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A massless pulley can rotate freely around a horizontal axis. Around the pulley is attached a non-stretching string connected to 2 identical blocks that are at rest. A ball of clay is dropped from height $h$ onto the left block. Then the left block and the clay starts to move as one entity.

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Question: What is the magnitude of the acceleration of the left block and clay after the collision?

My attempt

Due to energy conservation, the speed of the ball of clay must be $v_{clay}=\sqrt{2gh}$. We are dealing with a completely inelastic collision, so $mv_{clay}+3m \cdot 0 = (m+3m)u$.

Solving this equation with respect to $u$ gives $u=\frac{1}{4}\sqrt{2gh}$. The time it took the ball of clay to fall was $t=\sqrt{\frac{2h}{g}}$.

$$a=\frac{\Delta v}{\Delta t} =\frac{1}{4}\sqrt{2gh} \cdot \sqrt{\frac{g}{2h}} = \frac{1}{4}g$$

But it turns out the correct answer is actually $a=\frac{1}{7}g$. Can anyone explain to me why this is the case?

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The force of gravity on one block is cancelled by the opposing force of gravity on the other. So the overall force is just the gravitational force on the clay. That force has to accelerate the combined mass of the blocks and the clay, ie 7m. Therefore, the resulting acceleration will be g/7.

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The collision of the clay with the block is a distractor. After the collision, even though the system begins to move because of the collision, the acceleration after the collision depends only on the relative mass difference of the two sides and the gravitational field.

Your time calculation is irrelevant because that time interval of the fall is unrelated to any acceleration of the blocks after the collision. The acceleration during the collision will not be constant because the collision force will not be constant.

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You have forgotten the momentum change of the block on the right. It is connected and will follow along so it must also be accelerated. In the momentum conservation law you must include all momentum changes (you must have a closed system).

So, it is not just $m+3m$ that is accelerated, but $m+3m+3m$: $$mv_\text{clay}+3m \cdot 0+3m \cdot 0 = (m+3m+3m)u$$

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  • $\begingroup$ This doesn't answer the question, which is about acceleration after the collision, not new velocity of system. $\endgroup$
    – Bill N
    Commented May 5, 2021 at 12:30
  • $\begingroup$ @BillN I see. I simply noted an error in the procedure - but you are right, the question leads towards focusing on the collision even though the collision itself has no influence on the acceleration that happens later on. Good catch. $\endgroup$
    – Steeven
    Commented May 5, 2021 at 13:41
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After the collision, (with an x axis wrapped CCW around the pulley) you have: 4mg – 3mg = 7ma.

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