Chapter 7, Problem 57AP

(a)

To determine

The spring constant when ball is modelled as a spring.

Answer to Problem 57AP

The spring constant when ball is modelled as a spring is $8\times {10}^7\text{N/m}$.

Explanation of Solution

As steel ball is modelled as a spring this shows elastic behavior of the ball, so to calculate spring constant Hooke’s law is used.

Write the expression for force applied to the ball.

    $F=kx$

Here, $F$ is the force applied to the ball, $k$ is spring constant and $x$ is compression in the ball.

Rearrange above equation for $k$ .

    $k=\frac{F}{x}$                                                                                                             (I) 

Conclusion:

Substitute $16\text{kN}$ for $F$ and $0.2\text{mm}$ for $x$ in equation (I).

    $\begin{array}{c}k=\frac{\left(16\text{kN}\right)\left(\frac{1000\text{N}}{1\text{kN}}\right)}{\left(0.2\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)}\\ =8\times {10}^7\text{N/m}\end{array}$

Thus, the spring constant when ball is modelled as a spring is $8\times {10}^7\text{N/m}$.

(b)

To determine

The interaction of the ball during the collision.

Answer to Problem 57AP

The interaction of the ball during the collision is for a nonzero time interval.

Explanation of Solution

The interaction of the ball during the collision is for a time interval because if the interaction of the balls were for instant and not for some time then the force exerted by each ball on the other could be infinite and that is not possible.

Therefore, the interaction is for some time interval.

Conclusion:

Thus, the interaction of the ball during the collision is for a nonzero time interval

(c)

To determine

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

Answer to Problem 57AP

The kinetic energy of each of the balls before they collide is $0.8\text{J}$.

Explanation of Solution

The Kinetic energy for both the balls remains same as they have equal mass and they are moving with the same speed.

Consider iron as the main constituent in the density of steel to calculate mass of the balls.

Write the expression for mass in terms of density.

    $m=\rho V$                                                                                                        (II)

Here, $m$ is the mass of the steel ball, $\rho$ is the density of steel and $V$ is the volume of ball i.e. sphere.

Write the expression for volume of sphere.

    $V=\frac{4}{3}\pi r^3$

Here, $r$ is radius of the sphere.

Substitute $\frac{4}{3}\pi r^3$ for $V$ in equation (II).

    $m=\frac{4}{3}\rho \pi r^3$                                                                                                  (III)

Write the expression for Kinetic energy.

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

Here, $v$ is the speed of the ball before collision.

Write the expression for radius.

    $r=\frac{D}{2}$

Here, $D$ is the diameter and $r$ is the radius.

Substitute $\frac{D}{2}$ for $r$ in equation(III).

    $m=\rho \frac{4}{3}\pi {\left(\frac{D}{2}\right)}^3$                                                                                           (V)

Conclusion:

Substitute  $7860{\text{kg/m}}^{\text{3}}$ for $\rho$ and $25.4\text{mm}$ for $D$ in equation (V).

    $\begin{array}{c}m=\left(7860{\text{kg/m}}^{\text{3}}\right)\frac{4}{3}\pi {\left(\frac{\left(25.4\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)}{2}\right)}^3\\ =\frac{\left(7860{\text{kg/m}}^{\text{3}}\right)\left(4\pi \right){\left(0.0254\right)}^3}{24}\\ =0.0674\text{kg}\end{array}$

Substitute $0.0674\text{kg}$ for $m$ and $5\text{m/s}$ for $v$ in equation (IV).

    $\begin{array}{c}K=\frac{1}{2}\left(0.0674\text{kg}\right){\left(5\text{m/s}\right)}^2\\ =0.843\text{J}\\ \simeq 0.8\text{J}\end{array}$

Thus, the kinetic energy of each of the balls before they collide is $0.8\text{J}$.

(d)

To determine

The maximum amount of compression each ball undergoes when the balls collide.

Answer to Problem 57AP

The maximum amount of compression each ball undergoes when the balls collide is $0.15\text{mm}$.

Explanation of Solution

The maximum amount of elastic potential energy each ball has when the balls collide is equal to the kinetic energy they have before collision.

Write the expression for elastic potential energy.

    $U_s=\frac{1}{2}k{x}^2$

Here, $U_s$ is the elastic potential energy.

Write the expression for conservation of energy for this system.

    $K=U_s$

Substitute $\frac{1}{2}k{x}^2$ for $U_s$ in above equation.

    $K=\frac{1}{2}k{x}^2$                                                                                                   (VI)

Rearrange equation (VI) for $x$ .

    $x=\sqrt{\frac{2K}{k}}$                                                                                               (VII)

Conclusion:

Substitute $0.843\text{J}$ for $K$ and $8\times {10}^7\text{N/m}$ for $k$ in equation (VII).

    $\begin{array}{c}x=\sqrt{\frac{2\left(0.843\text{J}\right)}{\left(8\times {10}^7\text{N/m}\right)}}\\ =\left(1.45\times {10}^{-4}\text{m}\right)\left(\frac{1000\text{mm}}{1\text{m}}\right)\\ =0.145\text{mm}\\ \simeq 0.15\text{mm}\end{array}$

Thus, the maximum amount of compression each ball undergoes when the balls collide is $0.15\text{mm}$.

(e)

To determine

The time interval for which the balls are in contact.

Answer to Problem 57AP

The time interval for which the balls are in contact is nearly ${10}^{-4}\text{s}$.

Explanation of Solution

The balls are in contact for a very small time and distance covered by them is the maximum amount of compression the ball undergoes when it collides with the average speed.

Write the expression for average speed of the ball.

    $v_{avg}=\frac{v_i+v_f}{2}$                                                                                            (VIII)

Here, $v_{avg}$ is the average speed, $v_i$ is the initial speed by which they collide and $v_f$ is the final speed that is zero as they momentarily comes at rest.

Write the expression for time.

    $t=\frac{x}{v_{avg}}$                                                                                                        (IX)

Here, $t$ is the time interval and $x$ is the maximum compression.

Conclusion:

Substitute $5\text{m/s}$ for $v_i$ and $0$ for $v_f$ in equation (VIII).

    $\begin{array}{c}{v}_{avg}=\frac{5\text{m/s}+0}{2}\\ =2.5\text{m/s}\end{array}$

Substitute $2.5\text{m/s}$ for $v_{avg}$ and $0.15\text{mm}$ for $x$ .

    $\begin{array}{c}t=\frac{0.15\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)}{2.5\text{m/s}}\\ =6\times {10}^{-5}\text{s}\end{array}$

The order of magnitude estimate for time interval is ${10}^{-4}\text{s}$ .

Thus, the time interval for which the balls are in contact is nearly ${10}^{-4}\text{s}$.