Question

A ball moving with speed v makes a direct central impact with an identical stationary ball. If the total kinetic energy of the balls after the impacts remains 52% of the original, find the coefficient of restitution.

Answer 1

0.2

Answer 2

Explanation of the Solution:

Let $m$ be the mass of each identical ball.
Let $v$ be the initial speed of the moving ball and $0$ be the initial speed of the stationary ball.
Let $v_1$ and $v_2$ be the final speeds of the first and second ball, respectively.

1. Conservation of Linear Momentum:
   For a direct central impact, linear momentum is conserved.
   $m v + m (0) = m v_1 + m v_2$
   $v = v_1 + v_2$ (Equation 1)

2. Coefficient of Restitution (e):
   The coefficient of restitution is defined as the ratio of the relative velocity of separation to the relative velocity of approach.
   $e = \frac{v_2 - v_1}{v - 0}$
   $v_2 - v_1 = e v$ (Equation 2)
   From this, we can also write $v_1 - v_2 = -e v$.

3. Kinetic Energy Condition:
   The total kinetic energy after the impact is 52% of the original kinetic energy.
   Initial Kinetic Energy ($KE_i$) = $\frac{1}{2} m v^2 + \frac{1}{2} m (0)^2 = \frac{1}{2} m v^2$
   Final Kinetic Energy ($KE_f$) = $\frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2$

   Given $KE_f = 0.52 \times KE_i$:
   $\frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2 = 0.52 \times \frac{1}{2} m v^2$
   $v_1^2 + v_2^2 = 0.52 v^2$ (Equation 3)

4. Combining the Equations:
   We use the algebraic identity: $a^2 + b^2 = \frac{(a+b)^2 + (a-b)^2}{2}$.
   Applying this to $v_1^2 + v_2^2$:
   $v_1^2 + v_2^2 = \frac{(v_1+v_2)^2 + (v_1-v_2)^2}{2}$

   Substitute Equation 1 ($v_1+v_2 = v$) and the rearranged Equation 2 ($v_1-v_2 = -ev$) into this identity:
   $v_1^2 + v_2^2 = \frac{(v)^2 + (-ev)^2}{2}$
   $v_1^2 + v_2^2 = \frac{v^2 + e^2 v^2}{2}$
   $v_1^2 + v_2^2 = \frac{v^2(1+e^2)}{2}$

   Now, substitute this expression for $v_1^2 + v_2^2$ into Equation 3:
   $\frac{v^2(1+e^2)}{2} = 0.52 v^2$

   Assuming $v \neq 0$, we can cancel $v^2$ from both sides:
   $\frac{1+e^2}{2} = 0.52$
   $1+e^2 = 2 \times 0.52$
   $1+e^2 = 1.04$
   $e^2 = 1.04 - 1$
   $e^2 = 0.04$
   $e = \sqrt{0.04}$
   $e = 0.2$

The coefficient of restitution is 0.2.