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I am currently doing my Physics homework and am stuck on a problem that is giving me an issue.

A tennis ball has a mass of 0.057 kg. A professional tennis player hits the ball hard enough to give it a speed of 46 m/s (about 103 miles per hour.) The ball moves toward the left, hits a wall and bounces straight back to the right with almost the same speed (46 m/s). As indicated in the diagram below, high-speed photography shows that the ball is crushed about d = 2.7 cm at the instant when its speed is momentarily zero, before rebounding.

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Making the very rough approximation that the large force that the wall exerts on the ball is approximately constant during contact, determine the approximate magnitude of this force.

After solving a very similar problem correctly, I took the following approach to solve it.

$$|\vec{v_{avg}}|=\frac{46}{2}=23 \frac{m}{s}$$ This is the average velocity from the time that the ball first makes contact with the wall until it comes to rest, and this answer was marked as correct.

The next step I took was to find the time it took from first contact with the wall until the ball came to rest. The question this was intended to answer was

How much time elapses between first contact with the wall, and coming to a stop?

$$t=\frac{|\vec{r}|}{|\vec{v_{avg}}|}=\frac{0.027 m}{23 \frac{m}{s}}=.0012\, s$$ However, this answer was marked incorrect, along with my answer that followed which found the total force exerted on the ball by the wall using this time.

$$\Delta\vec{p}=\vec{F_{net}}\Delta t$$ $$|\vec{F_{net}}|=\frac{\Delta|\vec{p}|}{\Delta t}=\frac{2.622 \frac{kgm}{s}}{.0012\,s}=2185\,N$$

Where did I go wrong in solving this problem?

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  • $\begingroup$ Note, your expression for time is incorrect. The expression should give a vector as the result, but doesn't. You are missing the numerical symbols on $\vec r$. And again on force and momentum, you are finding the magnitudes, not the vectors, even though they are written with vector arrows. $\endgroup$ – Steeven Aug 28 '16 at 22:09
  • $\begingroup$ @Steeven Thank you, I believe I edited it correctly. $\endgroup$ – Chris Loonam Aug 28 '16 at 22:12
  • $\begingroup$ Your calculations are correct. Where you went wrong, it seems, is in choice of teacher. $\endgroup$ – Ben51 Jan 20 '18 at 17:10
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Edited after the mistake was pointed out by Ben51 :-

The time can be found out by using the equation $\vec{s}= \vec{u}t + \frac{1}{2}\vec{a}t^2$ where $s$ is displacement in time $t$, $u$ is the initial velocity and $a$ is acceleration :- $$ 0.027=46t - \frac{1}{2}at^2 ; a=\frac{46}{t} $$ Therefore $$0.027=46t - 23t$$ Thus $$t=\frac{0.027}{23}$$ So the answer is coming out to be the same.

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  • $\begingroup$ Velocity and acceleration are of opposite signs during the deceleration, so the first equation is wrong. $\endgroup$ – Ben51 Jan 20 '18 at 17:09

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