Two particles A and B have mass 0.12 kg and 0.08 kg respectively - Edexcel - A-Level Maths Mechanics - Question 2 - 2003 - Paper 1

Using the principle of conservation of momentum before and after the collision:

For particle A: Initial momentum of A = $0.12 imes 3 = 0.36$ kg m/s

Let the speed of B after the collision be $v$. Then, the total momentum after the collision for both particles is:

$0.12 imes 1.2 + 0.08 imes v = 0.36$

Solving for $v$:

$0.12 imes 1.2 + 0.08v = 0.36$

$0.144 + 0.08v = 0.36$

$0.08v = 0.36 - 0.144$

$0.08v = 0.216$

v = rac{0.216}{0.08} = 2.7 ext{ m s}^{-1}

Therefore, the speed of B immediately after the collision is 2.7 m/s.

The impulse exerted on particle A can be calculated by considering the change in momentum of particle A. Initial momentum of A before collision (moving at 3 m/s) was:

$I_{initial} = 0.12 imes 3 = 0.36 ext{ kg m/s}$

After the collision, the momentum of A (moving at 1.2 m/s) is:

$I_{final} = 0.12 imes 1.2 = 0.144 ext{ kg m/s}$

Thus, the impulse exerted on A is:

$ext{Impulse} = I_{final} - I_{initial} = 0.144 - 0.36 = -0.216 ext{ kg m/s}$

The magnitude of the impulse is therefore 0.216 Ns.