JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Conservation of Momentum

A block moving horizontally on a smooth surface with a speed of

40 m / s

splits into two parts with masses in the ratio of

1 : 2

. If the smaller part moves at

60 m / s

in the same direction, then the fractional change in kinetic energy is :-

(A) $\frac{2}{3}$
(B) $\frac{1}{3}$
(C) $\frac{1}{4}$
(D) $\frac{1}{8}$ ✓ Correct

Answer: (D)

Solution

Using conservation of momentum:

3 m \times 40 = m \times 60 + 2 m \times v

Solving:

v = 30 m/s

Initial KE:

K E_{i} = \frac{1}{2} (3 m) (40)^{2} = 2400 m

Final KE:

K E_{f} = \frac{1}{2} m (60)^{2} + \frac{1}{2} (2 m) (30)^{2} = 1800 m + 900 m = 2700 m

Fractional change:

\frac{K E _{f} - K E _{i}}{K E _{i}} = \frac{2700 m - 2400 m}{2400 m} = \frac{300 m}{2400 m} = \frac{1}{8}

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Single correct option

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