JEE Main · 2008 · 27 Apr 2008 · Offline · Conservation of Momentum

A block of mass 0.50 kg is moving with a speed of

2.00 m / s

on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

(A)0.16 J
(B)1.00 J
(C)0.67 J✓ Correct
(D)0.34 J

Answer: (C)

Solution

The two blocks stick together, so momentum is conserved while kinetic energy is not. Conservation of momentum gives

0.50 \times 2.00 = (0.50 + 1.00) v

, so the common speed after impact is

v = \frac{1.0}{1.5} = \frac{2}{3} m s^{- 1}

. The kinetic energy beforehand is

\frac{1}{2} (0.50) (2.00)^{2} = 1.0 J

and afterwards

\frac{1}{2} (1.50) (\frac{2}{3})^{2} = \frac{1}{3} J

, so the energy lost in the collision is

1.0 - \frac{1}{3} \approx 0.67 J

JEE Main · 2008 · 27 Apr 2008 · Offline · Single correct option

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