Solving Glancing Collisions: Velocity & Energy Loss

[MattF]

I have a bit of a problem with this question:

A 0.30-kg puck, initially at rest on a frictionless horizontal surface, is struck by a 0.20-kg puck that is intitially moving along the x-axis with a velocity of 2.0 m/s. After the collision, the 0.20-kg puck has a speed of 1 m/s at an angle of 53 degrees to the positive x-axis. (a) Determine the velocity of the 0.30-kg puck after the collision. (b) Find the fraction of the kinetic energy lost in the collision.

mass one= 0.20 kg
mass two= 0.30 kg

First off, is it possible to say that after the collision the two objects move off at an angle of 90 degrees from each other?

The initial momentum of the stationary puck is zero, so according to the conservation of momentum I get 0.667 m/s as the final velocity of the 0.30-kg puck. Is this correct? Do I have to factor in separate velocity along the x- and y-axis?

For (b), I'm not sure what to do. How do I find the energy loss?

 

[Diane_]

First off, is it possible to say that after the collision the two objects move off at an angle of 90 degrees from each other?

If I were your teacher, I'd require you to prove that before you could use it. Your mileage may vary.

Split the problem into components parallel to and perpendicular to the initial velocity of the puck. You know that the total momentum parallel to the initial velocity must be the same both before and after the collision, and that the total momentum perpendicular to that velocity should be zero. That should give you enough information to find the velocity (both parallel to and perpendicular to the initial velocity) of the second puck. From there, the total velocity is just a vector sum away.

For the second part - well, once you have all the velocities involved, just figure out what the total kinetic energy was before the collision (easy, since there's only one object moving) and what it is after the collision (not much harder - just get the kinetic energies for each object and add them), and Bob's your uncle.

 

[MattF]

Ok, I think I got it now. Final velocity of mass one in the x-direction is 0.6 m/s, and for the y-direction 0.8 m/s.

Conservation of momentum in the x-direction;

m_1v_1_x_i+m_2v_2_x_i=m_1v_1_x_f+m_2v_2_x_f

Solve for v_2_x_f, and I get 0.933 m/s.

Do the above, only now for velocity in the y-direction, and I get 0.8 m/s. Vector sum equals 1.23 m/s, which is the final velocity of mass two. Is this correct now?

For loss of kinetic energy I get 0.073 J.

Thanks for the tip!