Question

Two ice skaters, with masses of $50\mathrm{kg}$and $75\mathrm{kg}$, are at the center of a $60\mathrm{m}$diameter circular rink. The skaters push off against each other and glide to opposite edges of the rink. If the heavier skater reaches the edge in $20\mathrm{s}$, how long does the lighter skater take to reach the edge?

Short answer

It will take$13.3\mathrm{s}.$

Step-by-step solution

Given information

We have given,

Two ice skaters of masses =$50\mathrm{kg},75\mathrm{kg}$

Diameter of circular rink = $60\mathrm{m}$

We have to find that how much takes the lighter skater take to reach the edge.

Simplify

Using law of conservation of momentum which says that momentum before collision and after will be same.

${\mathrm{P}}_{\mathrm{I}}={\mathrm{P}}_{\mathrm{f}}$

Where this momentums will be zero since initially both in rest.

then,

$$\begin{gathered} {\mathrm{P}}_{\mathrm{i}}=0 \\ {\mathrm{P}}_f={\mathrm{m}}_1{\mathrm{v}}_1+{\mathrm{v}}_2{\mathrm{m}}_2 \\ {\mathrm{P}}_f=(50\mathrm{kg})\left({\mathrm{v}}_1\right)+(75\mathrm{kg})\left({\mathrm{v}}_2\right).........\left(1\right) \end{gathered}$$

and,

Since the distance travel by the each player will be equal to radius of the rink. then,

${\mathrm{v}}_2=\frac{30\mathrm{m}}{20\mathrm{s}}=1.5\mathrm{m}/\mathrm{s}$

Put this value in above equation (1),

$$\begin{gathered} {\mathrm{P}}_{\mathrm{f}}=0=(50\mathrm{kg})\left({\mathrm{v}}_1\right)+(75\mathrm{kg})(1.5\mathrm{m}.\mathrm{s}) \\ {\mathrm{v}}_1=-\frac{112.5}{50}=-2.25\mathrm{m}/\mathrm{s} \end{gathered}$$

Here opposite direction indicates by the negative sign.

Then time taken will be,

$$\begin{gathered} {\mathrm{t}}_1=\frac{30\mathrm{m}}{{\mathrm{v}}_1} \\ {\mathrm{t}}_1=\frac{30\mathrm{m}}{2.25\mathrm{m}/\mathrm{s}} \\ {\mathrm{t}}_1=13.3\mathrm{s} \end{gathered}$$