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Relative motion

A 40kg boy is standing on a plank of mass 160Kg. The plank originally at rest, is free to slide on a smooth frozen lake. The boy walks along the plank at a constant speed of 1.5m/s relative to the plank. What is the speed of the boy relative to the ice surface?

Solution:

Let the mass of the boy be \(m_b\) and that of the plank be \(m_p\). Let us consider the velocity of the boy with respect to the ice be \(v_i\), that of the plank with respect to ice be \(v_i\) and that of the boy with respect to the plank is \(v_{bp}\).
Now, using conservation of momentum,
$$m_bv_{bi}+(m_pv_{pi})=0$$
$$(40)v_{bi}+(160)v_{pi}=0$$
$$v_{bi}=-4v_{pi}….(i)$$
Also,
$$v_{bi}=v_{bp}+v_{pi}$$
From equation (i),
$$-4v_{pi}=v_{bp}+v_{pi}$$
$$-5v_{pi}=v_{bp}$$
$$v_{pi}=-\frac{1.5}{5}=0.3$$

Hence,

$$v_{bi}=1.2m/s$$

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