Question

A 60 kg boy and 40 kg girl stand on skateboards on a frictionless horizontal surface. The boy pushes the girl away from him. The girl gains a speed of 0.3 m/s during the 0.50 s the boy's hands are in contact with her.
A. What will be the boy's speed after?


B. Assuming that in each case the girl achieved the same speed, would it matter whether the boy pushed the girl, the girl pushed the boy, or they put their hands together and pushed each other away?

Answer 1

Here in the above situation when boy pushes the girl there will be equal and opposite force on boy as per Newton's III law.

Due to this action reaction law boy will also go back with some speed.

Since there is no external force on this girl + boy system so we can use momentum conservation principle here.

As per momentum conservation

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

$60*0 + 40*0 = 60*v_{1f} + 40*0.3$

$0 = 60*v_{1f}+ 12$

$v_{1f} = -0.2 m/s$

So boy will go back with speed 0.2 m/s

Part b)

Since the boy and girl will always exert same force on each other by Newton's III law so it has no it matter whether the boy pushed the girl, the girl pushed the boy, or they put their hands together and pushed each other away.

As in all above cases the as per Newton's III law the force on them is always equal and opposite.