JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Conservation of Momentum

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A rod of mass M and length L is lying on a horizontal frictionless surface. A particle of mass 'm' travelling along the surface hits at one end of the rod with a velocity 'u' in a direction perpendicular to the rod. The collision is completely elastic. After collision, particle comes to rest. The ratio of masses

(\frac{m}{M})

\frac{1}{x}

. The value of 'x' will be ____.

Answer: 4

Solution

Linear momentum along the particle's line of travel gives

m u = M v_{c m}

, so

v_{c m} = \frac{m u}{M}

. Angular momentum about the rod's centre gives

m u \frac{L}{2} = \frac{M L ^{2}}{12} ω

, so

ω = \frac{6 m u}{M L}

. The collision is elastic and the particle is left at rest, so all the kinetic energy now belongs to the rod:

\frac{1}{2} m u^{2} = \frac{1}{2} M v_{c m}^{2} + \frac{1}{2} (\frac{M L ^{2}}{12}) ω^{2}

Substituting the two results, the right side becomes

\frac{m ^{2} u ^{2}}{2 M} + \frac{3 m ^{2} u ^{2}}{2 M} = \frac{2 m ^{2} u ^{2}}{M}

, so

\frac{1}{2} m u^{2} = \frac{2 m ^{2} u ^{2}}{M}

and

\frac{m}{M} = \frac{1}{4}

. Hence

x = 4

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Integer answer

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JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Conservation of Momentum

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