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 after the collision, we use the principle of conservation of momentum:

Let the speed of A after the collision be $u$.

The initial momentum of the system is given by: $P_{initial} = m_A v_A + m_B (-v_B)$ where:

• $m_A = 2 ext{ kg}$,
• $v_A = 5 ext{ m s}^{-1}$
• $m_B = 3 ext{ kg}$,
• $v_B = 6 ext{ m s}^{-1}$ (moving in the opposite direction).

Substituting the values: $P_{initial} = 2 imes 5 + 3 imes (-6) = 10 - 18 = -8 ext{ kg m s}^{-1}$

The impulse imparted is also equal to the change in momentum, thus: $ext{Impulse} = m_A (u - v_A)$ (14 = 2(u - 5))

Changing the equation to solve for u: (14 = 2u - 10) (24 = 2u) (u = 12 ext{ m s}^{-1})

The speed of A immediately after the collision is 12 m s⁻¹.