On a frictionless surface, a 32 kg student pushes a 43 kg student. If the 32 kg student slides back at 2.4 m/s, how fast will the 43 kg student be sliding and in what direction?

The 43kg student will be sliding at 1.79m/s opposite the direction the 34kg student is going.

Explanation:

Conservation of linear momentum!

The law of conservation of momentum says that in an isolated system, the momentum before must equal the momentum after:

.

For our two students

(notice the - sign in -2.4m/s, this means going to the left)

since the students were not moving at first, , therefore we have

solving for gives

Hence the 43kg student will be sliding at 1.79m/s to the right.

the chart show the masses and velocities of two colliding objects that stick together after a collision.

according to the law of conservation of momentum, the momentum of the object after the collision is  

answer

1,500

kg • m/s.