JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Conservation of Momentum

Two particles, of masses M and 2M, moving as shown, with speeds of

10 m / s

and

5 m / s

, collide elastically at the origin. After the collision, they move along the indicated directions with speeds

v_{1}

and

v_{2}

respectively. The value of

v_{1}

and

v_{2}

are nearly:

(A) $3.2 m / s$ and $12.6 m / s$
(B) $6.5 m / s$ and $6.3 m / s$ ✓ Correct
(C) $6.5 m / s$ and $3.2 m / s$
(D) $3.2 m / s$ and $6.3 m / s$

Answer: (B)

Solution

An elastic collision conserves both momentum and kinetic energy, but the two momentum equations alone pin down the unknown speeds here. Resolving along

x

2 M v_{1} cos 3 0^{\circ} + M v_{2} cos 4 5^{\circ} = 10 M cos 3 0^{\circ} + 10 M cos 4 5^{\circ}

, which simplifies to

v_{1} 3 + \frac{v _{2}}{2} = 53 + 52

. Resolving along

y

2 M v_{1} sin 3 0^{\circ} - M v_{2} sin 4 5^{\circ} = - 10 M sin 3 0^{\circ} + 10 M sin 4 5^{\circ}

, i.e.

v_{1} - \frac{v _{2}}{2} = - 5 + 52

. Solving the pair gives

v_{1} = \frac{5 ( 3 - 1 ) + 10 2}{3 + 1} \approx 6.5

m/s and

v_{2} \approx 6.3

m/s.

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Conservation of Momentum

Solution

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