A body of mass 1 kg collides head on elastically with a stationary body of mass 3 kg. After collision, the smaller body reverses its direction of motion and moves with a speed of 2 m/s. The initial speed of the smaller body before collision is ___________ ms⁻¹.

Solution

Physics sir · 4 steps

The 1 kg body rebounds at exactly half its initial speed here, so double the 2 m/s.

1. Velocity after collision. In a head-on elastic collision with a body at rest, momentum and energy conservation together give the moving body's final velocity. Take the initial direction as positive and call the initial speed u.

   v₁ = (m₁ - m₂)/(m₁ + m₂) u = (1 - 3)/(1 + 3) u = -u/2

2. Read the sign. The minus sign shows the 1 kg body reverses, as the question says. Its speed after the collision is u/2 = 2 m s⁻¹, so u = 4 m s⁻¹.
3. Quick check. The 3 kg body moves off at v₂ = 2m₁/(m₁+m₂)u = 2 m s⁻¹. Momentum: 1 × 4 = 1 × (-2) + 3 × 2 = 4. Kinetic energy: 1/2(1)(16) = 8 J before, and 2 + 6 = 8 J after.
4. So. The smaller body was moving at 4 m s⁻¹ before the collision.

The common slip: 2. Treating the 1 kg body as bouncing off a wall, with no change in speed, gives 2. The 3 kg body moves too, so the rebound speed is only half of u.

The idea to keep. For a body hitting a stationary one elastically, v₁ = (m₁-m₂)/(m₁+m₂)u. A lighter body bounces back, but with its full speed only off a wall.

Answer from the NTA final answer key. Solution by PrepWiser.