Four particles A, B, C, D of mass m/2, m, 2 m, 4 m, have same momentum, respectively. The particle with maximum kinetic energy is :

Solution

Physics sir · 3 steps

Same momentum for all four, so write KE in terms of p, and the lightest particle wins.

1. KE from momentum. Put v = p/m into K = 1/2mv². This gives KE in terms of momentum alone.

   K = p²/2m

2. Compare the four. Here p is the same for every particle, so K ∝ 1/m. The smaller the mass, the larger the kinetic energy.

   Particle	Mass	K = p²/2m
   A	m/2	p²/m, largest
   B	m	p²/2m
   C	2m	p²/4m
   D	4m	p²/8m

3. So. Particle A, the lightest one, has the maximum kinetic energy. Its KE is 8 times that of D.

The idea to keep. When momentum is fixed, use K = p²/2m: KE goes as 1/m. When velocity is fixed, KE goes as m. Always check which quantity is held equal.

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