Chapter 7, Problem 37P

(a)

To determine

To show: The value of e is 1 for a perfectly elastic collision and $e=0$ for a completely inelastic collision.

Explanation of Solution

Given:

For a perfectly elastic collision, kinetic energy is conserved before and after the collision that is $(v_a-v_b)=-(v^{\text{'}}{}_a-v_b{}^{\text{'}})$ . For a completely inelastic collision, there is a loss in kinetic energy. This means kinetic energy is converted into other forms of energy after the collision that is $v^{\text{'}}{}_a=v_b{}^{\text{'}}$ .

Formula Used:

  $\text{Co-efficient of restitution (}e\text{)}=\frac{\text{Relative velocity after collision}}{\text{Relative velocity before collision}}$

Calculation:

Substitute the condition for perfectly elastic collision in the formula of co-efficient of restitution.

  $e=\frac{(v^{\text{'}}{}_a-v_b{}^{\text{'}})}{(v_b-v_a)}=-\frac{(v_a-v_b)}{(v_b-v_a)}=1$

Now, substitute the condition for completely inelastic collision in the formula of co-efficient of restitution.

  $e=\frac{(v^{\text{'}}{}_a-v_b{}^{\text{'}})}{(v_b-v_a)}=\frac{(v^{\text{'}}{}_a-v_a{}^{\text{'}})}{(v_b-v_a)}=0$

Conclusion:

Hence, for a perfectly elastic collision $e=1$ , and a completely inelastic collision $e=0$

(b)

To determine

A formula for $e$ in terms of the original height $h$ and the maximum height $h\text{'}$ reached after one collision.

Answer to Problem 37P

  $e=\sqrt{\frac{h\text{'}}{h}}$

Explanation of Solution

Given:

An object is dropped onto a heavy steel plate, as shown in the figure.

  

As seen in the figure, heavy steel plate will remain stationary before and after the collision. So, $v_b=0=v{\text{'}}_b$

Formula Used:

  $\text{Co-efficient of restitution }(e)=\frac{\text{Relative velocity after collision}}{\text{Relative velocity before collision}}$

During falling or rising, energy is conserved that is kinetic Energy is converted into Potential Energy and vice versa.

  $\begin{array}{l}mgh=\frac{1}{2}m{v}^2\\ v=\sqrt{2gh}\end{array}$

Calculation:

For falling on the steel plate, velocity will be positive $v_a=\sqrt{2gh}$

For moving up after bouncing from the steel plate, velocity will be negative $v{\text{'}}_a=-\sqrt{2gh}$

Now, substituting both of these velocities in the formula to calculate the co-efficient of restitution, we get

  $\begin{array}{l}e=-\frac{v_a\text{'}}{v_a}=-\frac{-\sqrt{2gh\text{'}}}{\sqrt{2gh}}\\ e=\sqrt{\frac{h\text{'}}{h}}\end{array}$

Conclusion:

Hence, formula for $e$ in terms of the original height $h$ and the maximum height $h\text{'}$ reached after one collision is $e=\sqrt{\frac{h\text{'}}{h}}$