On a frictionless, horizontal air table, puck A (with mass 0.25kg) is moving toward puck B (with mass 0.35kg), which is initially at rest.After the collision,puck A has a velocity of 0.12m/s to the left, and puck B has a velocity of 0.65m/s to the right.

Question

a.what is speed of puck A before collision?
b.calculate the change in the total kinetic energyof the system that occurs during the collision

Answer 1

Let A's initial velocity be in the + direction (to the right)
If the two initial velocities are Va and Vb, then to conserve momentum we need
.25Va + .35*0 = .25(-0.12) + .35(0.65)
solve for Va
Now you can figure the total KE (1/2 mv^2) before and after

Answer 2

Given:
M1 = 0.25kg, V1 = ?.
M2 = 0.35 kg, V2 = 0.
V3 = -0.12 m/s = velocity of M1 after collision.
V4 = 0.65 m/s = velocity of M2 after collision.

Momentum before = Momentum after
a. M1*V1 + M2*V2 = M1*V3 + M2*V4.
0.25*V1 + 0.35*0 = 0.25*(-0.12) + 0.35*0.65,
0.25V1 = -0.03 + 0.228 - 0.35,
0.25V1 = -0.153,
V1 = -0.61 m/s = 0.61 m/s to the left.

b. Before collision:
KE = 0.5M1*V1^2 + 0.5M2*V2^2 = 0.5*0.25*(-0.61)^2 + 0 = 0.047 J.

After collision:
KE = 0.5M1*V3^2 + 0.5M2*V4^2.
KE = 0.5*0.25*(-0.12)^2 + 0.5*0.35*0.65^2 = ?

Change in KE = KE after - KE before.