Chapter 7, Problem 7.57AP

Two identical steel balls, each of diameter 25.4 nun and moving in opposite directions at 5 m/s, run into each other head-on and bounce apart. Prior to the collision, one of the balls is squeezed in a vise while precise measurements are made of the resulting amount of compression. The results show that Hooke’s law is a fair model of the ball’s elastic behavior. For one datum, a force of 16 kN exerted by each jaw of the vise results in a 0.2-mm reduction in the diameter. The diameter returns to its original value when the force is removed, (a) Modeling the ball as a spring, find its spring constant. (b) Does the interaction of the balls during the collision last only for an instant or for a nonzero time interval? State your evidence, (c) Compute an estimate for the kinetic energy of each of the balls before they collide, (d) Compute an estimate for the maximum amount of compression each ball undergoes when the balls collide, (e) Compute an order-of-magnitude estimate for the time interval for which the balls are in contact. (In Chapter 15, you will learn to calculate the contact time interval precisely.)

To determine

The spring constant of ball when models the ball as a spring.

Answer to Problem 7.57AP

The spring constant of ball is $8\times {10}^7\text{N}/\text{m}$.

Explanation of Solution

Given info: The diameter of each steel ball is $25.4\text{mm}$, the speed of balls is $5\text{m}/\text{s}$, the force exert by each jaw is $16\text{kN}$ and the compression of the steel along diameter is $0.2\text{mm}$.

The ball models as a spring then from Hooke’s law, the force exerts on the spring is,

$F=kx$

Here,

$k$ is the spring constant of steel ball.

$x$ is the length of spring.

Rearrange the above equation.

$k=\frac{F}{x}$

Substitute $0.2\text{mm}$ for $x$ and $16\text{kN}$ for $F$ in the above equation.

$\begin{array}{c}k=\frac{16\times {10}^3\text{N}}{0.2\times {10}^{-3}\text{m}}\\ =8\times {10}^7\text{N}/\text{m}\end{array}$

Conclusion:

Therefore, the spring constant of ball is $8\times {10}^7\text{N}/\text{m}$.

To determine

The interaction of the balls during the collision.

Answer to Problem 7.57AP

The interaction of the balls during the collision cannot happen.

Explanation of Solution

The interaction of the balls during the collision is calculable through the time period.

From the Newton’s law, the force exerts by a jaw is,

$F=ma$ (1)

Here,

$m$ is the mass of the steel ball.

$a$ is the acceleration of the steel balls.

The expression for the acceleration of the steel balls is,

$a=\frac{v}{t}$ (2)

Here,

$v$ is the speed of the ball.

$t$ is the time interval during the collision.

The total velocity of the balls during the collision is,

$v=v_{\text{f}}-v_{\text{i}}$

Here,

$v_{\text{f}}$ is the final speed of the steel ball.

$v_{\text{i}}$ is the initial speed of the steel ball.

Substitute $v_{\text{f}}-v_{\text{i}}$ for $v$ in equation (2).

$a=\frac{v_{\text{f}}-v_{\text{i}}}{t}$

Substitute $\frac{v_{\text{f}}-v_{\text{i}}}{t}$ for $a$ in equation (1).

$F=m\left(\frac{v_{\text{f}}-v_{\text{i}}}{t}\right)$

The interaction of the balls during the collision lasts for a time interval if the interaction takes no time interval that means $t=0$.

Substitute $0$ for $t$ in the above equation.

$F=m\left(\frac{v_{\text{f}}-v_{\text{i}}}{0}\right)$

The zero time interval of interaction, the force exerted by each ball on the other would be infinite and that cannot happen.

Conclusion:

Therefore, the interaction of the balls during the collision cannot happen.

To determine

The kinetic energy of each of the balls before they collide.

Answer to Problem 7.57AP

The kinetic energy of each ball is $0.8\text{J}$.

Explanation of Solution

The value of density of the steel is $7860\text{kg}/{\text{m}}^3$.

The expression for the volume of steel ball is,

$V=\frac{4}{3}\pi r^3$ (3)

Here,

$r$ is the radius of ball.

The radius of ball is the half of its diameter so the radius of steel ball is,

$r=\frac{d}{2}$

Here,

$d$ is the diameter of steel ball.

Substitute $25.4\text{mm}$ for $d$ in the above equation.

$\begin{array}{c}r=\frac{25.4\times {10}^{-3}\text{m}}{2}\\ =0.0127\text{m}\end{array}$

Substitute $0.0127\text{m}$ for $r$ in equation (3).

