Two particles A and B, of mass 2 kg and 3 kg respectively; are moving towards each other in opposite directions along the same straight line on a smooth horizontal surface - Edexcel - A-Level Maths Mechanics - Question 1 - 2013 - Paper 1

To find the speed of particle A immediately after the collision, we can apply the principle of conservation of momentum, which states that the momentum before the collision is equal to the momentum after the collision.

Let the final velocity of A be denoted as $v_A'$. The initial momenta of A and B before the collision are:

• For A: $p_A = m_A v_A = 2 ext{ kg} imes 5 ext{ m s}^{-1} = 10 ext{ kg m s}^{-1}$
• For B: $p_B = m_B v_B = 3 ext{ kg} imes (-6 ext{ m s}^{-1}) = -18 ext{ kg m s}^{-1}$ (the negative sign indicates that B is moving in the opposite direction)

The total initial momentum is:

$p_{initial} = p_A + p_B = 10 - 18 = -8 ext{ kg m s}^{-1}$

After the collision, the final momenta are:

$p_{final} = m_A v_A' + m_B v_B'$

We know the impulse on B by A is 14 N s, therefore the change in momentum for B is:

$ext{Impulse} = m_B(v_B' - v_B) = 14 \Rightarrow 3(v_B' - (-6)) = 14$

Solving this gives:

$v_B' = \frac{14}{3} - 6 = \frac{14 - 18}{3} = -\frac{4}{3} ext{ m s}^{-1}$

Now, apply conservation of momentum:

$-8 = 2v_A' + 3(-\frac{4}{3})$

Calculating:

$-8 = 2v_A' - 4 \Rightarrow 2v_A' = -4 \Rightarrow v_A' = -2 ext{ m s}^{-1}$

Thus, the speed of A immediately after the collision is 2 m s⁻¹ in the opposite direction.