## Saturday, May 16, 2009

### IIT JEE Level Revision Questions - 9. Centre of mass, linear momentum, collision

Past JEE Questions

Centre of mass. Linear momentum collisions fill blanks

A particle of mass 4m which is at rest explodes into three fragments. Two of the fragments, each of mass m, are found to move with a speed v, each in mutually perpendicular directions. The total energy released in the process of explosion is _________ .

1982
1(iv)
Two particles initially at rest move toward each other under a mutual force of attraction. At the instant when the speed of A is V and the speed of B is 2V, then the speed at the centre of the mass of the system is

a. zero
b. 1.5 V
c. V
d. 3 V

Velocity of centre of mass due to motion of particles due to internal forces is zero.

1986
1.(ii)
Multiple alternatives are correct. Find them

A ball hits the floor and rebounds after an inelastic collision. In this case:

a. The momentum of the ball just after the collision is the same as that just before the collision.
b. the mechanical energy of the ball remains the same in the collision.
c. the total momentum of the ball and the earth is conserved.
d. the total energy of the ball and the earth is conserved.

In an inelastic collision, momentum of the system is conserved. Also the total energy of the system is conserved. But total kinetic energy is not conserved.

1(iii)
A shell is fired from a cannon with a velocity V (m/s) at an angle θ with the horizontal direction. At the highest point in its path, it explodes into two pieces of equal masses. One of the pieces retraces its path to the cannon and the speed (in m/s) of the other piece immediately after the explosion is:

At the highest point, the shell has only horizontal component of velocity. When the particle breaks into pieces, the momentum of the system is conserved. As one of the components retraces back its horizontal component at the time of explosion is V cos θ.
The systems momentum at the time of explosion is mV cos θ. The momentum of the piece that retraced the path is (m/2)V cos θ.

Therefore mV cos θ = -(m/2)V cos θ + (m/2)V’

(m/2)V’ = 3(m/2) V cos θ

V’ = 3 V cos θ.

1987
2(iii)
A particle of mass 4m which is at rest explodes into three fragments. Two of the fragments, each of mass m, are found to move with a speed v, each in mutually perpendicular directions. The total energy released in the process of explosion is ---------------.

The momentum of the fragment with mass 2m has to be equal to the sum of momentums of the other two fragments.

As the magnitude of momentum of other two vectors is mv and the angle between them is 90°, the resultant vector will have the magnitude 2mv cos 45° .
2mv cos 45° = [2/SQRT(2)] mv

2mv’ = [2/SQRT(2)] mv
v’ = v [1/SQRT(2)]

So energy released in the explosion = sum kinetic energies of the three fragments

(½)mv² + (1/2)mv² +(1/2) 2m [(v(1/SQRT(2)] ²

mv²+mv²(1/2) = (3/2)mv²

1. A ball hits the floor and rebounds after an inelastic collision. In this case

a. the momentum of the ball just after the collision is the same as that just before the collision.
b. the mechanical energy of the ball remains the same in the collision.
c. the total momentum of the ball and the earth is conserved.
d. the total energy of the ball and the earth is conserved.

(More than one alternative may be correct)
JEE 1986

Answer ( c ) and (d)

Note: total kinetic energy is not conserved but total energy is conserved.

2. A particle of mass 4m which is at rest explodes into three fragments. Two of the fragments, each of mass m, are found to move with a speed v, each in mutually perpendicular directions. The total energy released in the process of explosion is _________ .

(JEE 1987)

Solution;

As the two fragments went in mutual perpendicular directions with velocity v the resultant momentum

2mv cos 45° = 2mv(1/√2) = √2 mv

if the velocity of fragment with 2m mass is V

then 2 mV =/√2mv

V = v/√2

Hence energy released = ½ mv² + ½ mv² + ½ *2 m (v/√2) ²

= (3/2)mv²