What are the velocities and kinetic energy changes in an ice skating collision?

[MAPgirl23]

Two ice skaters, Daniel (mass 65.0 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 8.00 m/s at an angle of 53.1 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?

** I think it's an elastic collision assuming they stick together after the collision; I tried solving it by using the formula:
v_f = [m_1/(m_1 + m_2)v_1] + [m_2/(m_1 + m_2)v_2] is this right?

b) What is the direction of Daniel's velocity after the collision? (degrees from the Rebecca's original)

c) What is the change in total kinetic energy of the two skaters as a result of the collision?

** k_f = k_i using the formula for kinetic energy as k = 0.5*mass*(velocity)^2

Please help!

 

[MAPgirl23]

my mistake, if the colliding objects stick together then it's an inelastic collision therefore k_f != k_i and k_f = 0.5*(m_1+m_2)*v_f^2 now is that right?

 

[Doc Al]

** I think it's an elastic collision assuming they stick together after the collision; I tried solving it by using the formula:
v_f = [m_1/(m_1 + m_2)v_1] + [m_2/(m_1 + m_2)v_2] is this right?

The two skaters do not stick together, since they move off with different speeds and directions. (And if they did stick together, that would mean that the collision is perfectly inelastic, not elastic.)

Make no assumptions about energy conservation. (Especially since the third parts asks for the change in KE.)

Solve the problem using conservation of momentum. Assume that Daniel heads off with some speed (call it v) at some angle (call it \theta). Now write down what conservation of momentum tells you.

 

[Doc Al]

my mistake, if the colliding objects stick together then it's an inelastic collision therefore k_f != k_i and k_f = 0.5*(m_1+m_2)*v_f^2 now is that right?

That would be right, if they stuck together. But they don't!

 

[MAPgirl23]

so since they don't stick and use momentum (p): p_i = p1 + p2 --> m1*v1 + m2*v2
p_f = (m1+m2)(v_f)^2 now for v_f = [m1/(m1+m2)v1] + [m2/(m1+m2)v2]

 

[Doc Al]

so since they don't stick and use momentum (p): p_i = p1 + p2 --> m1*v1 + m2*v2
p_f = (m1+m2)(v_f)^2 now for v_f = [m1/(m1+m2)v1] + [m2/(m1+m2)v2]

Note that your equation for p_f again assumes that they stick together! No good.

Try this instead. Call the initial direction of Rebecca to be the +x direction. After the collision, assume she flies off at an angle of 53.1 degrees above the x-axis. Now assume that Daniel flies off with speed "v" at an angle of \theta below the x-axis. Write the conservation of momentum equations for vertical and horizontal components.