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# The Principle of Conservation of Linear Momentum

In a closed system of colliding bodies, the total momentum after the collision is equal to the total momentum before the collision provided there is no net external force acting on the system.

Case 1: Consider two bodies A and B of masses m1 and mmoving in the same direction with velocities u1 and urespectively. After collision their velocities were V1 and Vas shown below.

Applying the principle of conservation of linear momentum,

M1U1+M2U2=M1V1+M2V2

Case 2: Consider two bodies A and B of masses m1 and mmoving towards each other with velocities u1 and urespectively. After collision there velocities were V1 and V2 .

M1U1+M2U2=M1V1+M2V2

Assuming M1U1>M2U2,

Applying the principle of conservation of linear momentum,

M1U1+M2U2=(M1+M2)V

CALCULATIONS:

Example 1:

A trolley of mass 4kg moving on a smooth horizontal platform with a speed of 1.0ms-1 collides perfectly with a stationary trolley of the same mass on the same platform. Calculate the total momentum of the two trolleys immediately after the collision.

Solution:

M1 = 4kg,  U1 = 10m/s,   U2 = 0,   M2 = 4kg

Applying the principle of conservation of linear momentum,

Total momentum before the collision  = total momentum after the collision.

M1U1+M2U2=M1V1+M2V2SinceU2=0,M2U2=0

Momentum after collision (M1V1+M2V2)=M1U1=4×1=4.0kgms−1

Example 2:

A ball of mass 0.5kg moving at 10ms-1 collides with another ball of equal mass at rest. If the two balls move off together after the impact, calculate their common velocity.

M1 = 0.5kg,  U1 = 10m/s,   U2 = 0,   M2 = kg

Solution:

Applying the principle of conservation of linear momentum,

Total momentum before the collision  = total momentum after the collision.

M1U1+M2U2=(M1+M2)V0.5×10+0.5×0=(0.5+0.5)V5=1.0VV=5m/s

EVALUATION

1. State the following:

(a) Newton’s third law of motion

(b) Principle of conservation of linear momentum.

1. A ball P of mass 0.25kg loses one-third of its velocity when it makes a head-on collision with an identical ball Q at rest. After the collision, Q moves off with a speed of 2m/s in the original direction of P. Calculate the initial velocity of P.

# Types of Collisions

There are two major types of collisions, elastic and inelastic collisions.

Elastic Collision:

In an elastic collision both momentum and kinetic energy are conserved. This means that for two colliding bodies with masses m1 and m and initial velocities u1 and uand final velocities after collision V1 and V,

M1U1+M2U2=M1V1+M2V212M1U21+12M2U22=12M1V21+12M2V22

An example of perfectly elastic collision is a ball which bounces off the ground back to its original height.

Inelastic Collision:

In this case momentum is conserved but kinetic energy is not conserved. The energy lost is usually converted to heat, sound or elastic potential energy.

M1U1+M2U2=M1V1+M2V212M1U21+12M2U22≠12M1V21+12M2V22

Example 3:

A body of mass 5kg moving with a velocity of 20m/s due south hits a stationary body of mass 3kg. If they move together after collision with a velocity v due south, find the value of v.

Solution:

Applying the principle of conservation of linear momentum,

Total momentum before the collision = total momentum after the collision.

M1U1+M2U2=(M1+M2)V5×20+3×0=(5+3)V100=8VV=12.5m/s

EVALUATION

1. Distinguish between perfectly elastic collision and perfectly in elastic collision.
2. A tractor of mass kg is used to tow a car of mass kg. The tractor moves with a speed of 3.0m/s just before the towing rope becomes taut. Calculate the

(i) speed of the tractor immediately the rope becomes taut.

(ii) loss in kinetic energy of the system just after the car starts moving.

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