Question about momentum (elastic collision)

[RuthlessTB]

Homework Statement

A block with mass 4 kg and velocity of 3 m/s collides elastically with block B with mass 2 kg which is originally at rest. Find the velocities of the two blocks after the collision.

Homework Equations

P= m * v

The Attempt at a Solution

This is the first time I try to solve an elastic collision problem, not sure if its the right way or no.

(M1*V1) + (M2*V2)= (M1+M2) Vc
(4*3) + (2*0) = (6) Vc
Vc= 12/6 = 2 m/s

 

[Doc Al]

(M1*V1) + (M2*V2)= (M1+M2) Vc

You cannot assume that both blocks end up with the same velocity. That would be true for an inelastic collision.

Hint: What two quantities are conserved in an elastic collision? (What does elastic collision mean?)

 

[RuthlessTB]

In elastic collision, momentum before equals to momentum after (Pi=Pf).

(M1*V1) + (M2*V2) = (M1 * V1') (M2*V2')
1) 12= (4 * V1') + (2 * V2')

V1 - V2 = -(V1' - V2')
2) 3-0 = -(V1' + V2')

3 + V1' = V2'

By substitution to get V1'
12 = (4*V1') + (2 * (3+V1'))
Therefore, V1'= 1 m/s

And for V2'
12= 4+ (2*V2')
V2'= 4 m/s

 

[Doc Al]

In elastic collision, momentum before equals to momentum after (Pi=Pf).

(M1*V1) + (M2*V2) = (M1 * V1') (M2*V2')
1) 12= (4 * V1') + (2 * V2')

V1 - V2 = -(V1' - V2')
2) 3-0 = -(V1' + V2')

3 + V1' = V2'

By substitution to get V1'
12 = (4*V1') + (2 * (3+V1'))
Therefore, V1'= 1 m/s

And for V2'
12= 4+ (2*V2')
V2'= 4 m/s

Excellent.

Two comments:
(1) Momentum is conserved in any collision, not just elastic ones.
(2) Your second equation, that the relative velocities are reversed, is only true in elastic collision.

Good job.