1. ## Forces Question

Question:
A steel ball of mass $73g$ is held above a horizontal steel plate, as illustrated in the figure below.

The ball is dropped from rest and it bounces on the plate, reaching a height h.

(a) Calculate the speed of ball as it reaches the plate.

(b) As the ball loses contact with the plate after bouncing, the kinetic energy of the ball is $90\%$ of that just before bouncing. Calculate
(i) the height h to which the ball bounces.
(ii) the speed of the ball as it leaves the plate after bouncing.

(c) Using your answers to (a) and (b), determine the change in momentum of the ball during the bounce.

Attempt:
(a) $h = 1.6m \ , \ g = 9.81 ms^{-2}$

$\frac{1}{2}\not{m}v^2 = \not{m}gh$

$v = \sqrt{2gh}$
$v = \sqrt{2 \times 9.81 \times 1.6}$
$v = 5.6 ms^{-1}$

(b)(i) Don't know how to solve this this question... need help!

2. Originally Posted by looi76
(b)(i) Don't know how to solve this this question... need help!
Since when the ball just loses contact it has 90% of the kinetic energy, the ball will go up until all of this kinetic energy is converted to potential energy.

So

$0.9 \left(\frac12 mv^2 \right) = mgh$

$h = \frac{0.9v^2}{2g}$

Here v is the velocity when the ball just loses contact with the plank

3. Thnx Isomorphism,

(b)(i) $1.4m$
(ii) $5.3ms^{-1}$

Part C

$Change \ in \ Momentum = Final Momentum - Initial Momentum$
$Change \ in \ Momentum = mv - mu$
$Change \ in \ Momentum = (73 \times 10^{-3} \times 5.3) - (73 \times 10^{-3} \times 5.6)$
$Change \ in \ Momentum = -2.19 \times 10^{-3}Ns$

Where did I go wrong?

4. Originally Posted by looi76
Thnx Isomorphism,

(b)(i) $1.4m$
(ii) $5.3ms^{-1}$

Originally Posted by looi76
Part C

$Change \ in \ Momentum = Final Momentum - Initial Momentum$
$Change \ in \ Momentum = mv - mu$
$Change \ in \ Momentum = (73 \times 10^{-3} \times 5.3) - (73 \times 10^{-3} \times 5.6)$
$Change \ in \ Momentum = -2.19 \times 10^{-3}Ns$

Where did I go wrong?
I think its right.Why do you think its wrong?

5. I think its right.Why do you think its wrong?
The answer in the book is 0.80Ns

6. Here is the answer, I scanned it:

Can you help me now?

7. Originally Posted by looi76
Thnx Isomorphism,

(b)(i) $1.4m$
(ii) $5.3ms^{-1}$

Part C

$Change \ in \ Momentum = Final Momentum - Initial Momentum$
$Change \ in \ Momentum = mv - mu$
$Change \ in \ Momentum = (73 \times 10^{-3} \times 5.3) - (73 \times 10^{-3} \times 5.6)$
$Change \ in \ Momentum = -2.19 \times 10^{-3}Ns$

Where did I go wrong?
Oh yes , silly me...

The velocity of the ball is directed towards the plank. While it leaves the plank, it is directed away. So if you assume one is positive, the other is negative...

Now this means
$Change \ in \ Momentum = mv - mu$
$Change \ in \ Momentum = (73 \times 10^{-3} \times 5.3) - (73 \times 10^{-3} \times (-5.6))$
(That is because v = 5.3 and u = -5.6.)
$Change \ in \ Momentum = (73 \times 10^{-3} \times 5.3) + (73 \times 10^{-3} \times (5.6))$
$Change \ in \ Momentum = 0.8 Ns$