ .

The equation for the kinetic energy is given as,

$K=(11−u2/c2−1)mc2$

Here,

$k$ is the kinetic energy.

$m$ is the mass.

$u$ is the speed.

$c$ is the speed of light.

Rearrange the above equation for $u$ .

$Kmc2=(11−u2/c2−1)$

Let us assume $H=Kmc2$ .

$H=(11−u2/c2−1)11−u2/c2=1+H1−u2/c2=(11+H)2u2=c2(1−(11+H)2)$

Further solve the equation.

$u2=c2(1−(11+H)2)u2=c2(H2+2H+1−1(H2+2H+1))u=cH2+2H(H2+2H+1)$ (1)

Replace $H$ by $Kmc2$ in above equation.

$u=c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$ (2)

Conclusion:

Therefore, the equation for $u$ from the given equation is $c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$ .

(b)

To determine

The minimum possible value of speed and corresponding kinetic energy.

(b)

Expert Solution

Answer to Problem 39.74AP

The minimum possible value of speed can be zero and corresponding kinetic energy will also be zero.

Explanation of Solution

Given info: The given equation is $K=(11−u2/c2−1)mc2$ .

From equation (2), the expression for the speed is given as,

$u=c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$

From the above expression all the term is positive as well as the expression contains only positive sign so the minimum possible value that the speed can have according to the above expression is zero.

At zero speed the corresponding value of kinetic energy is also zero.

Conclusion:

Therefore, the minimum possible value of speed can be zero and corresponding kinetic energy will also be zero.

(c)

To determine

The maximum possible value of speed and corresponding kinetic energy.

(c)

Expert Solution

Answer to Problem 39.74AP

The maximum possible value of speed can be speed of light and corresponding kinetic energy will increases without any limit.

Explanation of Solution

Given info: The given equation is $K=(11−u2/c2−1)mc2$ .

From equation (2), the expression for the speed is given as,

$u=c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$

The maximum value of speed is equal to the speed of light according to relativistic concept if the speed becomes more than the speed of light then its energy become unstable that would not exist practically.

At this speed of light, the kinetic energy increases without any limit.

Conclusion:

Therefore, the maximum possible value of speed can be speed of light and corresponding kinetic energy will increases without any limit.

(d)

To determine

The equation for the acceleration of the particle as a function of kinetic energy and power input.

(d)

Expert Solution

Answer to Problem 39.74AP

The equation for the acceleration of the particle as a function of kinetic energy and power input is

$Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$ .

Explanation of Solution

Given info: The given equation is $K=(11−u2/c2−1)mc2$ .

From equation (1), the expression for the speed is given as,

$u=c(H)2+2(H)((H)2+2(H)+1)$

Write the expression for the acceleration of a particle.

$a=d2udt2$

Substitute $c(H)2+2(H)((H)2+2(H)+1)$ for $u$ to find $a$ .

$a=d2(c(H)2+2(H)((H)2+2(H)+1))dt2=c((H)2+2(H)((H)2+2(H)+1))−1/2(H+1(H+1)4)d(H)dt=cH1/2(H+2)1/2(H+1)2d(H)dt$

Replace $H$ by $Kmc2$ and $P$ by $dKdt$ .

$a=cH1/2(H+2)1/2(H+1)2d(Kmc2)dt=1mcH1/2(H+2)1/2(H+1)2dKdt=PmcH1/2(H+2)1/2(H+1)2$ (3)

Substitute $Kmc2$ for $H$ in above equation.

$a=Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$ (4)

Conclusion:

Therefore, the equation for the acceleration of the particle as a function of kinetic energy and power input is $Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$ .

(e)

To determine

The limiting form of the expression in part (d) at low energy and compare with the non-relativistic expression.

(e)

Expert Solution

Answer to Problem 39.74AP

The limiting form of the expression of acceleration at low energy is

$P(2mK)1/2$ and it is same as that of expression of acceleration of non-relativistic case.

Explanation of Solution

Given info: The non-relativistic expression for the acceleration is $a=P/(2mK)1/2$ .

