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 calculate the speed of Q immediately after the impact, we will use the principle of conservation of linear momentum (CLM). Before the collision, the momentum of particle P and Q can be expressed as:

1. Momentum of P before collision = mass × speed = $1.5 \, \text{kg} \times 3 \, \text{m/s}$
2. Momentum of Q before collision = mass × speed = $2.5 \, \text{kg} \times (-4) \, \text{m/s}$ (negative since it's in the opposite direction)

The total momentum before collision: $\text{Total Momentum} = (1.5 \times 3) + (2.5 \times (-4)) = 4.5 - 10 = -5.5 \, \text{kg m/s}$

After the collision, the momentum of particle P (with reversed direction and speed) is:

1. Momentum of P after collision = $1.5 \times (-2.5) = -3.75 \, \text{kg m/s}$

Let the speed of Q after the collision be $v$. The momentum of Q after collision will be: 2. Momentum of Q after collision = $2.5 \times v$

By conservation of momentum: $-5.5 = -3.75 + 2.5 v$

Solving for $v$ gives: $2.5 v = -5.5 + 3.75 \Rightarrow 2.5 v = -1.75 \Rightarrow v = -0.7 \, \text{m/s}$

Thus, the speed of Q immediately after the impact is 0.7 m/s.