News: Currently the LaTeX and hidden solutions on this blog do not work on Google Reader.
Email me if you have suggestions on how to improve this blog!

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?


Post a Comment

This webpage is LaTeX enabled. To type in-line formulae, type your stuff between two '$'. To type centred formulae, type '\[' at the beginning of your formula and '\]' at the end.