From equation (4), the expression for the acceleration is given as,

$a=Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$

At low energy the value of $Kmc2$ is very small that $Kmc2<<<1$ . So equation becomes,

$a=Pmc(Kmc2)1/2(2)1/2(1)2=P(2mK)1/2$

Thus, the limiting form of the expression of acceleration at low energy is $P(2mK)1/2$ and it is same as that of expression of acceleration of non-relativistic case.

Conclusion:

Therefore, the limiting form of the expression of acceleration at low energy is $P(2mK)1/2$ and it is same as that of expression of acceleration of non-relativistic case.

(f)

To determine

The limiting form of the expression in part (d) at high energy and compare with the non-relativistic expression.

(f)

Expert Solution

Answer to Problem 39.74AP

The limiting form of the expression of acceleration at high energy is

$Pm2c5(K)3$ .

Explanation of Solution

Given info: The non-relativistic expression for the acceleration is $a=P/(2mK)1/2$ .

From equation (4), the expression for the acceleration is given as,

$a=Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$

At high energy the value of $Kmc2$ is very small that $Kmc2>>>1$ . So equation becomes,

$a=Pmc(Kmc2)1/2(Kmc2)1/2(Kmc2)2=Pmc(Kmc2)3=Pm2c5(K)3$

Thus, the limiting form of the expression of acceleration at low energy is $P(2mK)1/2$ and it is same as that of expression of acceleration of non-relativistic case.

Conclusion:

Therefore, the limiting form of the expression of acceleration at high energy is $Pm2c5(K)3$ .

(g)

To determine

The reason that answer to part (f) help account for the answer to part (c) at constant input power.

(g)

Expert Solution

Answer to Problem 39.74AP

The acceleration of the particle is very less at high energy that gives the velocity of the particle a constant value.

Explanation of Solution

Given info: The non-relativistic expression for the acceleration is $a=P/(2mK)1/2$ .

From the answer of part (f) the expression for the acceleration is,

$a=Pm2c5(K)3$

Here,

$p$ is the power.

$m$ is the mass of a particle.

$K$ is the kinetic energy of the particle.

In part (c), the speed at high energy approaches to the speed of light. But from the acceleration equation if the energy is imparted to the particle at constant input power the acceleration is steeply decreases because the acceleration is inversely proportional to the cube root of the kinetic energy. So at high energy acceleration is very less and the velocity of the particle approaches to a constant value as indicate in part (c).

Conclusion:

Therefore, the acceleration of the particle is very less at high energy that gives the velocity of the particle a constant value.

Answer 6

Chapter 39, Problem 39.74AP

(a)

To determine

The equation for $u$ from the given equation.

(a)

Expert Solution

Answer to Problem 39.74AP

The equation for

$u$ from the given equation is

$c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$ .

Explanation of Solution

Given info: The given equation is $K=(11−u2/c2−1)mc2$ .

The equation for the kinetic energy is given as,

$K=(11−u2/c2−1)mc2$

Here,

$k$ is the kinetic energy.

$m$ is the mass.

$u$ is the speed.

$c$ is the speed of light.

Rearrange the above equation for $u$ .

$Kmc2=(11−u2/c2−1)$

Let us assume $H=Kmc2$ .

$H=(11−u2/c2−1)11−u2/c2=1+H1−u2/c2=(11+H)2u2=c2(1−(11+H)2)$

Further solve the equation.

$u2=c2(1−(11+H)2)u2=c2(H2+2H+1−1(H2+2H+1))u=cH2+2H(H2+2H+1)$ (1)

Replace $H$ by $Kmc2$ in above equation.

$u=c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$ (2)

Conclusion:

Therefore, the equation for $u$ from the given equation is $c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$ .

Answer 7

Chapter 39, Problem 39.74AP

(a)

To determine

The equation for $u$ from the given equation.

Answer 8

(a)

Expert Solution

Answer to Problem 39.74AP

The equation for

$u$ from the given equation is

$c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given info: The given equation is $K=(11−u2/c2−1)mc2$ .

The equation for the kinetic energy is given as,

$K=(11−u2/c2−1)mc2$

Here,

$k$ is the kinetic energy.

$m$ is the mass.

$u$ is the speed.

$c$ is the speed of light.

Rearrange the above equation for $u$ .

