5 * 0.1 + 10 * 0 = 5 u + 10 v conservation of momentum
(1/2)(5)(0.1)^2 = (1/2)(5)(u^2) + (1/2)(10)(v^2) conservation of energy
A 5kg object travelling at 0.1m/s collides head on with 10kg object initially at rest. Determine the velocity of each object after the impact if the collision is elastic
2 answers
Given:
M1 = 5kg, V1 = 0.1 m/s.
M2 = 10kg, Vz = 0.
V3 = Velocity of M1 after collision.
V4 = Velocity of M2 after collision.
Momentum before = Momentum after.
M1*V1 + M2*V2 = M1*V3 + M2*V4.
5 * 0.1 + 10 * 0 = 5*V3 + 10*V4,
Eq1: 5V3 + 10V4 = 0.5.
V3 = ((V1(M1-M2) + 2M2*V2)/(M1+M2).
V3 = ((0.1(5-10) + 20*0)/(5+10) = (-0.5 + 0)/15 = -0.0333 m/s. In opposite direction.
In Eq1, replace V3 with -0.0333 m/s and solve for V4.
M1 = 5kg, V1 = 0.1 m/s.
M2 = 10kg, Vz = 0.
V3 = Velocity of M1 after collision.
V4 = Velocity of M2 after collision.
Momentum before = Momentum after.
M1*V1 + M2*V2 = M1*V3 + M2*V4.
5 * 0.1 + 10 * 0 = 5*V3 + 10*V4,
Eq1: 5V3 + 10V4 = 0.5.
V3 = ((V1(M1-M2) + 2M2*V2)/(M1+M2).
V3 = ((0.1(5-10) + 20*0)/(5+10) = (-0.5 + 0)/15 = -0.0333 m/s. In opposite direction.
In Eq1, replace V3 with -0.0333 m/s and solve for V4.
1 answer