# 6 - Centre of mass and momentum conservation Questions Answers

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

a gun is mounted on a railroad car .the mass of the car,the gun,the shells, and the operator is 50m where m is the mass of one shell.if velocity of the shell w.r.t. The gun(in its state before firing) is 200m/s,what is the recoil of the car after the second shot?neglect friction.
Joshi sir comment

in first firing

m200 = 49mv so velocity of gun = 200/49

now apply momentum conservation again by taking 200/49 as the velocity of gun