$Kmc2=(11−u2/c2−1)$

Let us assume $H=Kmc2$ .

$H=(11−u2/c2−1)11−u2/c2=1+H1−u2/c2=(11+H)2u2=c2(1−(11+H)2)$

Further solve the equation.

$u2=c2(1−(11+H)2)u2=c2(H2+2H+1−1(H2+2H+1))u=cH2+2H(H2+2H+1)$ (1)

Replace $H$ by $Kmc2$ in above equation.

$u=c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$ (2)

Conclusion:

Therefore, the equation for $u$ from the given equation is $c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$ .

Answer 13

(b)

To determine

The minimum possible value of speed and corresponding kinetic energy.

(b)

Expert Solution

Answer to Problem 39.74AP

The minimum possible value of speed can be zero and corresponding kinetic energy will also be zero.

Explanation of Solution

Given info: The given equation is $K=(11−u2/c2−1)mc2$ .

From equation (2), the expression for the speed is given as,

$u=c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$

From the above expression all the term is positive as well as the expression contains only positive sign so the minimum possible value that the speed can have according to the above expression is zero.

At zero speed the corresponding value of kinetic energy is also zero.

Conclusion:

Therefore, the minimum possible value of speed can be zero and corresponding kinetic energy will also be zero.

Answer 14

(b)

To determine

The minimum possible value of speed and corresponding kinetic energy.

Answer 15

(b)

Expert Solution

Answer to Problem 39.74AP

The minimum possible value of speed can be zero and corresponding kinetic energy will also be zero.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given info: The given equation is $K=(11−u2/c2−1)mc2$ .

From equation (2), the expression for the speed is given as,

$u=c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$

From the above expression all the term is positive as well as the expression contains only positive sign so the minimum possible value that the speed can have according to the above expression is zero.

At zero speed the corresponding value of kinetic energy is also zero.

Conclusion:

Therefore, the minimum possible value of speed can be zero and corresponding kinetic energy will also be zero.

Answer 20

(c)

To determine

The maximum possible value of speed and corresponding kinetic energy.

(c)

Expert Solution

Answer to Problem 39.74AP

The maximum possible value of speed can be speed of light and corresponding kinetic energy will increases without any limit.

Explanation of Solution

Given info: The given equation is $K=(11−u2/c2−1)mc2$ .

From equation (2), the expression for the speed is given as,

$u=c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$

The maximum value of speed is equal to the speed of light according to relativistic concept if the speed becomes more than the speed of light then its energy become unstable that would not exist practically.

At this speed of light, the kinetic energy increases without any limit.

Conclusion:

Therefore, the maximum possible value of speed can be speed of light and corresponding kinetic energy will increases without any limit.

Answer 21

(c)

To determine

The maximum possible value of speed and corresponding kinetic energy.

Answer 22

(c)

Expert Solution

Answer to Problem 39.74AP

The maximum possible value of speed can be speed of light and corresponding kinetic energy will increases without any limit.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given info: The given equation is $K=(11−u2/c2−1)mc2$ .

From equation (2), the expression for the speed is given as,

$u=c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)$

The maximum value of speed is equal to the speed of light according to relativistic concept if the speed becomes more than the speed of light then its energy become unstable that would not exist practically.

At this speed of light, the kinetic energy increases without any limit.

Conclusion:

Therefore, the maximum possible value of speed can be speed of light and corresponding kinetic energy will increases without any limit.

Answer 27

(d)

To determine

The equation for the acceleration of the particle as a function of kinetic energy and power input.

(d)

Expert Solution

Answer to Problem 39.74AP

The equation for the acceleration of the particle as a function of kinetic energy and power input is

$Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$ .

Explanation of Solution

Given info: The given equation is $K=(11−u2/c2−1)mc2$ .

From equation (1), the expression for the speed is given as,

$u=c(H)2+2(H)((H)2+2(H)+1)$

Write the expression for the acceleration of a particle.

$a=d2udt2$

Substitute $c(H)2+2(H)((H)2+2(H)+1)$ for $u$ to find $a$ .

