Question

On an air track, a 403.0 glider moving to the right at 1.70 collides elastically with a 500.5 glider moving in the opposite direction at 3.00 .

1. Find the velocity of first glider after the collision.

2.
Find the velocity of second glider after the collision.

Answer 1

M1 = 0.403kg, V1 = 1.70 m/s.
M2 = 0.500kg, V2 = -3.0 m/s.

Conservation of Momentum:
M1*V1 + M2*V2 = M1*V3 + M2*V4.
0.403*1.7 + 0.50*(-3.0) = 0.403*V3 + 0.5*V4,
0.6851 - 1.50 = 0.403*V3 + 0.50*V4
Eq1: 0.403*V3 + 0.50*V4 = -0.815.

Conservation of Kinetic Energy Eq:
1. V3 = (V1(M1-M2) + 2M2*V2)/(M1+M2).
V3 = (1.7(0.403-0.5) + 1*(-3))/(0.403+0.5) = -3.50 m/s = Velocity of first glider.

2. In Eq1, replace V3 with (-3.50) and solve for V4.