Question

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?

Answer 1

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

Step-by-step 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:

`mv_1+m_1v_2=m_1_v_(1f)+m_2v_(2f)`.

For our two students

`(32kg)(v_1)+(43kg)(v_2)= (32kg)+(43kg)(-2.4m/s)+(43kg)(v_(2f))` (notice the - sign in -2.4m/s, this means going to the left)

since the students were not moving at first,
`v_1=v_2= 0`, therefore we have

`0= (32kg)(-2.4m/s)+(43kg)(v_(2f))`

solving for
`v_(2f)` gives

`76.8=(43kg)(v_(2f))`

`\boxed{v_(2f) = 1.79m/s}`

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