Post by: Stormer on January 11, 2010, 01:37:43 am
A student throws a ball horizontally at a wall. As it leaves her hand the ball has 12 joules of kinetic energy. Use this value of kinetic energy to estimate the average horizontal force that the student's hand exerts on the ball during the throw. You need to estimate a piece of information to determine this.
Post by: Akirus on January 11, 2010, 02:41:08 am
Well, for starters:
We know
Now, change in momentum/impulse:
Impulse is defined by the product of force and time applied. Now, since it doesn't specify the time, I'm supposing we need another estimate.
Where did you get this question from, anyway? I've never seen a physics question that needs you to make up half of the figures...
Post by: superflya on January 11, 2010, 02:43:12 am
Post by: Akirus on January 11, 2010, 02:48:54 am
And I swear, the note at the end was enough, did you have to remind me? >.>
Post by: superflya on January 11, 2010, 02:59:33 am
Post by: Stormer on January 11, 2010, 11:26:28 am
Post by: superflya on January 11, 2010, 11:57:59 am
Post by: QuantumJG on January 11, 2010, 12:50:36 pm
Hey, need one for physics too lol. This question is proving to be quite difficult.
A student throws a ball horizontally at a wall. As it leaves her hand the ball has 12 joules of kinetic energy. Use this value of kinetic energy to estimate the average horizontal force that the student's hand exerts on the ball during the throw. You need to estimate a piece of information to determine this.
Strange, I've never seen a question where you have to estimate the values... I suppose I can try.
Well, for starters:
We knowand for this problem we'll need the velocity so we can find the average horizontal force after calculating change in momentum (change in momentum = m * v = force * change in time). Lets assume the ball weighs 100g (I think this is what it wants you to do when it says "estimate"?), i.e.
:
Now, change in momentum/impulse:
Impulse is defined by the product of force and time applied. Now, since it doesn't specify the time, I'm supposing we need another estimate. For simplicity, lets assume the throw takes one second.
Where did you get this question from, anyway? I've never seen a physics question that needs you to make up half of the figures...
\\Dammit, forgot about gravity.
Actually if you think about it your hand is counteracting this gravity when you are holding the ball, also you are finding the horizontal force component so gravity doesn't need to be considered. Your hand will provide ~2N of force to throw a 100g ball which seems about right.
I wouldn't worry about these questions.
Post by: superflya on January 11, 2010, 01:45:38 pm
Post by: Akirus on January 11, 2010, 01:46:31 pm
Either way, the question is stupid, I wouldn't bother with it.
Just read the question again, it says kinetic energy only. I should probably pay more attention.
Post by: Mao on January 11, 2010, 02:12:59 pm
i.e. The poor girl would either have to do a runner-up or she'll need ~3.8m long arms. This seems slightly ridiculous, and this is caused by a poor estimation of the time.
I'd like to say this is a prime example of over-analysis. The VCE mindset make us link the phrase 'kinetic energy' automatically with
For a typical throw over 1 meter, a force of 12N is required, and the contact time is roughly 0.13 seconds (assuming 100g ball).
Post by: Akirus on January 11, 2010, 03:44:34 pm
For the amount of kinetic energy and assumed 100g mass, the velocity would be ~15.5 m/s, or ~60 km/h. Assuming force is constant, we can use the constant acceleration formula to see the distance over which the force was applied:
i.e. The poor girl would either have to do a runner-up or she'll need ~3.8m long arms. This seems slightly ridiculous, and this is caused by a poor estimation of the time.
I'd like to say this is a prime example of over-analysis. The VCE mindset make us link the phrase 'kinetic energy' automatically with. The formula you really want is
, where W is the work done (change in mechanical energy in this case).
For a typical throw over 1 meter, a force of 12N is required, and the contact time is roughly 0.13 seconds (assuming 100g ball).
If you use my working and change the time used in the impulse calculation to 0.13s instead of 1s, those are the figures you'd get, aren't they?
Edited my initial post to reflect what you just said (changed the assumed contact time).
Post by: Stormer on January 11, 2010, 04:17:47 pm
Post by: Mao on January 11, 2010, 05:08:07 pm
For the amount of kinetic energy and assumed 100g mass, the velocity would be ~15.5 m/s, or ~60 km/h. Assuming force is constant, we can use the constant acceleration formula to see the distance over which the force was applied:
i.e. The poor girl would either have to do a runner-up or she'll need ~3.8m long arms. This seems slightly ridiculous, and this is caused by a poor estimation of the time.
I'd like to say this is a prime example of over-analysis. The VCE mindset make us link the phrase 'kinetic energy' automatically with
For a typical throw over 1 meter, a force of 12N is required, and the contact time is roughly 0.13 seconds (assuming 100g ball).
If you use my working and change the time used in the impulse calculation to 0.13s instead of 1s, those are the figures you'd get, aren't they?
Edited my initial post to reflect what you just said (changed the assumed contact time).
yes
Can you insert the 7.7 into the W = Fd equation?
yes
Post by: Stormer on January 11, 2010, 05:13:31 pm
Change in Ek = 12J?
Post by: superflya on January 11, 2010, 05:14:41 pm
Post by: QuantumJG on January 11, 2010, 05:17:53 pm
ΔW = ΔK => 0.5mΔv2 = F x Δs
let m = 0.1kg => Δv = sqrt(240)m/s (~56m/s - reasonable)
let Δs = 1m (I would say an overarm throw is ~1m) => F = 12N
So the other parts are really redundant.
Post by: Stormer on January 11, 2010, 05:53:05 pm
Post by: QuantumJG on January 11, 2010, 06:08:52 pm
What do you mean by the s? Is that the length of the arm or is that how far the ball has been thrown?
It's how far your arm has moved.
Post by: Stormer on January 11, 2010, 06:32:01 pm
So 12J = F x ΔS?
Gives the answer of 12 still. I don't see why it's particular necessary to estimate the mass. This would mean that only one estimation is needed as well.
Post by: Akirus on January 11, 2010, 08:18:46 pm
Post by: Mao on January 11, 2010, 11:03:37 pm
Can you just go ΔK = 12J
So 12J = F x ΔS?
Gives the answer of 12 still. I don't see why it's particular necessary to estimate the mass. This would mean that only one estimation is needed as well.
exactly :)
Post by: Akirus on January 13, 2010, 10:58:37 pm