JEE Main · 2015 · 04 Apr 2015 · Offline · Conservation of Momentum

A particle of mass

m

moving in the

x

direction with speed

2 v

is hit by another particle of mass

2 m

moving in the y direction with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to

Answer: (B)

Solution

Before the hit the two particles carry

\frac{1}{2} m (2 v)^{2} + \frac{1}{2} (2 m) v^{2} = 2 m v^{2} + m v^{2} = 3 m v^{2}

of kinetic energy. Momentum is conserved along each axis, so the

x

-momentum

2 m v

and the

y

-momentum

2 m v

are now shared by the combined mass

3 m

, giving velocity components

v_{x} = v_{y} = \frac{2 v}{3}

. The lump therefore keeps

\frac{1}{2} (3 m) (\frac{4 v ^{2}}{9} + \frac{4 v ^{2}}{9}) = \frac{4 m v ^{2}}{3}

. The fraction of energy lost is

\frac{3 m v ^{2} - \frac{4}{3} m v ^{2}}{3 m v ^{2}} = \frac{5}{9} \approx 56%

JEE Main · 2015 · 04 Apr 2015 · Offline · Single correct option