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 principle of conservation of momentum, we have:

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

Where:

• m_A = mass of particle A = 0.12 kg
• u_A = initial velocity of A = 3 m/s
• m_B = mass of particle B = 0.08 kg
• u_B = initial velocity of B = 0 m/s
• v_A = final velocity of A = 1.2 m/s
• v_B = final velocity of B (unknown)

Substituting the known values into the equation:

$0.12 imes 3 + 0.08 imes 0 = 0.12 imes 1.2 + 0.08 v_B$

This simplifies to:

$0.36 = 0.144 + 0.08 v_B$

Solving for $v_B$:

0.216 = 0.08 v_B\ v_B = rac{0.216}{0.08} = 2.7 m/s$$ Thus, the speed of B immediately after the collision is 2.7 m/s.

The impulse exerted on A can be calculated using the formula:

$I_A = m_A(v_{A_{final}} - v_{A_{initial}})$

Here:

• $m_A$ = mass of particle A = 0.12 kg
• $v_{A_{initial}}$ = initial velocity of A = 3 m/s
• $v_{A_{final}}$ = final velocity of A = 1.2 m/s

Plugging in the values:

$I_A = 0.12 imes (1.2 - 3) = 0.12 imes (-1.8) = -0.216 ext{ Ns}$

The magnitude of the impulse exerted on A is 0.216 Ns.