A body A, of mass m = 0.1 kg has an initial velocity of 3î ms⁻¹ . It collides elastically with another body, B of the same mass which has an initial velocity of 5ĵ ms⁻¹. After collision, A moves with a velocity v⃗ = 4( î + ĵ ). The energy of B after collision is written as x/10. The value of x is ___________.

Solution

Physics sir · 4 steps

Elastic means total kinetic energy is conserved, so B gets whatever energy A does not keep.

1. Energy before. The speeds are 3 m s⁻¹ for A and 5 m s⁻¹ for B, each of mass 0.1 kg. Total kinetic energy is 1/2(0.1)(9) + 1/2(0.1)(25) = 0.45 + 1.25 = 1.7 J.
2. Energy of A after. A moves with 4î + 4ĵ, so v_A² = 16 + 16 = 32 m²s⁻². Its kinetic energy is 1/2(0.1)(32) = 1.6 J.

   K_B = 1.7 - 1.6 = 0.1 J = 1/10 J

3. Check with momentum. Momentum conservation gives v⃗_B = (3î + 5ĵ) - (4î + 4ĵ) = -î + ĵ. Then K_B = 1/2(0.1)(2) = 0.1 J, which matches.
4. So. K_B = x/10 = 1/10, so x = 1.

The common slip: 9. Taking the speed of A as 4 instead of 4√2 gives K_A = 0.8 J, so K_B = 0.9 J and x = 9, not 1.

The idea to keep. Kinetic energy is a scalar, so add the energies of all the bodies directly, with no vector components. For a velocity like aî + bĵ, always use v² = a² + b².

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