Question

Two objects moving in the positive x direction have a perfectly inelastic collision. The first object has a mass of 12 kg and a velocity of 4 m/s. The second has a mass of 7 kg and a velocity of 1 m/s. (a) What is the velocity of the center of mass of the system? (b) What are the velocities of the masses in the center of mass frame? (I.e., imagine that you are moving with a velocity vcm. What are the velocities you observe for the two masses?) (c) What is the velocity of the combined masses after the collision? (Hint: You shouldn’t have to do much here.)

Answer 1

a)
`V_(CM)=2.89 m/s`

b)
`v_(1)'=1.11 m/s`
`v_(2)'=-1.89 m/s`

c)
`v=2.89 m/s`

Step-by-step explanation:

The equatiom of velocity of the center mass it is given by:

`V_(CM)=(m_(1)v_(1)+m_(2)v_(2))/(m_(1)+m_(2))`

m₁=12 kg, v₁=4 m/s.

m₂=7 kg, v₂=1 m/s.

`V_(CM)=(55)/(19)=2.89 m/s`

b) The velocity of the first and second particle in the center of mass reference frame is given by:

`v_(1)'=v_(1)-V_(CM)` (1)

`v_(2)'=v_(2)-V_(CM)` (2)

So we will have:

`v_(1)'=4-2.89=1.11 m/s`

`v_(2)'=1-2.89=-1.89 m/s`

c) Using the conservation of momentum in a perfectly inelastic collision:

`m_(1)v_(1)+m_(2)v_(2)= (m_(1)+m_(2))v` (3)

Solving (3) for v:

`v=(m_(1)v_(1)+m_(2)v_(2))/(m_(1)+m_(2))` (4)

Let's recall a perfectly inelastic collision means that the colliding particles stick together afte the collision.

If we see, the equation 4 is the same as the velocity of the center of mass. So v = 2. 89 m/s.

I hope it helps!

Have a nice day!