JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Conservation of Momentum

A body of mass M at rest explodes into three pieces, in the ratio of masses

1 : 1 : 2

. Two smaller pieces fly off perpendicular to each other with velocities of

30 ms^{- 1}

and

40 ms^{- 1}

respectively. The velocity of the third piece will be:

(A) $15 ms^{- 1}$
(B) $25 ms^{- 1}$ ✓ Correct
(C) $35 ms^{- 1}$
(D) $50 ms^{- 1}$

Answer: (B)

Solution

The masses are

\frac{M}{4}

\frac{M}{4}

, and

\frac{M}{2}

. By conservation of momentum,

P_{1} + P_{2} + P_{3} = 0

. Since

P_{1}

and

P_{2}

are perpendicular,

P_{3} = P_{1}^{2} + P_{2}^{2}

. With

P_{1} = \frac{M}{4} \times 30

and

P_{2} = \frac{M}{4} \times 40

, we get

P_{3} = \frac{M}{4} 3 0^{2} + 4 0^{2} = \frac{M}{4} \times 50

. Since

P_{3} = \frac{M}{2} v

, solving gives

v = 25

m/s.

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Single correct option

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