Ice tennis

Classical Mechanics Level 2

I am standing on an ice rink, holding a tennis ball.

I threw the tennis horizontally such that the ball moves with speed v v relative to me. With what speed did I move (relative to the ground) after throwing the ball?

Details and assumptions

  • My mass is M M , the ball's mass is m m

  • There is no friction between me and the ground.

v v m v M \dfrac{mv}{M} m v m + M \dfrac{mv}{m + M} I will remain at rest

Pranshu Gaba by Pranshu Gaba , 22, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Pranshu Gaba
Feb 23, 2017

The total momentum of the ball + me = 0.

Suppose I am moving with velocity V V relative to the ground. My momentum is M V MV . The ball's velocity relative to the ground is − v + V -v + V . Its momentum is m ( − v + V ) m(-v+V) .

M V − m v + m V = 0 ( M + m ) V = m v V = m v M + m \begin{aligned}MV - mv + mV & = 0 \\ (M+m) V & = mv\\ V & = \frac{mv}{M + m} \end{aligned}

My speed relative to the ground is m v M + m \frac{mv}{M + m} .

3 Helpful 1 Interesting 1 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...