JEE Main · 2018 · 08 Apr 2018 · Offline · Conservation of Momentum

In a collinear collision, a particle with an initial speed

v_{0}

strikes a stationary particle of the same mass. If the final total kinetic energy is

50%

greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is :

Answer: (C)

Solution

Momentum conservation gives

v_{1} + v_{2} = v_{0}

, and the kinetic energy being

50%

larger gives

\frac{1}{2} m (v_{1}^{2} + v_{2}^{2}) = \frac{3}{2} \times \frac{1}{2} m v_{0}^{2}

, so

v_{1}^{2} + v_{2}^{2} = \frac{3}{2} v_{0}^{2}

. Squaring the momentum relation,

v_{1}^{2} + v_{2}^{2} + 2 v_{1} v_{2} = v_{0}^{2}

, hence

2 v_{1} v_{2} = v_{0}^{2} - \frac{3}{2} v_{0}^{2} = - \frac{1}{2} v_{0}^{2}

. The relative speed after the collision is

∣ v_{2} - v_{1} ∣ = (v_{1}^{2} + v_{2}^{2}) - 2 v_{1} v_{2} = \frac{3}{2} v_{0}^{2} + \frac{1}{2} v_{0}^{2} = 2 v_{0}

JEE Main · 2018 · 08 Apr 2018 · Offline · Single correct option