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## Saturday, 20 August 2011

### Physics GRE - #19

As shown below, a ball of mass $m$, suspended on the end of a wire, is released from height $h$ and collides elastically, when it is at its lowest point, which a block of mass $2m$ at rest on a frictionless surface. After the collision, the ball rises to a final height equal to:

• $\dfrac{h}{9}$
• $\dfrac{h}{8}$
• $\dfrac{h}{3}$
• $\dfrac{h}{2}$
• $\dfrac{2h}{3}$

Solution :

Using conservation of energy, we know that: $2mgh = mv_0^2 = mv_1^2+2mv_2^2$ where $v_0$  is the initial velocity and $v_1$, $v_2$ are the resulting velocities after the collision.

Using conservation of momentum, we also know that $mv_0 = mv_1+2mv_2.$

Solving these equations give $v_1 = -\frac{v_0}{3}, v_2 = \frac{2v_0}{3}$ which make sense as the ball must bounce back after the collision. Now using conservation of energy again, we can determine the height that the ball will rise to: $mgh_{new} = \frac{1}{2}mv_1^2 = \frac{1}{2}m\frac{v_0^2}{9}= \frac{mgh}{9}\implies h_{new}=\frac{h}{9}.$

There is a faster way to find $v_1$ using the concept of a centre of mass. Can you see it?

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