Question

A ball of mass \(5\,\text{kg}\) moving with a kinetic energy of \(90\,\text{J}\) collides head-on with another ball of mass \(4\,\text{kg}\) at rest. If the relative velocity of separation between the two balls after collision is \(3\,\text{m s}^{-1}\), then the loss of kinetic energy due to the collision is

• A. \(30\,\text{J}\)
• B. \(60\,\text{J}\)
• C. \(90\,\text{J}\)
• D. \(45\,\text{J}\)

Hint:

For one-dimensional collisions: \[ \boxed{\text{Momentum is always conserved}.} \] Also, \[ \boxed{\text{Relative speed of separation}=e\times\text{Relative speed of approach}.} \] Use these relations together to determine the final velocities.

Answer 1

The Correct Option is A

Step 1: Find the initial velocity of the moving ball. The initial kinetic energy is \[ \frac12mu^2=90. \] Substituting \[ m=5\,\text{kg}, \] \[ \frac12(5)u^2=90, \] \[ u^2=36, \] \[ u=6\,\text{m s}^{-1}. \] The second ball is initially at rest.

Step 2: Apply conservation of momentum. Let the final velocities be \(v_1\) and \(v_2\). Then, \[ 5v_1+4v_2=5\times6=30. \] Since the relative velocity of separation is \[ v_2-v_1=3, \] we have \[ v_2=v_1+3. \] Substituting, \[ 5v_1+4(v_1+3)=30, \] \[ 9v_1=18, \] \[ v_1=2\,\text{m s}^{-1}, \] \[ v_2=5\,\text{m s}^{-1}. \]

Step 3: Calculate the final kinetic energy. The final kinetic energy is \[ \frac12(5)(2^2)+\frac12(4)(5^2). \] Thus, \[ =10+50 =60\,\text{J}. \] Initially, \[ K_i=90\,\text{J}. \] Therefore, \[ \text{Loss in kinetic energy} = 90-60 = 30\,\text{J}. \] Hence, \[ \boxed{\text{Loss in kinetic energy}=30\,\text{J}.} \] Therefore, the correct option is \(\boxed{(A)}\).