Question

A ball P of mass \( 0.5 \) kg moving with a velocity of \( 10 \, ms^{-1} \) collides with another ball Q of mass \( 1 \) kg at rest. If the coefficient of restitution is \( 0.4 \), the ratio of the velocities of the balls P and Q after the collision is?

• A. \( 1:7 \)
• B. \( 2:7 \)
• C. \( 2:5 \)
• D. \( 5:6 \) \

Hint:

When solving collision problems, always use the conservation of momentum and the coefficient of restitution equation. Solve for the velocities systematically.

Answer 1

The Correct Option is A

We use the principles of conservation of momentum and the coefficient of restitution to solve this problem. Step 1: Apply conservation of linear momentum
The total momentum before and after the collision must be the same: \[ m_P u_P + m_Q u_Q = m_P v_P + m_Q v_Q \] Given: \[ m_P = 0.5 \text{ kg}, \quad m_Q = 1 \text{ kg}, \quad u_P = 10 \text{ ms}^{-1}, \quad u_Q = 0 \] \[ 0.5 \times 10 + 1 \times 0 = 0.5 v_P + 1 v_Q \] \[ 5 = 0.5 v_P + v_Q \quad \text{(Equation 1)} \] Step 2: Apply the coefficient of restitution formula
\[ e = \frac{v_Q - v_P}{u_P - u_Q} \] \[ 0.4 = \frac{v_Q - v_P}{10} \] \[ v_Q - v_P = 4 \quad \text{(Equation 2)} \] Step 3: Solve for \( v_P \) and \( v_Q \)
From Equation 2: \[ v_Q = v_P + 4 \] Substituting into Equation 1: \[ 5 = 0.5 v_P + (v_P + 4) \] \[ 5 = 1.5 v_P + 4 \] \[ 1 = 1.5 v_P \] \[ v_P = \frac{2}{3} \] \[ v_Q = v_P + 4 = \frac{2}{3} + 4 = \frac{14}{3} \] Step 4: Find the ratio of velocities
\[ \frac{v_P}{v_Q} = \frac{2/3}{14/3} = \frac{2}{14} = \frac{1}{7} \] Thus, the ratio of velocities of P and Q after the collision is \( 1:7 \). \bigskip