Question

m1 is released from rest at a height of 0.6 meters and has a totally elastic collision with m2 which is initially at rest. Solve for velocities of m1 and m2 right after collision and maximum height of m2.

m1 = 3kg

m2 = 9kg

What was the change in momentum of m2?

What was the impulse acting on m2?

If the collision time was 0.2 seconds, what was the force on m2?

What was the change in momentum of m1?

What was the impule acing on m1?

If the collision time was 0.2 seconds, what was the force on m1?

Answer 1

using energy conservation to find speed just bfore collision,

mgh = mv^2 / 2

v = sqrt(2 x 9.81 x 0.6) = 3.43 m/s

collision is elastic,

so velocity of approach = velocity of seperation

3.43 = v1 + v2

v1 = 3.43 - v2   ....(i)

using momentum conservation,

m1v = m2v2 - m1v1

3 x 3.43 = 9v2 - 3v1

- v1 + 3v2 = 3.43

putting v1 from (i) .

-3.43 + v2 + 3v2 = 3.43

v2 =1.715 m/s

v1 = 3.43 - 1.715 = 1.715 m/s

change in momentum of m2 = 9 x 1.715 = 15.435 kg m/s

Impulse = change in momentum = 15.435 N /s

Impulse = F x t

15.435 = F x 0.2

F = 77.175 N

change in momentum of m1 = 3 ( 3.43 - (-1.715)) = 15.435 kg m/s

Impulse = change in momentum = 15.435 N /s

Impulse = F x t

15.435 = F x 0.2

F = 77.175 N