Question

A 1 kg ball moving at 12 ms^-1 collides with a 2 kg ball moving in opposite direction at 24 ms^-1. If the coefficient of restitution is \(\frac{2}{3}\), then the velocities after the collision are

• A. -4 ms^-1, - 28 ms^-1
• B. -28 ms^-1, -4 ms^-1
• C. 4 ms^-1, 28 ms^-1
• D. 28 ms^-1, 4 ms^-1

Answer 1

The Correct Option is B

1. Define variables:

• $m_1 = 1$ kg (mass of the first ball)
• $u_1 = 12$ m/s (initial velocity of the first ball)
• $m_2 = 2$ kg (mass of the second ball)
• $u_2 = -24$ m/s (initial velocity of the second ball - negative because it's moving in the opposite direction)
• $e = \frac{2}{3}$ (coefficient of restitution)
• $v_1$ (final velocity of the first ball)
• $v_2$ (final velocity of the second ball)

2. Apply the principle of conservation of momentum:

The total momentum before the collision equals the total momentum after the collision:

$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$

$(1)(12) + (2)(-24) = (1)v_1 + (2)v_2$

$12 - 48 = v_1 + 2v_2$

$v_1 + 2v_2 = -36$ (Equation 1)

3. Apply the definition of the coefficient of restitution:

$e = \frac{\text{relative velocity of separation}}{\text{relative velocity of approach}} = \frac{v_2 - v_1}{u_1 - u_2}$

$\frac{2}{3} = \frac{v_2 - v_1}{12 - (-24)}$

$\frac{2}{3} = \frac{v_2 - v_1}{36}$

$24 = v_2 - v_1$

$v_2 - v_1=24$ (Equation 2)

4. Solve the simultaneous equations:

Adding Equation 1 and Equation 2:

$3v_2 = -12$

$v_2 = -4$ m/s

Substituting $v_2$ back into Equation 2:

$-4 - v_1 = 24$

$v_1 = -28$ m/s

The correct answer is (B) $-28$ ms^-1, $-4$ ms^-1.