Question

Two skaters, initially at rest, are $5\text{m}$ apart. They each have one end of a single rope and each pull on the rope with a force of $50\text{N}$ for a period of $1s$. One skater weighs $80\text{kg}$ and the other weighs $45\text{kg}$. (Assume no friction between the skates and the ice.)

• A. The two skaters meet at a distance of $1.8\text{m}$ from the initial position of the heavy skater
• B. The two skaters meet at a distance of $3.2\text{m}$ from the initial position of the heavy skater
• C. The relative velocity of the skaters when they meet is $\frac{125}{72}\text{m/s}$
• D. The relative velocity of the skaters when they meet is $\frac{25}{72}\text{m/s}$

Answer 1

Answer: (A), (C)

The correct options are
A The two skaters meet at a distance of $1.8\text{m}$ from the initial position of the heavy skater
C The relative velocity of the skaters when they meet is $\frac{125}{72}\text{m/s}$


Let $S_1,S_2$ be displacement of $80kg,45kg$ respectively, as they move towards each other.
As no external force is acting on the system COM does not shift, so
$80\times S_1=45S_2$
$S_2+S_1=5$
$80\times S_1=45(5-S_1)$
$125S_1-225=0\Rightarrow S_1=1.8\text{m}$

We know that
$\therefore a=\frac{F}{m}$,
For $80kg$ skater
$\frac{50}{80}=a_1$
Now from $v=u+at$
$v_1=\frac{5}{8}\text{m/s}$
In $1s$, this skater travels distance of $S=\frac{a{t}^2}{2}=0.3125m$. Thus they meet at a time greater than $1s$, after $1s$ their velocities remain constant.
For $45kg$ skater
$\frac{50}{45}=a_2$
$v_2=\frac{10}{9}\text{m/s}$

Relative velocity as skaters are moving towards one another is
$v_{\text{rel}}=\frac{5}{8}+\frac{10}{9}=\frac{125}{72}$