Conservation of Momentum

Let two objects ( two balls A and B ) of masses m_A and m_B are travelling in the same direction along a straight line at different velocities u_A and u_B, respectively. And there are no other external unbalanced forces acting on them. Let u_A>u_B and the two balls collide with each other . During collision which lasts for a time t, the ball A exerts a force F_AB on ball B and the ball B exerts a force F_BA on ball A. Suppose v_A and v_B are the velocities of the two balls A and B after the collision, respectively.

The momenta of ball A before and after the collision are m_Au_A and m_Av_A, respectively. The rate of change of its momentum during the collision will be m_A (v_A – u_A)/t.

Similarly, the rate of change of momentum of ball B during the collisison will be m_B (v_B -u_B)/t.

According to the third law of momentum of ball B during the collision will be m_B exerted by ball A on ball B and the force F_AB exerted by the ball b on ball A must be equal and opposite to each other. Therefore,

F_AB = −F_BA

or, m_A ( v_A − u_A)/t = − m_B (v_B − u_B)/t

m_Au_A + m_Bu_B = m_Av_A + m_Bv_B

Since (m_Au_A + m_Bu_B) is the total momentum of the two balls A and B before the collision and (m_Av_A + m_Bv_B) is their total momentum after the collision, we observe that the total momentum of the two balls remains unchanged or conserved provided no other external force acts.

As a result of this ideal collision experiment, we say that the sum of momenta of the two objects before collision provided there is no external unbalanced force acting on them. This is known as law of conservation of momentum.

Example:- A bullet of mass 20g is horizontally fired with a velocity 150 m s^-1 from a pistol of mass 2 kg. What is the recoil velocity of the pistol?

Sol:- Mass of bullet = 20 g = 20/1000 = 0.02 kg.

Mass of pistol = 2 kg

Initial velocities of bullet (u₁) and pistol (u₂) = 0

Final velocity of bullet (v₁) = + 150 m s^-1

Let the recoil of velocity of pistol = v

Total momenta of the pistol and bullet before fire, when the gun is at rest = (2 + 0.02) × 0 m s^-1

= 0 kg m s^-1

Total momenta of the pistol and bullet after it is fired = 0.02 kg × ( + 150 m s^-1) + 2 kg × v m s^-1

= (3 + 2v) kg m s^-1

According to law of conservation of momentum,

Total momenta after the fire = Total momenta before the fire

3 + 2v = 0

v = – 1.5 m s^-1.

Negative sign indicates that the direction in which the pistol would recoil is opposite to that of bullet, that is, right to left.