# 11 - Centre of mass and momentum conservation Questions Answers

In a gravity free space a small ball is placed at the centre of a box of the same mass. The ball and the box both are moving with velocity u towards a fixed wall. Assume all collisions are perfectly elastic.

A) How many collisions are possible?

B) find velocities of the box and the ball and the position of the ball relative to the box after all possible collisions.

Joshi sir comment
Joshi sir comment
Joshi sir comment

Updated:

Initially there is no external force. Centre of mass of the system accelerates when insect moves with acceleration.

Now suppose displacement of insect at t = t is x with respect to rod, and y is the acceleration of rod in downward direction with respect to ground. Thus acceleration of counterweight is y with respect to ground. NowNow solve for x = l,

For dynamics, take friction upward in free body diagram of insect and downward in the free body diagram of rod.

Now solve

Inform for any issue.

Joshi sir comment

a particle moving in a straight line with an initial velocity u is subjected to a constant acceleration,a. Show that the distance traveled by the particle in the nth second is; Sn=u+a(n-0.5)
Joshi sir comment

distance covered in n sec. = un+an2/2   (1)

distance covered in n-1 sec. = u(n-1)+a(n-1)2/2    (2)

so sn = (1) - (2)

put equation (1) and (2) and get the answer

the moment of inertia of a thin uniform rod about an axis through its end and perpendicular to its length is ML²/3 . The moment of inertia of the rod about an axis parallel to given axis and at a distance L from the center of the rod is

Asked By: AVANTIKA PURI Solved By: SARIKA SHARMA
Joshi sir comment

for an axis parellel to the axis of rod and at distance l, rod will act as a set of equidistance particles. so answer will be Ml2

answer given by sarika is not completely correct

a ball falls on an inclined plane of inclination @ from a height h above the point of impact and makes a perfectly elastic collisiom.where will it hit the plane again?
Joshi sir comment

velocity of impact = √2gh

so components of velocity after rebound = √2ghcosθ perpendicular to the plane and √2ghsinθ parallel to the plane for the collision is perfectly elestic

now with perpendicular component take net displacement 0 and calculate time

then find displacement along the plane by using s = ut + 1/2at2

here remember that you should divide gravity in comnents too