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Here is a strange test question I really don't get:

A ball falls freely from a certain height, and due to air resistance it will reach a certain terminal velocity (assuming there is a normal gravitational acceleration going on all the time). Then it hits the ground, and bounces back without loss of energy.

Question : what is the acceleration on the ball immediately after it bounced?

The answer is $2g$ (2 times gravitational acceleration), but the explanation is not included.

Note: This is a question from the Physics Olympiad!

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Right before impact, there is no acceleration, such that forces by air resistance and gravity cancel. Right after impact, the velocity has changed sign (no energy loss), such that also the air resistance changes sign. Instead of $a=g-g$ you get $a=-g-g=-2g$.

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  • $\begingroup$ Now you tell me, is it really this easy?? So when I bounce without energy loss, I just change the sign of my acceleration? It's pretty hard to imagine this... $\endgroup$ – user209347 Aug 19 '13 at 7:11
  • $\begingroup$ This is the result when you apply conservation of energy. $\endgroup$ – Bernhard Aug 19 '13 at 7:45
  • $\begingroup$ oh I get it: so the air resistence is equal to g, but now you simply turn the vector velocity upside down and so does air resistence! what a typical Olympiad question: very easy when explained, but hard to think of. $\endgroup$ – user209347 Aug 19 '13 at 18:30
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Bernhard's answer is correct. But his shortcuts may be a bit too short. This would be a more extended answer.

First by doing a free-body-diagram (FBD) of the ball just before impact (with down being the positive direction), it'll make obvious that the only two components of forces are air resistance, $F_a$, and the force by gravity, $F_g = mg$, where $m$ is the mass of the ball and $g$ is the gravitational acceleration. By the simplified Newton's second law,

$$-F_a + F_g = ma$$

Because the ball has reached terminal velocity, we know that $a = 0$. Thus,

$$F_a = F_g \rightarrow F_a = mg \tag{1}$$

A FBD of the ball right after impact, will make clear that now both $F_a$ and $F_g$ point in the same direction. Again using Newton's second law,

$$ma = F_a + F_g \rightarrow ma = F_a + mg \rightarrow a = \frac{F_a}{m} + g \tag{2}$$

From (1), we know that $\frac{F_a}{m} = g$, and using this in (2) we get the answer

$$a = 2g$$

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  • $\begingroup$ As you are extending, why not add a sketch of de FBD? $\endgroup$ – Bernhard Aug 19 '13 at 21:05
  • $\begingroup$ Hello Bernhard, thank you for the edits. I'm pretty new to stackexchange as my "rept" would suggest; I don't know how to include non-internet figures. I might get back to it once it becomes second nature! $\endgroup$ – Esteban Aug 19 '13 at 21:08

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