Start Free Mock LoginAlready registered? Login here

Physics - Center of Mass Question with Solution | TestHub

PhysicsCenter of MassMomentum conservationMedium3 minQB

PhysicsMediumsingle choice

Passage / Comprehension

A horizontal frictionless rod is threaded through a bead of mass m. The length of the cart is L and the radius of the bead, r, is very small in comparison with L (r $<<$ L). Initially at (t = 0) the bead is at the right edge of the cart. The cart is struck and as a result, it moves with velocity $v_{0}$ . When the bead collides with the cart's walls, the collisions are always completely elastic. (Given : m $>$ M).

The first collision takes place at time $t_{1}$ and the second collision takes place at $t = t_{2}$ . Find $t_{2} - t_{1}$ .

Options:

Answer:

Solution:

After the first collision: bead moves right at $\frac{2 M v _{0}}{m + M}$ , cart moves left at $\frac{( m - M ) v _{0}}{m + M}$ . Closing speed on the cart's far wall $= \frac{2 M v _{0} + ( m - M ) v _{0}}{m + M} = v_{0}$ over distance L . $t_{2} - t_{1} = \frac{L}{v _{0}}$ . Answer: (B)

Stream:JEESubject:PhysicsTopic:Center of MassSubtopic:Momentum conservation

⏱ 3mℹ️ Source: QB

Doubts & Discussion

Loading discussions...