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 13.0 m/s before she collides with him. After the collision, Rebecca has a velocity of magnitude 6.00 m/s at an angle of 54.1 ∘ from her initial direction. Both skaters move on the frictionless, horizontal surface of the rink.

Answer 1

Answer:

The magnitude of  Daniel's velocity after collision is 6.84 m/s.

Explanation:

Given that,

Mass of Daniel = 70.0 Kg

Mass of Rebecca = 45.0 kg

Initial speed of Rebecca= 13.0 m/s

Final speed of Rebecca = 6.00 m/s

Angle = 54.1°

Suppose we need to calculate the magnitude of Daniel's velocity after the collision

We need to calculate the velocity of Daniel along x axis

Using conservation of momentum

$m_{d}u_{d}+m_{r}u_{r}=m_{d}v_{d}+m_{r}v_{r}\cos\theta$

Put the value into the formula

$0+45\times13.0=70\times v_{x}+45\times6\cos54.1$

$v_{x}=\dfrac{45\times13-45\times6\cos54.1}{70}$

$v_{x}=6.09\ m/s$

We need to calculate the velocity of Daniel along y axis

Using conservation of momentum

$m_{d}u_{d}+m_{r}u_{r}\sin\theta=m_{d}v_{d}+m_{r}v_{r}$

Put the value into the formula

$0+45\times6.00\sin54.1=70\times v_{y}+0$

$v_{y}=\dfrac{45\times6.00\sin54.1}{70}$

$v_{y}=3.12\ m/s$

We need to calculate the magnitude of velocity after collision

Using formula of velocity

$v=\sqrt{v_{x}^2+v_{y}^2}$

Put the value into the formula

$v=\sqrt{(6.09)^2+(3.12)^2}$

$v=6.84\ m/s$

Hence, The magnitude of  Daniel's velocity after collision is 6.84 m/s.