$\begin{array}{c}V=\frac{4}{3}\pi {\left(0.0127\text{m}\right)}^3\\ \approx 8.580\times {10}^{-6}{\text{m}}^3\end{array}$

The expression for the mass of the steel ball is,

$m=\rho V$

Substitute $7860\text{kg}/{\text{m}}^3$ for $\rho$ and $8.580\times {10}^{-6}{\text{m}}^3$ for $V$ in the above equation.

$\begin{array}{c}m=\left(7860\text{kg}/{\text{m}}^3\right)\left(8.580\times {10}^{-6}{\text{m}}^3\right)\\ \approx 0.0674\text{kg}\end{array}$

The expression for the kinetic energy of the steel ball before they collide is,

$K=\frac{1}{2}m{v}^2$

Substitute $0.0674\text{kg}$ for $m$ and $5\text{m}/\text{s}$ for $v$ in the above equation.

$\begin{array}{c}K=\frac{1}{2}\left(0.0674\text{kg}\right){\left(5\text{m}/\text{s}\right)}^2\\ \approx 0.84300\text{J}\\ \approx 0.8\text{J}\end{array}$

Conclusion:

Therefore, the kinetic energy of each ball is $0.8\text{J}$.

To determine

The maximum amount of compression each balls when balls collide.

Answer to Problem 7.57AP

The maximum amount of compression of each balls is $0.15\text{mm}$.

Explanation of Solution

From part (c), the kinetic energy before collision of balls is,

$K=0.8\text{J}$

The expression for the kinetic energy of spring after the collision of balls is,

$K=\frac{1}{2}k{x}_{\text{max}}^2$

Here,

$x_{\text{max}}$ is the maximum compression in length of spring.

The ball is models as a spring so the kinetic energy before the collisions of balls is converted into spring energy.

From part (a), the spring constant of balls that models as a spring is,

$k=8\times {10}^7\text{N}/\text{m}$

Substitute $0.8\text{J}$ for $K$ and $8\times {10}^7\text{N}/\text{m}$ for $k$ in the above equation.

$\begin{array}{c}0.8\text{J}=\frac{1}{2}\left(8\times {10}^7\text{N}/\text{m}\right)x_{\text{max}}^2\\ x_{\text{max}}=\sqrt{\frac{2\left(0.8\text{J}\right)}{8\times {10}^7\text{N}/\text{m}}}\\ \approx 0.1414\times {10}^{-3}\text{m}\\ \approx 0.15\text{mm}\end{array}$

Conclusion:

Therefore, the maximum amount of compression of each balls is $0.15\text{mm}$.

To determine

The time interval for which the balls are in contact.

Answer to Problem 7.57AP

The time interval is ${10}^{-4}\text{s}$.

Explanation of Solution

From part (a), the force exerts on the both the balls that models as spring is,

$F=kx$

The average force of both the balls is,

$\begin{array}{c}F=\frac{kx+kx}{2}\\ =\frac{1}{2}\left(kx\right)\end{array}$

From part (b), the expression for the time interval is,

$t=\frac{m\left(v_{\text{f}}-v_{\text{i}}\right)}{F}$

Substitute $\frac{1}{2}\left(kx\right)$ for $F$ in the above equation.

$t=\frac{m\left(v_{\text{f}}-v_{\text{i}}\right)}{\frac{1}{2}\left(kx\right)}$ (4)

The steel balls move in opposite direction of each other so the velocity has opposite sign after they bounce apart to each other.

From part (a), the spring constant of balls that models as a spring is,

$k=8\times {10}^7\text{N}/\text{m}$

From part (c), the mass of steel ball is,

$m=0.0674\text{kg}$

From part (d), the spring constant of the ball is,

$x=0.15\text{mm}$

Substitute $0.0674\text{kg}$ for $m$ and $5\text{m}/\text{s}$ for $v_{\text{f}}$, $-5\text{m}/\text{s}$ for $v_{\text{i}}$, $0.15\text{mm}$ for $x$ and $8\times {10}^7\text{N}/\text{m}$ for $k$ in equation (4).

$\begin{array}{c}t=\frac{2\left(0.0674\text{kg}\right)\left(5\text{m}/\text{s}-\left(-5\text{m}/\text{s}\right)\right)}{\left(8\times {10}^7\text{N}/\text{m}\right)\left(0.15\times {10}^{-3}\text{m}\right)}\\ \approx 1.0\times {10}^{-4}\text{s}\\ \approx {10}^{-4}\text{s}\end{array}$

Conclusion:

Therefore, the time interval is ${10}^{-4}\text{s}$.