Chapter 7, Problem 76GP

(a)

To determine

The velocity of the lighter ball before impact.

Answer to Problem 76GP

The velocity of the lighter ball before impact is $1.7\text{m}/\text{s}$ .

Explanation of Solution

Given:

The mass of the ball A is $m_1=40\text{ g}$ .

The mass of the ball B is $m_2=60\text{ g}$ .

The angle is $\theta =60°$ .

Formula used:

The expression for velocity is,

  $v=\sqrt{2gh}$

Here, g is the acceleration due to gravity and h is the height.

Calculation:

Consider the acceleration due to gravity as $g=9.8\text{m}/{\text{s}}^{\text{2}}$ .

Sketch the representation of balls as shown below.

  

The height h is,

  $\begin{array}{l}\cos 60°=\frac{L-h}{L}\\ h=\left(30\text{ cm}\right)-0.5\left(30\text{ cm}\right)\\ h=15\text{ cm}\end{array}$

The velocity of the lighter ball before impact is,

  $\begin{array}{l}v=\sqrt{2gh}\\ v=\sqrt{2\times \left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(15\text{ cm}\times \frac{1\text{ m}}{100\text{ cm}}\right)}\\ v=1.7\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the velocity of the lighter ball before impact is $v=1.7\text{m}/\text{s}$ .

(b)

To determine

The velocity of each ball after the elastic collision.

Answer to Problem 76GP

The velocity of ball A after the elastic collision is $v_1=-0.34\text{m}/\text{s}$ .

The velocity of ball B after the elastic collision is $v_2=1.36\text{m}/\text{s}$ .

Explanation of Solution

Given:

The mass of the ball A is $m_1=40\text{ g}$ .

The mass of the ball B is $m_2=60\text{ g}$ .

The angle is $\theta =60°$ .

Formula used:

The expression for relative velocity is,

  $u_1-u_2=-\left(v_1-v_2\right)$

Here, $u_1$ is the initial velocity of object 1, $u_2$ is the initial velocity of the object 2, $v_1$ is the final velocity of object 1, and $v_2$ is the final velocity of the object 2.

The expression for law of conservation of momentum is,

  $m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$

Here, $m_1$ is the mass of the object 1, $m_2$ is the mass of object 2, $u_1$ is the initial velocity of object 1, $u_2$ is the initial velocity of the object 2, $v_1$ is the final velocity of object 1, and $v_2$ is the final velocity of the object 2.

Calculation:

From part (a), the velocity of the lighter ball before impact is $v=1.7\text{m}/\text{s}$ .

The relative velocity is,

  $\begin{array}{l}{u}_1-u_2=-\left(v_1-v_2\right)\\ \left(\text{1}\text{.7 }\text{m}/\text{s}\right)-\left(0\right)=v_2-v_1\\ v_1=v_2-1.7\end{array}$ …… (1)

The speed of each object using the relation of conservation of momentum is,

  $\begin{array}{l}{m}_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2\\ \left(40\text{ g}\right)\left(\text{1}\text{.7 }\text{m}/\text{s}\right)+\left(60\text{ g}\right)\left(0\right)=\left(40\text{ g}\right)v_1+\left(60\text{ g}\right)v_2\\ 68=40\left(v_2-1.7\right)+60v_2\\ 136=100v_2\\ v_2=1.36\text{m}/\text{s}\end{array}$

The speed of the ball A using equation (1) is,

  $\begin{array}{c}{v}_1=v_2-1.7\\ v_1=\left(1.36\text{m}/\text{s}\right)-1.7\\ v_1=-0.34\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the velocity of ball A after the elastic collision is $v_1=-0.34\text{m}/\text{s}$ and the velocity of ball B after the elastic collision is $v_2=1.36\text{m}/\text{s}$ .

(c)

To determine

The maximum height of each ball after elastic collision.

Answer to Problem 76GP

The maximum height of ball A after elastic collision is $h_A=6\times {10}^{-3}\text{ m}$ .

The maximum height of ball B after elastic collision is $h_B=0.09\text{ m}$ .

Explanation of Solution

Given:

The mass of the meteor is $m=1.\times {10}^8\text{ kg}$ .

The mass of the Earth is $m_E=6.0\times {10}^{24}\text{ kg}$ .

The speed of meteor is $v=15\text{km}/\text{s}$ .

Formula used:

The expression for the height is,

  $h=\frac{v^2}{2g}$

Here, v is the velocity and g is the acceleration due to gravity.

Calculation:

Consider the acceleration due to gravity as $g=9.8\text{m}/{\text{s}}^{\text{2}}$ .

From part (b), the velocity of ball A after the elastic collision is $v_1=-0.34\text{m}/\text{s}$ .

The velocity of ball B after the elastic collision is $v_2=1.36\text{m}/\text{s}$ .

The height of the ball A is,

  $\begin{array}{l}{h}_{\text{A}}=\frac{v_1^2}{2g}\\ h_{\text{A}}=\frac{{\left(-0.34\text{m}/\text{s}\right)}^2}{2\times \left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}\\ h_A=6\times {10}^{-3}\text{ m}\end{array}$

The height of the ball A is,

  $\begin{array}{l}{h}_{\text{B}}=\frac{v_2^2}{2g}\\ h_{\text{B}}=\frac{{\left(\text{1}\text{.36 }\text{m}/\text{s}\right)}^2}{2\times \left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}\\ h_B=0.09\text{ m}\end{array}$

Conclusion:

Thus, the maximum height of ball A after elastic collision is $h_A=6\times {10}^{-3}\text{ m}$ and the maximum height of ball B after elastic collision is $h_B=0.09\text{ m}$ .