A ball of mass A moving with velocity V, collides with another B of the same mass at rest. If the coefficient of restriction between the balls is e, the ratio of final and initial velocity of A is

$$m_A V_A + m_B V_B = m_A V_A' + m_B V_B'$$

Substitute the given values:

$$m V + m (0) = m V_A' + m V_B'$$ $$m V = m (V_A' + V_B')$$ $$V = V_A' + V_B' \quad \text{(Equation 1)}$$

$$e = -\frac{V_B' - V_A'}{V_B - V_A}$$

Substitute the given values:

$$e = -\frac{V_B' - V_A'}{0 - V}$$ $$e = \frac{V_B' - V_A'}{V}$$ $$e V = V_B' - V_A' \quad \text{(Equation 2)}$$

We have:

1. $V_A' + V_B' = V$
2. $-V_A' + V_B' = e V$

Subtract Equation 2 from Equation 1:

$$(V_A' + V_B') - (-V_A' + V_B') = V - e V$$ $$2 V_A' = V(1 - e)$$ $$V_A' = \frac{V(1 - e)}{2}$$

$$\frac{V_A'}{V_A} = \frac{\frac{V(1 - e)}{2}}{V}$$ $$\frac{V_A'}{V_A} = \frac{1 - e}{2}$$

The ratio of final and initial velocity of A is $\boxed{\frac{1-e}{2}}$.

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A block of mass m moving with a speed v collides with another object of mass 2m at rest. The lighter block comes to rest after the collision. The coefficient of restitution is

• Given:

  • Mass of lighter block (A): $m_A = m$
  • Initial speed of lighter block (A): $V_A = v$
  • Mass of heavier block (B): $m_B = 2m$
  • Initial speed of heavier block (B): $V_B = 0$ (at rest)
  • Final speed of lighter block (A): $V_A' = 0$ (comes to rest)

• To find: The coefficient of restitution, $e$

$$m_A V_A + m_B V_B = m_A V_A' + m_B V_B'$$

Substitute the given values:

$$m(v) + (2m)(0) = m(0) + (2m)V_B'$$ $$mv = 2m V_B'$$ $$V_B' = \frac{mv}{2m}$$ $$V_B' = \frac{v}{2}$$

$$e = -\frac{V_B' - V_A'}{V_B - V_A}$$

Substitute the known values:

$$e = -\frac{\frac{v}{2} - 0}{0 - v}$$ $$e = -\frac{\frac{v}{2}}{-v}$$ $$e = \frac{\frac{v}{2}}{v}$$ $$e = \frac{1}{2}$$

The coefficient of restitution is $\boxed{\frac{1}{2}}$.