Question
A trolley of mass 240 kg is initially at rest on the frictionless horizontal straight rails lying in east-west direction. The length of the trolley is 20 meters. A man of mass 60 kg is standing at the eastern end of the trolley. The man starts moving on the trolley from eastern end towards west with velocity 1.0 m/s relative to the trolley. Assume the displacement and the velocity in eastward direction to be positive.
Options
Solution
Sol. Key concept: Relative velocity and conservation of horizontal momentum.
Take east as positive. Let $v_{T}$ be the trolley velocity and $v_{m}$ the man’s velocity, both measured from the ground. The man walks west relative to the trolley at 1 m s⁻¹.
$$v_{m} - v_{T} = - 1\ \ \Rightarrow \ \ \ v_{m} = v_{T} - 1$$
$$60v_{m} + 240v_{T} = 0$$
$$60(v_{T} - 1) + 240v_{T} = 0\ \ \Rightarrow \ \ \ v_{T} = 0.2\ m\ s^{- 1}$$
$$v_{m} = - 0.8\ m\ s^{- 1},\ \ \ \ \ t = \frac{20}{1} = 20\ s$$
$$\Delta\ x_{T} = 0.2(20) = + 4.0\ m$$
Thus a is false, while b, c and d are true.
Correction / clarification: The earlier intermediate equality that changed $v_{T}$ − 1 into −1 − $v_{T}$ had an incorrect sign. The equations above use one consistent east-positive convention.
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