Conservation of Linear Momentum

Introduction

Statement: The sum of linear momentum before collision is equal to the sum of linear momentum after collision, provided that the system is isolated.

When there are a number of bodies colliding against one another then, during collision, they share some of their energies with each other. It can be realized from the fact that when a footballer kicks a football, it starts to move – the person provided it the required initial kinetic energy for take off. In this event, to remember, the ball starts from rest and gains motion due to the acquired Kinetic Energy. The velocity of the football increased while it is also noticed that the velocity of the footballer has decreased. In this, the sum of linear momentum of the football and footballer before kicking is equal to the sum of linear momentum after kicking. This is an illustration of the law of conservation of linear momentum.

Mathematical Expression for Conservation of Linear Momentum

Let us consider a body A of mass m_A moves to the right with an initial velocity ‘u_A‘ and body B of mass m_B moves in the same direction with an initial velocity ‘u_B‘. In the beginning body A was ahead of body B. The sum of linear momentum of the bodies before meeting is given by $$\text{m}_\text{A}\text{u}_\text{A}\text{ + m}_\text{B}\text{u}_\text{B}$$Since, u_B > u_A, the bodies A and B meet at a certain point in the same direction and exchange energies. From then onwards, there are two possible cases as explained below:

Case I: The bodies move separately after collision.

Let us assume, the body A moves with a final velocity ‘v_A‘ after collision and body B moves with ‘v_B‘, both in the same direction as previous. Then the sum of final linear momentum is given by $$\text{m}_\text{A}\text{v}_\text{A}\text{ + m}_\text{B}\text{v}_\text{B}$$By principle or law of conservation of linear momentum, we have,

Sum of linear momentum before collision = Sum of linear momentum after collision

$$\implies \text{m}_\text{A}\text{u}_\text{A}\text{ + m}_\text{B}\text{u}_\text{B} \text{ = m}_\text{A}\text{v}_\text{A}\text{ + m}_\text{B}\text{v}_\text{B} \ — (1)$$

Case II: The bodies stick together after collision.

For this case, let us imagine a person of mass ‘3m’ carrying a load of ‘m’. Then, the effective mass of the person is ‘4m’, i.e. the sum of mass of the person and the load. Likewise, when the bodies stick together after collision, their effective mass is the sum of their individual masses. This means, mass in this case would be ‘ m_A + m_B‘. Another thing to be noticed, the bodies, as sticked together, would now be moving with a uniform velocity, say, ‘v’. In this case, the sum of final linear momentum is $$\left ( \text{m}_\text{A}\text{ + m}_\text{B} \right ) \text{v}$$

Therefore, by principle of conservation of linear momentum that states

Sum of linear momentum before collision = Sum of linear momentum after collision

$$\implies \text{m}_\text{A}\text{u}_\text{A}\text{ + m}_\text{B}\text{u}_\text{B} \ = \ \left ( \text{m}_\text{A}\text{ + m}_\text{B} \right ) \text{v} \ — (2)$$

Thus, equations (1) and (2) are the mathematical expressions for conservation of linear momentum in One-Dimension.

Cases of Conservation of Linear Momentum

• This law only holds true for motion in one-dimension. Both the velocities should be along X-axis, Y-axis, or Z-axis. In case, the bodies are moving in a plane, the momentum is conserved in separate axes. Following is a solved example emphasizing this point:

SOLVED EXAMPLE: A car of mass 4kg is moving due East with a velocity of 4 m/s and another car of mass 2kg is moving due North with a velocity of 5 m/s. After collision, the cars stick together and continue their motion. Find the final velocity of the system.

Solution: Generally, we consider East direction as positive X-axis and the North direction as positive Y-axis. Then total linear momentum is conserved in each axes separately. The bodies stick together after collision means that the effective mass would be (4+2 = 6) kg. Let the component of their common velocity ‘v’ in X-axis be v_x and that in Y-axis be v_y.

We know, Linear Momentum (P) = mv

Initial linear momentum in X-axis due to the car of mass 4kg $\text{4 x 4 = 16kgm/s}$

Initial linear momentum in Y-axis due to the car of mass 2kg $\text{2 x 5 = 10kgm/s}$

Final linear momentum in X-axis due to the combined mass of cars $\text{6 x v}_\text{x}$

Final linear momentum in Y-axis due to the combined mass of cars $\text{6 x v}_\text{y}$

By principle of conservation of linear momentum in X-axis

$$16 \ = \ 6 \text{v}_\text{x}$$

$$\text{v}_\text{x} \ = \ 2.67 \text{m/s}$$

By principle of conservation of linear momentum in Y-axis

$$10 \ = \ 6 \text{v}_\text{y}$$

$$\text{v}_\text{y} \ = \ 1.67 \text{m/s}$$

Hence, the required common velocity of the cars ‘v’ is given by

$$\sqrt{ \text{v}_\text{x}^2 \ + \ \text{v}_\text{y}^2}$$

$$= \ \sqrt{2.67^2 \ + \ 1.67^2}$$

$$= \ \sqrt{9.9178}$$

$$= \ 3.15 \text{m/s}$$

The direction of resultant velocity is $\arctan \left ( \dfrac{1.67}{2.67} \right )$.

• When the bodies have velocities in opposite directions, you can consider either direction to be positive. Then, the other direction shall be taken as negative.

• GUN-BULLET Problem: Problems involving firing of a bullet from a gun with certain initial velocity and the aim to find the velocity of recoil of the gun can be calculated by using the conservation of linear momentum. As the motion of the bullet and recoil of the gun are in opposite direction, the velocities should have opposite signs.

• ROCKET OR BOMB Problem: Problems invovling the splitting of a bomb or a rocket into smaller pieces also involves the idea of linear momentum conservation. In such cases, the masses split out in mutually opposite directions. This is done to obey the law we discussed above.