Question

At the center of a 50-m-diameter circular ice rink, a 75 kg skater traveling north at 2.5 m/s collides with and holds on to a 60 kg skater who had been heading west at 3.5 m/s.

a. How long will it take them to glide to the edge of the rink?

b. Where will they reach it? Give your answer as an angle north of west

Short answer

a. The time taken to glide to the edge is 12s$12.01s$

b. The angle at which it will reach is,$41.{49}^o$

Step-by-step solution

Part (a). Step 1. Given information

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.

Step 2. Evaluating the x-component

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,

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

Step 3. Evaluating the y-component

For y-component,

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

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.

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

Part (b) Step 4. Evaluating the angle after collision

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

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