Question

A uniform rod of mass M and length ℓ is placed on a smooth horizontal surface with its one end pivoted to the surface. A small ball of mass m moving along the surface with a velocity v₀, perpendicular to the rod, collides elastically with the free end of the rod. Find the impulse applied by the pivot on the rod during collision. (Take M/m = 2)

• A. $\frac{1}{5m{v}_0}$
• B. $\frac{3}{5m{v}_0}$
• C. $\frac{2}{5m{v}_0}$
• D. None of these

Answer 1

The correct option is C

$\frac{2}{5m{v}_0}$

If we take the ball + rod as one system then not external torque about the pivot would be zero [As the only external force acting is the normal reaction]

Applying conservation of angular momentum,

$L_i=m{v}_0\ell$ [as only the ball was moving]

$L_f=mv\ell +\frac{m{l}^2}{3}\omega$

$Li=Lf\Rightarrow m{v}_0\ell =mv\ell +\frac{m{l}^2}{3}\omega$

$\Rightarrow v_0=v+$ - - - - - - (1)

For elastic collision, c = 1

i.e., velocity of approach = velocity of separation

$v_0=\omega \ell -v$ - - - - - - (2)

from (1) and (2) $\omega$ =

angular impulse about the pivot = change in angular momentum about pivot

[Note: I represents impulse, not moment of Inertia]

Lets say $I_0$ is the impulse on the pivot then total linear impulse on the rod = $I+I_0$.

Also total linear impulse = change in linear momentum.