A particle P of mass 1.5 kg is moving along a straight horizontal line with speed 3 m s⁻¹ - Edexcel - A-Level Maths Mechanics - Question 1 - 2005 - Paper 1

To determine the speed of Q after the impact, we can apply the principle of conservation of linear momentum.

The total momentum before the collision is given by: $p_{initial} = m_P imes v_P + m_Q imes (-v_Q)$ Substituting the values: $p_{initial} = (1.5 ext{ kg} imes 3 ext{ m s}^{-1}) + (2.5 ext{ kg} imes -4 ext{ m s}^{-1})$ $= 4.5 - 10 = -5.5 ext{ kg m s}^{-1}$

After the collision, particle P has its motion reversed and is moving with speed 2.5 m s⁻¹, thus: $p_{final} = m_P imes (-v_P') + m_Q imes v_Q'$ Where $v_P' = 2.5 ext{ m s}^{-1}$, hence: $p_{final} = (1.5 ext{ kg} imes -2.5 ext{ m s}^{-1}) + (2.5 ext{ kg} imes v_Q')$ Setting the initial momentum equal to the final momentum gives:

$-5.5 = -3.75 + 2.5v_Q'$

Solving for $v_Q'$: $2.5v_Q' = -5.5 + 3.75$ $2.5v_Q' = -1.75$ v_Q' = rac{-1.75}{2.5} = -0.7 ext{ m s}^{-1}

Thus, the speed of Q immediately after the impact is 0.7 m s⁻¹, but it is moving in the opposite direction.