A uniform rod of mass $m_{1}$ and length L lies on the frictionless horizontal plane. A particle of mass $m_{2}$ moving at a speed $v_{0}$ perpendicular to the length of the rod strikes it at a distance L/3 from the centre and stops after the collision, Calculate the velocity of the centre of the rod .

Open in App

Solution

Centre of the mass of rod and particle of mass $m_{2}$ .
Linear momentum is conserved.
$m_{2} v_{0} = m_{1} v^{'} v^{'} = \frac{m_{2}}{m_{1}} v_{0}$
= speed of centre of mass of rod.

Suggest Corrections

Similar questions

Q. A uniform rod of mass

M

and length

a

lies on a smooth horizontal plane. A particle of mass

m

moving at a speed

v

perpendicular to the length of the rod strikes it at a distance

a / 4

from the centre and stops after the collision. Find
a. The velocity of the centre of the rod and
b. the angular velocity of the rod about its centre just after the collision.

A uniform rod of mass 'M' and length 'a' lies on a smooth horizontal plane. A particle of mass m moving at a speed 'v' perpendicular to the length of the rod strikes it at a distance $\frac{a}{4}$ from the center and stops after the collision. Find (a) the linear velocity of the center of the rod and (b) the angular velocity of the rod about its center just after the collision.

Q. A uniform rod of mass M and length a lies on a smooth horizontal plane. A particle of mass m, moving at a speed v, perpendicular to the length of the rod strikes it at a distance