From Newton’s law, we know that the time rate change of the momentum of a particle is equal to the net force acting on the particle and is in the direction of that force.
This clearly makes us understand that linear momentum can be changed only by a net external force. If there is no net external force, p cannot change.
Implies that there is no change in the momentum or the momentum is conserved. i.e
In order to apply conservation of momentum, you have to choose the system in such a way that the net external force is zero.
In the example given below, the two cars of masses m1 and m2 are moving with velocities v1 and v2 respectively before the collision. And their velocities change to
A bullet of mass m leaves a gun of mass M kept on a smooth horizontal surface. If the speed of the bullet relative to the gun is v, the recoil speed of the gun will be_______________________.
No external force acts on the system (gun + bullet) during their impact (till the bullet leaves the gun). Therefore the momentum of the system remains constant. Before the impact (gun + bullet) was at rest. Hence its initial momentum is zero. Therefore just after the impact, the momentum of the system (gun+bullet will be zero).
Conservation of Linear Momentum Video Lesson
Types Of Collision
The common normal at the point of contact between the bodies is known as the line of impact. If the mass centres of both the colliding bodies are located on the line of impact, the impact is called central impact and if mass centres of both or any one of the colliding bodies are not on the line of impact, the impact is called eccentric impact.
If velocities vectors of the colliding bodies are directed along the line of impact, the impact is called a direct or head-on impact;
If velocity vectors of both or of any one of the bodies are not along the line of impact, the impact is called an oblique impact.
Consider a ball of mass m1 moving with velocity v1 collides with another ball of mass m2 approaching with velocity v2. After a collision, they both move away from each other making an angle with the line of impact.
Coefficient of restitution
When two objects collide head-on, their velocity of separation after impact is in a constant ratio to their velocity of approach before impact.
The constant e is called as the coefficient of restitution.
An elastic collision between two objects is one in which total kinetic energy (as well as total momentum) is the same before and after the collision.
On a billiard board, a ball with velocity v collides with another ball at rest. Their velocities are exchanged as it is an elastic collision.
An inelastic collision is one in which total kinetic energy is not the same before and after the collision (even though momentum is constant).
When the colliding objects stick together after the collision, as it happens when a meteorite collides with the Earth, the collision is called perfectly inelastic.
In other inelastic collision, the velocities of the objects are reduced and they move away from each other.
Velocities of colliding bodies after collision in 1- dimension
Let there be two bodies with masses m1 and m2 moving with velocity u1 and u2. They collide at an instant and acquire velocities v1 and v2 after the collision. Let the coefficient of restitution of the colliding bodies be e. Then, applying Newton’s experimental law and the law of conservation of momentum, we can find the value of velocities v1and v2.
Conserving momentum of the colliding bodies before and the after the collision
Applying Newton’s experimental law
Putting (ii) in (i), we obtain
When the collision is elastic; e = 1
Having derived the velocities of colliding objects in a 1-dimensional collision (elastic and inelastic), let us try to set conditions and draw useful conclusions.
1. If m1 = m2
v1 = u2 and v2 = u1
When the two bodies of equal mass collide head-on elastically, their velocities are mutually
2. If m1 = m2 and u2 = 0, then
v1 = 0, v2 = u1
3. If the target particle is massive;
(a) m2 >> m1 and u2 = 0
v1 = -u1 and v2= 0
The light particle recoils with the same speed while the heavy target remains practically at rest.
(b) m2 >> m1 and u2 ≠ O
v1 ≈ – u1+ 2u2 and v2 ≈ u2
4. If the particle is massive: m1 >> m2
v1 = u1 and v2= 2u1— u2
If the target is initially at rest, u2 = O
v1 = u1 and v2 = 2 u1
The motion of the heavy particle is unaffected, while the light target moves apart at a speed twice that of the particle.
5. When the collision is perfectly inelastic, e = O
Elastic Collision in Two Dimension
A particle of mass m1 moving with velocity v1 along x-direction makes an elastic collision with another stationary particle of mass m2. After the collision, the particles move in different directions with different velocities.
Applying law of conservation of momentum,
Along y axis:
Conservation of kinetic energy,
Given the values of
Variable Mass System
Newton’s Second Law for the Rocket:
Let m be the instantaneous mass and mo be the initial mass of the rocket
⇒ m = mo – λt
Assuming g to be uniform,
Ignoring the effect of gravity, the instantaneous velocity and acceleration of the rocket (attributed only to the
thrust by the ejected mass) are;
m = mo – λt ……………..(3)
At any instant of time momentum of the mass (λdt) relative to the ground before ejection =
After ejection =
Change in momentum =
Rate of change of momentum of the ejected mass =
Force exerted on the ejected mass =
Force exerted on the system (Rocket). i.e.
Thrust F =
After perfectly inelastic collision between two identical particles moving with same speed in different directions, the speed of the particles becomes (2/3)rd of the initial speed. The angle between the two particles before collision is
Hence,(C) is the correct choice.