A particle A of mass 2 kg is moving along a straight horizontal line with speed 12 ms⁻¹ - Edexcel - A-Level Maths Mechanics - Question 1 - 2010 - Paper 1

To find the mass m, we need to apply the conservation of momentum principle. The momentum before the collision for particle B is:

$p_{B} = m_{B} v_{B} = m ext{ kg} imes (-8) ext{ ms}^{-1}$

(Note that it is negative because it is in the opposite direction). The momentum of B after the collision:

$p'_{B} = m_{B} v'_{B} = m ext{ kg} imes 4 ext{ ms}^{-1}$

Using the conservation of momentum in the system, we have:

$m_{A} v_{A} + m_{B} (-v_{B}) = m_{A} v'_{A} + m_{B} v'_{B}$

Substituting the values:

$2 imes 12 - m imes 8 = 2 imes 3 + m imes 4$

This simplifies to:

$24 + 8m = 6 + 4m$

Solving for m gives:

$24 - 6 = 4m - 8m$

$18 = -4m$

Thus,

$m = 1.5 ext{ kg}$