Law of Conservation of Momentum

→ When two (or more) bodies act upon one another, their total momentum remains constant (or conserved) provided no external forces are acting.

• Initial momentum = Final momentum

Suppose, two objects A and B each of mass m1 and mass m2 are moving initially with velocities u1 and u2, strike each other after time t and start moving with velocities v1 and v2 respectively.

Now, Initial momentum of object A = m1u1

Initial momentum of object B = m2u2

Final momentum of object A = m1v1

Final momentum of object B = m2v2

So, Rate of change of momentum in A,

F1 = (m1v1 - m1u1)t = m1(v1 - u1)/t  ....(i)

Rate of change of momentum in B,

F2 = (m2v2 – m2u2)t =  m2(v2 - u2)/t  ....(ii)

We know from 3rd law of motion,

F1 = −F2

So, m1(v1 - u1)/t = - m2(v2 - u2)/t

⇒ m1v1 – m1u1  = -m2v2 + m2u2

⇒ m1u1 + m2u=  m1v1 + m2v2

Thus, Initial momentum = Final momentum

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

Solution

Given, Mass (m1) of bullet = 20 g = 0.02 kg

Mass (m2) of pistol = 2 kg

Initially bullet is inside the gun and it is not moving.

∵ Mass = m1+m2 = (0.02 + 2) kg = 2.02 kg

and u1 = 0

∴ Initial momentum = 2.02 × 0 = 0  ....(i)

Let the velocity of pistol be v2 and v1 for bullet = 150

∴ Final momentum = m1v1 + m2v2 = 0.02×150 + 2v2 ...(ii)

We know that,

Initial momentum = Final momentum

∴ (0.02×150)/100 + 2v2 = 0   [From equations (i) and (ii)]

⇒ 3 + 2v2 = 0

⇒ 2v2 = −3

⇒ v2 = −1.5 m/s

(−)ve sign indicates that gun recoils in direction opposite to that of the bullet.

Example 2: Two hockey players viz A of mass 50 kg is moving with a velocity of 4 m/s and another one B belonging to opposite team with mass 60 kg is moving with 3 m/s, get entangled while chasing and fall down. Find the velocity with which they fall down and in which direction?

Solution

Given,

mA = 50 kg, uA = 4 m/s

mB = 60 kg, uB = 3 m/s

Initial momentum A = mAuA = 50 × 4 = 200 kgm/s

Initial momentum B = MBuB = 60 × 3 = 180 kgm/s

∴ Total initial momentum = 200 + 180 = 380 kgm/s ....(i)

Final momentum = (mA + mB)v = (50 + 60)v = 110v  ....(ii)

According to the law of conservation of momentum,

380  = 110v

⇒ v = 380/110 = 3.454 m/s

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