Question

a ball of 20 kg moves on a plank of mass 60 kg. find the distance moved by the plank when ball reaches another end of plank.

Answer 1

Explanations & Calculations

• Assume the plank is on a frictionless surface & the length of the plank is $\small L\,\,(m)$ .

To evaluate this question with respect to the Earth's frame of refence

• Take the velocity of the ball relative to the plank as $\small \overrightarrow { V_1}$
• Velocity of the plank relative to earth as $\small \overleftarrow{V}$
• Now the velocity of the ball relative to earth is

$\qquad\qquad \begin{aligned} \small \overrightarrow{V_{b,e}}&= \small\overrightarrow{ V_{b,p}} +\overrightarrow{V_{p,e}}\\ &= \small V_1+ (-V)\\ &= \small \overrightarrow{V_1-V} \end{aligned}$

• By conservation of linear momentum,

$\qquad\qquad \begin{aligned} \small 0 &= \small 20\times (V_1-V) - 60\times V\\ \small \frac{V_1}{V}&= \small 4 \end{aligned}$

• Now if the plank moves distance $\small x$, the ball may move $\small L-x$ then,

$\qquad\qquad \begin{aligned} \small t = \frac{x}{V}&= \small \frac{L-x}{V_1-V}\\ \small \frac{x}{L-x} &= \small \frac{V}{V_1-V}= \frac{1}{3}\\ \therefore x &= \small \bold{\frac{L}{4}\,\,(m)} \end{aligned}$

To evaluate with respect to the plank's frame of reference

• Consideration of the velocities are the same as above & that of plank's is always the same.
• By conservation of linear momentum to the right side

$\qquad\qquad \begin{aligned} \small 0 &= \small 20V_1-(20+60)V\\ \small \frac{V_1}{V}&= \small 4\\ \end{aligned}$

• Now time taken for the ball to reach the other end of the plank (this is relative to the plank) is the same for plank to move a distance of $\small x_1$
• Then,

$\qquad\qquad \begin{aligned} \small t_1= \frac{L}{V_1}&= \small \frac{x}{V}\\ \small x &= \small L\frac{V}{V_1}=L\times\frac{1}{4}\\ &=\small \bold{\frac{L}{4} \,\,(m)} \end{aligned}$