Trolley A of mass 7.2 kg moves to the right at 0.4 m s⁻¹ in a straight line on a horizontal floor - NSC Physical Sciences - Question 4 - 2023 - Paper 1

To find the velocity of the trolleys immediately after the collision, we apply the conservation of momentum:

Before collision:

• Momentum of A: $m_A v_{A_i} = (7.2 ext{kg})(0.4 ext{m/s}) = 2.88 ext{kg m/s}$
• Momentum of B: $m_B v_{B_i} = (5.3 ext{kg})(0) = 0 ext{kg m/s}$
• Total momentum before collision: $p_{initial} = 2.88 ext{kg m/s}$

After collision:

• Both trolleys move together, so: $p_{final} = (m_A + m_B) v_f$ $(7.2 ext{kg} + 5.3 ext{kg}) v_f = 2.88$ $12.5 v_f = 2.88$ Thus, v_f = rac{2.88}{12.5} = 0.2304 ext{ m/s} \\ \approx 0.23 ext{ m/s}

To calculate the average net force exerted by trolley A on trolley B during the collision, we use the formula:

$F_{net} = \frac{\Delta p}{\Delta t}$

Where:

• $\Delta p = p_{final} - p_{initial}$

• $\Delta t = 0.02 ext{ s}$

• Initial momentum of trolley B before the collision: $p_{initial} = 0$

• Final momentum of trolley B after the collision: $p_{final} = m_B v_f = (5.3 ext{kg})(0.23 ext{m/s}) = 1.219 ext{kg m/s}$

Thus:

\approx 61 ext{ N}$$