Question

Two identical steel balls, each of diameter 25.4 mm 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.)

Short answer

(a) The spring constant of ball is $8\times {10}^7\mathrm{N}/\mathrm{m}$.

(b) The collision lasts for a non-zero time interval.

(c) The kinetic energy for each ball is $0.8\mathrm{J}$.

(d) The maximum compression for each ball in the collision is 0.15 mm.

(e) The order of magnitude of time interval for contact between balls is ${10}^{-4}\mathrm{s}$.

Step-by-step solution

Identification of given data

The diameter of each steel ball is $d=25.4\mathrm{mm}$

The speed of each ball in opposite direction is $v=5\mathrm{m}/\mathrm{s}$

The force exerted by each jaw of vise is $F=16\mathrm{kN}$

The reduction in diameter of ball is$∆d=0.2\mathrm{mm}$

Determination of spring constant of ball

(a)

The spring constant of ball is given as:

$k=\frac{F}{∆d}$

Substitute all the values in the above equation.

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

Whether collision occurs for time interval or lasts only at instant

(b)

The diameter of ball reduced in the collision and it is only possible for collision with a non-zero time interval. If collision takes place for an instant then force exerted by each ball is same after collision and it value also will be very high. The deformation also becomes impossible.

Therefore, the collision lasts for a non-zero time interval.

Determination of kinetic energy for each ball

(c)

The kinetic energy for each ball is given as:

$$\begin{gathered} K=\frac{1}{2}k{\left(∆d\right)}^2+\frac{1}{2}k{\left(∆d\right)}^2 \\ K=k{\left(∆d\right)}^2 \end{gathered}$$

Substitute all the values in the above equation.

$$\begin{gathered} K=\left(8\times {10}^7\mathrm{N}/\mathrm{m}\right){\left[\left(0.2\mathrm{mm}\right)\left(\frac{1\mathrm{m}}{{10}^3\mathrm{mm}}\right)\right]}^2 \\ K=0.8\mathrm{J} \end{gathered}$$

Therefore, the kinetic energy for each ball is 0.8 J.

Determination of maximum amount of compression for each ball in collision

(d)

The maximum compression for each ball in the collision is given as:

$x_m=\sqrt{\frac{2K}{k}}$

Substitute all the values in the equation.

$$\begin{gathered} x_m=\sqrt{\frac{2\left(0.8\mathrm{J}\right)}{8\times {10}^7\mathrm{N}/\mathrm{m}}} \\ {x}_m=1.412\times {10}^{-4}\mathrm{m} \\ {\mathrm{x}}_m=\left(1.412\times {10}^{-4}\mathrm{m}\right)\left(\frac{{10}^3\mathrm{mm}}{1\mathrm{m}}\right) \\ {x}_m\approx 0.15\mathrm{mm} \end{gathered}$$

Therefore, the maximum compression for each ball in the collision is 0.15 mm.

Determination of order of magnitude of time interval for contact between balls

(d)

The order of magnitude of time interval for contact between balls is given as:

$∆t=\frac{2\pi x_m}{v}$

Substitute all the values in the above equation.

$$\begin{gathered} ∆t=\frac{2\mathrm{\pi }\left(0.15\mathrm{mm}\right)\left({\displaystyle \frac{1\mathrm{m}}{{10}^3\mathrm{mm}}}\right)}{5\mathrm{m}/\mathrm{s}} \\ ∆t=1.9\times {10}^{-4}\mathrm{s} \end{gathered}$$

Therefore, the order of magnitude of time interval for contact between balls is ${10}^{-4}\mathrm{s}$.