$a=d2(c(H)2+2(H)((H)2+2(H)+1))dt2=c((H)2+2(H)((H)2+2(H)+1))−1/2(H+1(H+1)4)d(H)dt=cH1/2(H+2)1/2(H+1)2d(H)dt$

Replace $H$ by $Kmc2$ and $P$ by $dKdt$ .

$a=cH1/2(H+2)1/2(H+1)2d(Kmc2)dt=1mcH1/2(H+2)1/2(H+1)2dKdt=PmcH1/2(H+2)1/2(H+1)2$ (3)

Substitute $Kmc2$ for $H$ in above equation.

$a=Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$ (4)

Conclusion:

Therefore, the equation for the acceleration of the particle as a function of kinetic energy and power input is $Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$ .

Answer 28

(d)

To determine

The equation for the acceleration of the particle as a function of kinetic energy and power input.

Answer 29

(d)

Expert Solution

Answer to Problem 39.74AP

The equation for the acceleration of the particle as a function of kinetic energy and power input is

$Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$ .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Given info: The given equation is $K=(11−u2/c2−1)mc2$ .

From equation (1), the expression for the speed is given as,

$u=c(H)2+2(H)((H)2+2(H)+1)$

Write the expression for the acceleration of a particle.

$a=d2udt2$

Substitute $c(H)2+2(H)((H)2+2(H)+1)$ for $u$ to find $a$ .

$a=d2(c(H)2+2(H)((H)2+2(H)+1))dt2=c((H)2+2(H)((H)2+2(H)+1))−1/2(H+1(H+1)4)d(H)dt=cH1/2(H+2)1/2(H+1)2d(H)dt$

Replace $H$ by $Kmc2$ and $P$ by $dKdt$ .

$a=cH1/2(H+2)1/2(H+1)2d(Kmc2)dt=1mcH1/2(H+2)1/2(H+1)2dKdt=PmcH1/2(H+2)1/2(H+1)2$ (3)

Substitute $Kmc2$ for $H$ in above equation.

$a=Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$ (4)

Conclusion:

Therefore, the equation for the acceleration of the particle as a function of kinetic energy and power input is $Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$ .

Answer 34

(e)

To determine

The limiting form of the expression in part (d) at low energy and compare with the non-relativistic expression.

(e)

Expert Solution

Answer to Problem 39.74AP

The limiting form of the expression of acceleration at low energy is

$P(2mK)1/2$ and it is same as that of expression of acceleration of non-relativistic case.

Explanation of Solution

Given info: The non-relativistic expression for the acceleration is $a=P/(2mK)1/2$ .

From equation (4), the expression for the acceleration is given as,

$a=Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$

At low energy the value of $Kmc2$ is very small that $Kmc2<<<1$ . So equation becomes,

$a=Pmc(Kmc2)1/2(2)1/2(1)2=P(2mK)1/2$

Thus, the limiting form of the expression of acceleration at low energy is $P(2mK)1/2$ and it is same as that of expression of acceleration of non-relativistic case.

Conclusion:

Therefore, the limiting form of the expression of acceleration at low energy is $P(2mK)1/2$ and it is same as that of expression of acceleration of non-relativistic case.

Answer 35

(e)

To determine

The limiting form of the expression in part (d) at low energy and compare with the non-relativistic expression.

Answer 36

(e)

Expert Solution

Answer to Problem 39.74AP

The limiting form of the expression of acceleration at low energy is

$P(2mK)1/2$ and it is same as that of expression of acceleration of non-relativistic case.

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Given info: The non-relativistic expression for the acceleration is $a=P/(2mK)1/2$ .

From equation (4), the expression for the acceleration is given as,

$a=Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$

At low energy the value of $Kmc2$ is very small that $Kmc2<<<1$ . So equation becomes,

$a=Pmc(Kmc2)1/2(2)1/2(1)2=P(2mK)1/2$

Thus, the limiting form of the expression of acceleration at low energy is $P(2mK)1/2$ and it is same as that of expression of acceleration of non-relativistic case.

Conclusion:

Therefore, the limiting form of the expression of acceleration at low energy is $P(2mK)1/2$ and it is same as that of expression of acceleration of non-relativistic case.

Answer 41

(f)

To determine

The limiting form of the expression in part (d) at high energy and compare with the non-relativistic expression.

