Question

A body of mass \(4\,kg\) moving with a velocity of \(12\,ms^{-1}\) collides head-on with a stationary body of mass \(2\,kg\). If the relative velocity of separation of the two bodies after collision is \(6\,ms^{-1}\), then the percentage loss of kinetic energy of the body of mass \(4\,kg\) is:

• A. \(75\)
• B. \(25\)
• C. \(15\)
• D. \(50\)

Hint:

Always use momentum conservation together with the restitution equation to determine velocities after collision.

Answer 1

The Correct Option is A

Concept: For one-dimensional collisions: \[ e=\frac{\text{relative velocity of separation}} {\text{relative velocity of approach}} \] and momentum is always conserved.

Step 1: Calculate coefficient of restitution. Given: \[ u_1=12\,ms^{-1} \] \[ u_2=0 \] Relative velocity of approach: \[ 12\,ms^{-1} \] Relative velocity of separation: \[ 6\,ms^{-1} \] Thus \[ e=\frac{6}{12} \] \[ e=\frac12 \]

Step 2: Apply momentum conservation. Let final velocities be \(v_1\) and \(v_2\). \[ 4(12)+2(0)=4v_1+2v_2 \] \[ 48=4v_1+2v_2 \] \[ 24=2v_1+v_2 \] \[ v_2=24-2v_1 \]

Step 3: Apply restitution equation. \[ v_2-v_1=6 \] Substituting: \[ 24-2v_1-v_1=6 \] \[ 24-3v_1=6 \] \[ 3v_1=18 \] \[ v_1=6\,ms^{-1} \]

Step 4: Calculate percentage loss in kinetic energy of the 4 kg body. Initial kinetic energy: \[ K_i=\frac12(4)(12)^2 \] \[ K_i=288\,J \] Final kinetic energy: \[ K_f=\frac12(4)(6)^2 \] \[ K_f=72\,J \] Loss: \[ 216\,J \] Percentage loss: \[ \frac{216}{288}\times100 \] \[ 75\% \]

Step 5: Final answer. \[ \boxed{75\%} \]