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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?

2 comments:

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[hide] Matt H said...

Sorry for the bad LaTeX formatting on my part.

One method I learned than solving the slightly complicated system of equations is to consider the center of mass frame (CM). If mass M falls down with velocity v in the CM frame, then it must return with velocity -v. Thus v_{CM}=\frac{Mv_1+2Mv_2}{M+2M} where v_1=\sqrt{2gh} by energy conservation. We thus see the return velocity,v_1, is -\frac{v}{3}. Using energy conservation again, the return height h_{new}=\frac{h}{9}

on 22 August 2011 at 11:25
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[hide] Paul Liu said...

Yup. That's exactly the method I was referring to in the solutions. Good job, Matt. =]

on 22 August 2011 at 12:14
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