Stormer's Question Thread

[Stormer]

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.

[Akirus]

Strange, I've never seen a question where you have to estimate the values... I suppose I can try.

Well, for starters:



We know and 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. We'll go with Mao's 0.13s.











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...

[superflya]

has to be gravity :p

[Akirus]

This is what I get for trying to do an ambiguous projectile motion question without any prior practice in the middle of the night: fifteen minutes of my life well wasted.

And I swear, the note at the end was enough, did you have to remind me? >.>

[superflya]

cmon, t'was ONLY 15 mins

[Stormer]

It was actually in the school homework book. :p

[superflya]

heinemann 2?

[QuantumJG]

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 know and 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.

[superflya]

Yea true, velocity is constant throughout the motion for the horizontal component

[Akirus]

When it leaves her hand it has 12J, which should account for both the gravitational potential (for the height it is from the ground) AND the kinetic energy from its movement, no?

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.

[Mao]

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).

[Akirus]

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).

[Stormer]

Can you insert the 7.7 into the W = Fd equation?

[Mao]

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).

yes

Can you insert the 7.7 into the W = Fd equation?

yes

[Stormer]

So can I say that W = Change in Ek?

Change in Ek = 12J?