right is +x
initial momentum = .160*0.9 - .291*2.27
final momentum = .160*V1 + .291*V2
set initial momentum = final momentum
initial kinetic energy = (1/2).160*.9^2 + (1/2)(.291)*2.27^2
final kinetic energy = (1/2).160*V1^2 + (1/2)(.291)*V2^2
since elastic set final Ke = initial Ke
now you have two equations, two unknowns, solve by substitution
A 0.160 kg glider is moving to the right with a speed of 0.9 m/s on a frictionless horizontal air track. The glider has a collision with a 0.291 kg glider that is moving to the left with a speed of 2.27 m/s. Find the final velocity of each glider if the collision is elastic.
2 answers
Given:
M1 = 0.160 kg, V1 = 0.9 m/s.
M2 = 0.291 kg, V2 = -2.27 m/s.
V3 = Velocity of M1 after collision.
V4 = Velocity of M2 after collision.
Momentum before = Momentum after.
M1*V1 + M2*V2 = M1*V3 + M2*V4.
0.160 * 0.9 + 0.291*(-2.27) = 0.160 * V3 + 0.291*V4,
Eq1: 0.160V3 + 0.291V4 = -.0.517.
V3 = (V1(M1-M2) + 2M2*V2)/(M1+M2).
V3 = (0.9(0.160-0.291) + 0.582*(-2.27))/(0.160+0.291),
V3 = (-0.118 + (-1.32)) / 0.451 = -3.19 m/s.
In Eq1, replace V3 with -3.19 and solve for V4.
M1 = 0.160 kg, V1 = 0.9 m/s.
M2 = 0.291 kg, V2 = -2.27 m/s.
V3 = Velocity of M1 after collision.
V4 = Velocity of M2 after collision.
Momentum before = Momentum after.
M1*V1 + M2*V2 = M1*V3 + M2*V4.
0.160 * 0.9 + 0.291*(-2.27) = 0.160 * V3 + 0.291*V4,
Eq1: 0.160V3 + 0.291V4 = -.0.517.
V3 = (V1(M1-M2) + 2M2*V2)/(M1+M2).
V3 = (0.9(0.160-0.291) + 0.582*(-2.27))/(0.160+0.291),
V3 = (-0.118 + (-1.32)) / 0.451 = -3.19 m/s.
In Eq1, replace V3 with -3.19 and solve for V4.