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 12.0 m/s before she collides with him. After the collision, Rebecca has a velocity of magnitude 7.00 m/s at an angle of 52.1 ? 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?

Answer 1

a). 6.09 m/s

b). α=54.35

Step-by-step explanation:

The collision is an elastic so both skaters return their mass but change their velocities

`m_(d)*v_(id)+m_(r)*v_(ir)=m_(d)*v_(fd)+m_(r)*v_(fr)`

The initial velocity of Daniel is zero because the before the collision he is in rest

`70kg*0+45kg*12(m)/(s)=70kg*v_(fdx)+45kg*7(m)/(s)*cos(52.1)`

To determine the final velocity of Daniel

`540=70*v_(fdx)*cos(\alpha)+193.49`

`v_(fdx)*cos(\alpha)=4.95`

So in the axis 'y'

`70*v_(fdy)*sin(\alpha)=45*7*sin(52.1)`

`v_(fdy)*sin(\alpha)=3.55`

a).

The magnitude of the velocity Daniels after the collision is

`V=\sqrt{v_(fdx)^2+v_(fdy)^2}`

`V=√(4.95^2+3.55^2)`

`v=6.09(m)/(s)`

b).

The direction of the velocity after the collision is

`=(cos(\alpha))/(sin(\alpha))`

`tan\alpha =(v_(dfx))/(v_(dfy))`

`tan(\alpha )=(4.95)/(3.55)`

`\alpha =tan^-1*(1.39)`

`\alpha =54.35`