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(C) Let $v_0$ be the linear velocity of the centre of mass after the collision and $\omega_0$ be the angular velocity of the composite body about the

Vedclass · Accepted Answer

$\frac{h}{R}=\frac{10 m+7 M}{5(m+M)}$

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(C) Let $v_0$ be the linear velocity of the centre of mass after the collision and $\omega_0$ be the angular velocity of the composite body about the centre of mass.Linear momentum is conserved:$m v = (m+M) v_0 \quad \dots(i)$Angular momentum about the centre of mass is conserved:$m v (h-R) = I \omega_0$Where $I$ is the moment of inertia of the composite body about the centre of mass: $I = \frac{2}{5} M R^2 + m R^2 = (\frac{2}{5} M + m) R^2$.Substituting $mv$ from Eq. $(i)$ into the angular momentum equation:$(m+M) v_0 (h-R) = (\frac{2}{5} M + m) R^2 \omega_0$For pure rolling,$v_0 = R \omega_0$. Substituting this:$(m+M) R \omega_0 (h-R) = (\frac{2}{5} M + m) R^2 \omega_0$$(m+M) (h-R) = (\frac{2}{5} M + m) R$Dividing by $R$:$(m+M) (\frac{h}{R} - 1) = \frac{2}{5} M + m$$\frac{h}{R} - 1 = \frac{2M + 5m}{5(m+M)}$$\frac{h}{R} = \frac{2M + 5m}{5(m+M)} + 1 = \frac{2M + 5m + 5m + 5M}{5(m+M)} = \frac{10m + 7M}{5(m+M)}$

$A$ bullet of mass $m$ is fired horizontally into a large sphere of mass $M$ and radius $R$ resting on a smooth horizontal table. The bullet hits the sphere at a height $h$ from the table and sticks to its surface. If the sphere starts rolling without slipping immediately on impact,then

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