91Ó°ÊÓ

A $75$-kilogram skater moving north at $2.5$ m/s collides with and clings on to a $60$ kg skater traveling west at 3.5 m/s in the center of a 50-m-diameter circular ice rink.

Skater 1 travels north (in the positive direction of the $y$axis), whereas skater 2 travels west (in the negative direction of the $x$axis). The law of conservation of momentum asserts that each component's momentum is conserved. So, for the $x-$component,

$$\begin{gathered} m_1.v_{1,x}+m_2.v_{2,x}=(m_1+m_2)v_{f,x} \\ 0+60×(-3.5)=(60+75)v_{f,x} \\ {v}_{f,x}=-1.56m/s \end{gathered}$$

For y-component,

$$\begin{gathered} m_1.v_{1,y}+m_2.v_{2,y}=(m_1+m_2)v_{f,y} \\ 75×2.5+0=(60+75)v_{f,x} \\ {v}_{f,x}=1.38m/s \end{gathered}$$

Now the final velocity is, $v_f=\sqrt{v_{f,y}^2+v_{f,x}^2}=2.08m/s$

Hence, From there, we can simply figure out how much time they'll need to go to the edge.

$t=\frac{R_{\text{disc}}}{v_f}=\frac{25m}{2.08m/s}=12.01s$

The skater traveling after the collision is found at an angle,

$$\begin{gathered} \tan \mathrm{θ}=\frac{1.38}{1.56} \\ \mathrm{θ}=41.{49}^o \end{gathered}$$