Consider the collision of two hockey pucks on a smooth ice rink. Various but not
ID: 3893320 • Letter: C
Question
Consider the collision of two hockey pucks on a smooth ice rink. Various but not all components of the velocities of the two pucks before and after the collision have been determined and are recorded in the table below.
Before:vx (cm/s)
Puck 1: 21
Puck 2: -46
Before:vy
Puck 1: 35
Puck 2: 15
After:vx
Puck 1: -13
After:vy 10
What is vx for puck 2 after collision? What is vy for puck 2 after collision?
What percentage of kinetic e is lost in the collision?
Please show your work so I can figure out how to do these problems.
Thank you!
Explanation / Answer
initial momentm along s direction = 21*m -46*m
after momentum = -13*m + m*V
by using conservation of momentum
21*m -46*m= -13m + mV
Vx of puck 2 after collision = -12
similary equating momentum along y direction
35*m + 15*m = 10*m + mVy
vy = 40
total initial KE = .5m(21^ + 46^2 + 35^2 + 15^2) = 2003.5m
total final KE = .5m(13^2 + 10^2 + 12^2 + 40^2) = 1006.5m
% lost = 2003.5 - 1006.5 / 2003.5 * 100 = 49.76 % lost