Question

Two ice skaters, Daniel (mass 70.0 kg) and Rebecca (mass 45.0 kg), are practicing. Daniel stops to tie his shoelace and, while at rest, is struck by Rebecca, who is moving at 14.0 m/s before she collides with him. After the collision, Rebecca has a velocity of magnitude 8.00 m/s at an angle of 55.1 degrees from her initial direction. Both skaters move on the frictionless, horizontal surface of the rink.
(a) What is the magnitude of Daniel's velocity after the collision?
(b) What is the direction of Daniel's velocity after the collision?
(c) What is the change in the total kinetic energy of the two skaters as a result of the collision?

Answer 1

Answer:

a) $v=7.32m/s$

b) $\alpha =-35$º

c) $ΔK=-1094.62J$

Explanation:

From the exercise we know that the collision between Daniel and Rebecca is elastic which means they do not stick together

So, If we analyze the collision we got

$p_{1x}=p_{2x}$

To simplify the problem, lets name D for Daniel and R for Rebecca

a) $p_{D1x}+p_{R1x}=p_{D2x}+p_{R2x}$

Since Daniel's initial velocity is 0

$m_{R}v_{Rx}=m_{D}v_{D2x}+m_{R}v_{R2x}$

$v_{D2x}=\frac{m_{R}*v_{R1x}-m_{R}*v_{R2x}}{m_{D}}$

$v_{D2x}=\frac{(45kg)(14m/s)-(45kg)(8cos(55.1)m/s)}{(70kg)}=6m/s$

Now, lets analyze the movement in the vertical direction

$p_{1y}=p_{2y}$

Since $p_{1y}=0$

$0=m_{D}v_{D2y}+m_{R}v_{R2y}$

$v_{D2y}=-\frac{m_{R}v_{2Ry}}{m_{D}}=-\frac{(45kg)(8sin(55.1)m/s)}{(70kg)}=-4.21m/s$

Now, we can find the magnitude of Daniel's velocity after de collision

$v_{D}=\sqrt{(6m/s)^2+(-4.21m/s)^2}=7.32m/s$

b) To know whats the direction of Daniel's velocity we need to solve the arctan of the angle

$\alpha =tan^{-1}(\frac{v_{y}}{v_{x}})=tan^{-1}(\frac{-4.21}{6})=-35$º

c) The change in the total kinetic energy is:

ΔK=$K_{2}-K_{1}$

ΔK=$\frac{1}{2}[(45kg)(8m/s)^2+(70kg)(7.32m/s)^2-(45kg)(14m/s)^2]=-1094.62J$

That means that the kinetic energy decreases