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

Solution :

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?

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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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 $h_{new}=\frac{h}{9}$

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

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