I'm doing mechanics and I came across this question:

A ball of mass 0.2kg is dropped from a height of 2.5m above horizontal ground.
After hitting the ground it rises to a height of 1.8m above the ground.
Find the magnitude of the impulse received by the ball from the ground

The answer in the book I'm using says 2.59Ns. I first calculated the speed at which the ball hits the ground using v^2= u^2 +2as, which is 7m/s. The momentum when hitting the ground is therefore -1.4kgm/s.

The final speed is the bit which stumps me. If the answer really is 2.59Ns, then the speed has to be 5.9m/s. I successfully managed to compute this, when I solve the problem in terms of mgh and 1/2mv^2, however, I haven't actually covered that yet in the Mechanics course I'm doing, so I feel I shouldn't do it that way.

However, doing it using SUVAT, and assuming acceleration to be -9.8 and u as 7m/s, my final speed works out to be 3.7m/s, so clearly my impulse cannot be 2.59Ns. I then thought I should work out final speed using u as 0, which would make the total distance travelled 2.5-1.8=0.7m, and acceleration being 9.8, but I end up with the same answer. The only way I can see to get 5.9 as the speed, is to take u as 49 AND s as 0.7, although this isn't correct at all, as if we get the net distance, we can't then use any speed that wasn't the definitive initial speed.

So how do I calculate the final speed of this ball, to then calculate impulse, without having to go into GPE and Kinetic energy?

  • $\begingroup$ You said you got the right answer when using when using kinetic and potential energy, but you want to do it without those concepts. So just divide everything by $m$. Now you are just using "SUVAT". $\endgroup$ – Brian Moths Sep 29 '15 at 18:44
  • $\begingroup$ @NowIGetToLearnWhatAHeadIs No, because if i did 3.7 by m I get 18.5, so that doesn't seem to work. I'm looking for an answer along the lines of the same way I worked out initial speed, as that would also make the most sense to me. $\endgroup$ – Logan545 Sep 29 '15 at 19:10
  • $\begingroup$ Whoever downvoted could you please give a reason, as I've given this question a lot of thought. $\endgroup$ – Logan545 Sep 29 '15 at 19:12
  • $\begingroup$ You have to divide both sides of the equation by $m$. If you start out with an equality, and you divide both sides by $m$, you will still have an equality. $\endgroup$ – Brian Moths Sep 29 '15 at 19:44
  • $\begingroup$ Wow, ok I turned it into an answer. $\endgroup$ – Brian Moths Sep 29 '15 at 19:50

I'll risk moderatorial opprobium with a partial answer because you have come so close.

You correctly use the SUVAT equation $v^2 = u^2 + 2as$ to find that the velocity of the ball just before it strikes the ground is $v_i = -7$ m/s (using the sign convention that upwards is positive). So far so good.

Now you know the ball rises back up to a height of 1.8m, so you can use the SUVAT equation again. This time $v = 0$ and $s = 1.8$ m so you can solve for $u$. What you've now got is the upwards (positive) velocity of the ball, $v_f$, just after it leaves the ground again. Some may say $v_f$ works out to be $+5.94$ m/s but I couldn't possibly comment.

The change in momentum is then just $mv_f - mv_i$ but remember the sign convention - $v_i$ is negative.

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  • $\begingroup$ Yeah I remembered it's negative as I set momentum as -1.4. Ok, why can't I solve for v and assume u to be 7? Is it because during that time of deflection, u would no longer be 7? $\endgroup$ – Logan545 Sep 29 '15 at 19:58
  • $\begingroup$ Because stuff generally loses energy when it bounces of the ground. Also, if it bounced back with the same speed it hit the ground with, it would have gone back up the same height. $\endgroup$ – Brian Moths Sep 29 '15 at 20:01
  • $\begingroup$ @Logan545: the bounceback speed has to be less than 7m/s because the ball doesn't get as high after the bounce - only 1.8m when it started at 2.5m. That means the bounce is inelastic. $\endgroup$ – John Rennie Sep 30 '15 at 4:40
  • $\begingroup$ I don't understand that the velocity when the ball hits the ground is $7m/s$ since: if $m=0.2kg$, $s=2.5$, $u=0$ and $a=0.2g$ due to gravity then $v^2=u^2+2as$ does not give $v=7$... i.e. $v^2=0^2+2(0.2\times 9.8)2.5=\frac{49}{5}\neq 49$. $\endgroup$ – Pixel May 11 '17 at 12:27
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    $\begingroup$ @Antinous: you've written $v^2=u^2+2mas$ instead of $v^2=u^2+2as$ i.e. you have a rogue factor of $m$ in the equation. The acceleration is simply $a=g$ not $a=mg$. $\endgroup$ – John Rennie May 11 '17 at 12:49

You already said if you have $h$ as 1.8 m, then $\dfrac{1}{2} m v^2 = m g h$ implies $v$ is 5.9 m/s. However, you were unhappy becuase this involves energy. So just divide both sides of the equation by $m$. Now you have $\dfrac{1}{2} v^2 = g h$. The solution must still be 5.9 m/s so you get the right answer, this time using only SUVAT.

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  • $\begingroup$ In this case, h would actually be 0.7 if we're taking u to be 0. It would only be 1.8, if we include u^2 which is 49, and in that scenrio g would be -9.8 $\endgroup$ – Logan545 Sep 29 '15 at 19:53

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