Linear momentum

Linear momentum

Linear momentum

Linear momentum is a product of the body’s mass and its velocity.

The S1 unit of momentum is kgms-1

When a force F is applied to a body, it changes the body’s velocity from u to v, the size of the force and the time for which it acts on a body.

From F = ma

Impulse of force is the product of force and the duration of its action or impulse is the change in momentum of the body which is acted on by the force

Example 1

 A body of mass 3kg initially moving with a velocity of 5ms-1 is acted on by a horizontal force of 15N for 2s. Find the impulse and final speed.


Impulse = Ft

              = 15 x 2

              = 30N

Impulse = change in momentum

         30 = m(v-u)

          30 = 3(v-5)

           v = 15ms-1

Example 2

A tennis ball has a mass of 0.07kg. it approaches a racket with a speed of 5ms-1 and bounces off and returns to the way it come with a speed of 4ms-1. The ball is in contact with the racket for 0.2 seconds. Calculate

(i) The impulse given to the ball

(ii) The average force exerted on the ball by the racket.


(i) Impulse = Ft = m(v-u)

                     = 0.07(-4-5)

                     = -0.63Ns

Collisions and principles of conservation of linear momentum

When two or more bodies collide, the total momentum of the system is conserved provided there is no external force on the system.

Consider a body of mass m1 moving with a velocity u1 to the right. Suppose the body makes a head on collision with a nother body of mass m2 moving with velocity u2 in the same direction

Let v1 and v2 be the velocities of the 2 bodies respectively after collision

Let F1 be the force exerted on m2 by m1 and F2 the force exerted on m1 by m2 using Newton’s 2nd law.

Hence, total momentum before collision = total momentum after collision, in other words momentum is conserved.

When two bodies collide, there is a short period of contact during which each exerts a force on each other at that instant, the force which each exert on each other is equal and opposite.

Types of collision

Collisions can be categorized as inelastic collision, perfectly inelastic, elastic or perfectly elastic collisions.

Elastic or perfectly elastic collisionInelastic collisionPerfectly inelastic collision
Kinetic energy is conservedKinetic energy is not conservedKinetic energy is not conserved
Linear momentum is conservedLinear momentum is conservedLinear momentum is conserved
Bodies separate after collision, e.g. collision of gas moleculesBodies separate after collision e.g. a ball bouncing from a  concrete floorBodies stick together and move with a common velocity. E.g.  a trailer colliding with a saloon car.

Elastic collision

Momentum is conserved

m1v1 + m2v2 = m1u1+m2u2

m1(u1 –v1) = m2(v2 – u2) ……………………………(i)

Kinetic energy is conserved

Example 3

A 200g block moves to the right at a speed of 100cms-1 and meets a 400g block moving to the left with a speed of 80cms-1. Find the final velocity of each block if the collision is elastic.


(v2 – v1) = -(-0.8 – 1)

(v2 – v1) = -1.8 …………………… (i)

Using conservation of momentum

m1v1 + m2u2 = m1v1 + m2v2

(0.2 x 1) + (-0.4 x 0.8)  = 0.2v1 + 0.4v2

                                    -0.12 = 0.2v1 + 0.4v2

                              -0.6 = v1 + 2v2 …………………….(ii)

Eqn (i) and Eqn (ii)

 v2 = -1.8 + v1

-0.6 = v1 + 2(-1.8 +v1)

    V1 = 1ms-1

    V2 = -0.8ms-1

Example 4

A particle of mass m1, travelling with velocity u1 makes a perfectly elastic collision with a stationary particle of mass m2. After the collision, the first particle moves a velocity v1 while the second particle moves in the same direction with velocity, v2. Show that

Example 5

A particle P of mass m1, travelling with a speed u1 makes a head on collision with a stationary particle Q of mass m2. If the collision is elastic and speed of P and Q after impact

Example 6

An object X of mass m moving with velocity 10ms-1 collides with a stationary object Y of equal mass. After collision, x, moves with speed u at an angle 300 to its initial direction, while Y moves with a speed of V at an angle 900 to the new direction of x.

(i) Calculate the speed u and v.

(ii) Determine whether the collision is inelastic or not.


  1.  A bullet of mass 300g travelling at a speed of 8ms-1hits a body of mass 450g moving in the same direction as the bullet at 1.5ms-1. The bullet and the body move together after collision. Find the loss in kinetic energy. [Ans. 3.8025J]
  2. A particle A of mass 4kg is incident with velocity V on a stationary helium nucleus B of mass 4kg. After collision, A moves in direction BC with velocity v/2 where BC makes an angle of 600 with the initial direction of AB and the helium nucleus moves along BD. Calculate the angle made in direction AB and the velocity of the helium along BD. [θ = 300,  velocity =  ]

Sponsored by The Science Foundation College + 256 753 80 27 09

By Dr. Bbosa Science

Share This


Wordpress (0)
Disqus (0 )