(f)

Expert Solution

Answer to Problem 39.74AP

The limiting form of the expression of acceleration at high energy is

$Pm2c5(K)3$ .

Explanation of Solution

Given info: The non-relativistic expression for the acceleration is $a=P/(2mK)1/2$ .

From equation (4), the expression for the acceleration is given as,

$a=Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$

At high energy the value of $Kmc2$ is very small that $Kmc2>>>1$ . So equation becomes,

$a=Pmc(Kmc2)1/2(Kmc2)1/2(Kmc2)2=Pmc(Kmc2)3=Pm2c5(K)3$

Thus, the limiting form of the expression of acceleration at low energy is $P(2mK)1/2$ and it is same as that of expression of acceleration of non-relativistic case.

Conclusion:

Therefore, the limiting form of the expression of acceleration at high energy is $Pm2c5(K)3$ .

Answer 42

(f)

To determine

The limiting form of the expression in part (d) at high energy and compare with the non-relativistic expression.

Answer 43

(f)

Expert Solution

Answer to Problem 39.74AP

The limiting form of the expression of acceleration at high energy is

$Pm2c5(K)3$ .

Answer 44

(f)

Expert Solution

Answer 45

(f)

Expert Solution

Answer 46

Expert Solution

Answer 47

Explanation of Solution

Given info: The non-relativistic expression for the acceleration is $a=P/(2mK)1/2$ .

From equation (4), the expression for the acceleration is given as,

$a=Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2$

At high energy the value of $Kmc2$ is very small that $Kmc2>>>1$ . So equation becomes,

$a=Pmc(Kmc2)1/2(Kmc2)1/2(Kmc2)2=Pmc(Kmc2)3=Pm2c5(K)3$

Thus, the limiting form of the expression of acceleration at low energy is $P(2mK)1/2$ and it is same as that of expression of acceleration of non-relativistic case.

Conclusion:

Therefore, the limiting form of the expression of acceleration at high energy is $Pm2c5(K)3$ .

Answer 48

(g)

To determine

The reason that answer to part (f) help account for the answer to part (c) at constant input power.

(g)

Expert Solution

Answer to Problem 39.74AP

The acceleration of the particle is very less at high energy that gives the velocity of the particle a constant value.

Explanation of Solution

Given info: The non-relativistic expression for the acceleration is $a=P/(2mK)1/2$ .

From the answer of part (f) the expression for the acceleration is,

$a=Pm2c5(K)3$

Here,

$p$ is the power.

$m$ is the mass of a particle.

$K$ is the kinetic energy of the particle.

In part (c), the speed at high energy approaches to the speed of light. But from the acceleration equation if the energy is imparted to the particle at constant input power the acceleration is steeply decreases because the acceleration is inversely proportional to the cube root of the kinetic energy. So at high energy acceleration is very less and the velocity of the particle approaches to a constant value as indicate in part (c).

Conclusion:

Therefore, the acceleration of the particle is very less at high energy that gives the velocity of the particle a constant value.

Answer 49

(g)

To determine

The reason that answer to part (f) help account for the answer to part (c) at constant input power.

Answer 50

(g)

Expert Solution

Answer to Problem 39.74AP

The acceleration of the particle is very less at high energy that gives the velocity of the particle a constant value.

Answer 51

(g)

Expert Solution

Answer 52

(g)

Expert Solution

Answer 53

Expert Solution

Answer 54

Explanation of Solution

Given info: The non-relativistic expression for the acceleration is $a=P/(2mK)1/2$ .

From the answer of part (f) the expression for the acceleration is,

$a=Pm2c5(K)3$

Here,

$p$ is the power.

$m$ is the mass of a particle.

$K$ is the kinetic energy of the particle.

In part (c), the speed at high energy approaches to the speed of light. But from the acceleration equation if the energy is imparted to the particle at constant input power the acceleration is steeply decreases because the acceleration is inversely proportional to the cube root of the kinetic energy. So at high energy acceleration is very less and the velocity of the particle approaches to a constant value as indicate in part (c).

Conclusion:

Therefore, the acceleration of the particle is very less at high energy that gives the velocity of the particle a constant value.

Answer 55

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