I'm doing mechanics and I came across this question:
A ball of mass 0.2 kg is dropped from a height of 2.5 m above horizontal ground.
After hitting the ground it rises to a height of 1.8 m 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 $2.59$ Ns. I first calculated the speed at which the ball hits the ground using $v^2= u^2 +2as$, which is $7$ m/s. The momentum when hitting the ground is therefore $-1.4$ kgm/s.
The final speed is the bit which stumps me. If the answer really is $2.59$ Ns, then the speed has to be $5.9$ m/s. I successfully managed to compute this, when I solve the problem in terms of $mgh$ and $\frac{1}{2}mv^2$, 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 $-9.8$ m/s$^2$ and $u$ as $7$ m/s, my final speed works out to be $3.7$ m/s, so clearly my impulse cannot be $2.59$ Ns. I then thought I should work out final speed using $u$ as $0$, which would make the total distance travelled $2.5-1.8=0.7$ m, and acceleration being $9.8$ m/$s^2$, but I end up with the same answer. The only way I can see to get $5.9$ m/s as the speed, is to take $u$ as $49$ m/s AND s as $0.7$ m, 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?
Best Answer
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.