Two particles A and B have mass 0.12 kg and 0.08 kg respectively - Edexcel - A-Level Maths Mechanics - Question 2 - 2003 - Paper 1

Using the law of conservation of momentum:

$m_A u_A + m_B u_B = m_A v_A + m_B v_B$

Here, both particles start at rest, so:

$0 = 0.12 \times 1.2 + 0.08 \times v_B$

Solving for $v_B$ gives:

$0.12 \times 1.2 = 0.08 v_B$

Calculating:

$0.144 = 0.08 v_B$

$v_B = \frac{0.144}{0.08} = 1.8 \text{ m s}^{-1}$

Thus, the speed of B immediately after the collision is 1.8 m s^-1.

The impulse exerted on A can be calculated as the change in momentum:

$I_A = m_A (v_{A_f} - v_{A_i})$

where:

• $v_{A_f} = 1.2 \text{ m s}^{-1}$ (final velocity of A)
• $v_{A_i} = 3 \text{ m s}^{-1}$ (initial velocity of A)

Substituting values, we have:

$I_A = 0.12 \times (1.2 - 3) = 0.12 \times (-1.8) = -0.216 \text{ Ns}$

The magnitude of the impulse is 0.216 Ns.