Two trolleys A and B of mass 3.2 kg and 2.6 kg respectively are held at rest on a straight horizontal, frictionless track, with a compressed spring between them, as shown in the diagram below - NSC Physical Sciences - Question 4 - 2024 - Paper 1

To calculate the distance travelled by trolley B, we can use the formula:

$ext{Distance} = ext{Speed} imes ext{Time}$

Trolley B's speed is unknown, but we know that trolley A moves at 0.4 m s⁻¹ to the left. Given that trolley B reaches the end of the track after 1.3 s, we first find its speed using the relation of momentum conservation:

$m_A v_A + m_B v_B = 0$

Substituting the known values:

$3.2 imes (-0.4) + 2.6 v_B = 0$

From this, we have:

v_B = rac{(3.2)(0.4)}{2.6} = 0.49 ext{ m/s to the right}

Now we calculate:

$ext{Distance} = 0.49 imes 1.3 = 0.637 ext{ m} ext{ or approximately } 0.64 ext{ m}$

Using the force exerted by the spring and the mass of trolley A:

The average force exerted is 4.2 N. For trolley A:

$F_{net} = m_a a$

where

a = rac{F}{m} = rac{4.2}{3.2} = 1.31 ext{ m/s}^2

Now, using kinematics:

ext{Acceleration} = rac{ ext{Change in Velocity}}{ ext{Time}}

So, we can solve for time when initial speed is 0:

Assuming the spring extends to its natural length at peak acceleration, we have:

LESS THAN. Trolley C has a larger mass compared to trolley B, resulting in a scenario where the force exerted remains constant, but due to the higher mass, the acceleration of trolley C will be lower than that of trolley B. Hence, the final velocity of trolley B after the spring has fallen will be less than its initial maximum potential velocity.