Relative motion Problem for Physics Olympiad
Try this problem useful for Physics Olympiad based on Relative motion.
The Problem: Relative motion Problem
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,
$$v_{bi}=1.2m/s$$
Relative Velocity of Canoe in River
A canoe has a velocity of 0.40m/s southeast relative to the earth. The canoe is on a river that is flowing 0.50m/s east relative to the earth. Find the velocity (magnitude and direction ) of the canoe relative to the river.
Discussion:
We apply the relative velocity relation. The relative velocities are (\vec{v_{CE}}), the canoe relative to the earth, (\vec{v_{RE}}), the velocity of the river with respect to Earth and (\vec{v_{CR}}), the velocity of the canoe relative to the earth.
$$\vec{v_{CE}}=\vec{v_{CR}}+\vec{v_{RE}}$$
Hence $$
\vec{v_{CR}}=\vec{v_{CE}}-\vec{v_{RE}}$$
The velocity components of (
\vec{v_{CR}}) are $$ -0.5+\frac{0.4}{\sqrt{2}}=-0.217m/s$$( in the east direction)
Now, for the velocity component in the south direction
$$ \frac{0.4}{\sqrt{2}}=0.28$$ (in the south direction)
Now, the magnitude of the velocity of canoe relative to river $$ \sqrt{(-0.217)^2+(0.28)^2}=0.356m/s$$
If we consider (\theta) as the angle between the canoe and the river,the direction of the canoe with respect to the river can be given by
$$ \theta=52.5^\circ$$ ( in south west direction)