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Byju's Answer

Standard XII

Physics

Angular Analogue of Linear Momentum

A rod AB of...

Question

A rod $A B$ of mass $M$ and length $8 l$ lies on a smooth horizontal surface. A particle of mass ' $m$ ' and velocity $v_{0}$ strikes as shown in figure. As a result of the collision, the centre of mass of rod attains a speed of $v_{0} / 8$ and the particle rebounds back with a speed of $v_{0} / 4$ . The ratio $\frac{M}{m}$ is :

\frac{M}{m} = 10

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\frac{M}{m} =

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\frac{M}{m} = 8

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\frac{M}{m} = 5

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Solution

The correct option is A $\frac{M}{m} = 10$
As there is no external force, thus linear momentum is conserved.
$P_{i} = P_{f}$
$m v_{o} + 0 = M \frac{v_{o}}{8} - m \frac{v_{o}}{4}$
$⟹ \frac{M}{m} = 10$

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A rod AB of mass M and length 8l lies on a smooth horizontal surface. A particle of mass 'm' and velocity v0 strikes as shown in figure. As a result of the collision, the centre of mass of rod attains a speed of v0/8 and the particle rebounds back with a speed of v0/4. The ratio Mm is :

The correct option is A Mm=10As there is no external force, thus linear momentum is conserved.Pi=Pfmvo+0=Mvo8−mvo4⟹Mm=10

The correct option is A $\frac{M}{m} = 10$
As there is no external force, thus linear momentum is conserved.
$P_{i} = P_{f}$
$m v_{o} + 0 = M \frac{v_{o}}{8} - m \frac{v_{o}}{4}$
$⟹ \frac{M}{m} = 10$