Applied Maths Question (Impacts & Collisions)

Closet Monkey

Stuck on this question. Anyone intrested in solving it. The answer is 1/3. I got the answer but I doubt I did it the right way.




Here is my solution:




I suspect that I made a mathematical error on the last part. Also to get the ratio you have to say K.E before - K.E after/ K.E before if you are wondering what I did at the end.

aarond280

are you sure the answer is 1/3 and not 5/9 ?

Closet Monkey

Yeah, it says that in the back of the book. I got 5/9 as well but I did it again because I thought it was wrong. Can anyone confirm the book is wrong?

Also you may have noticed that I put a +u Sin A for the before but it's actually meant to be a -u Sin A same goes for the after, it's a +VSinQ.

aarond280

is it Oliver Murphys new book ?

finality

You've made a mistake at the end there, with your substitution (you forgot to square the 2/3). Also, when you were calculating the kinetic energy, you only used the vertical velocity. When you rotate the whole diagram you can no longer say that the ball falls vertically- as it now has velocity in both the i and j directions.

Here's how I'd do it:


Though it's been a long time since I've done this kind of question, I could've done something crazy.

finality

But for the A1 you need a little more detail than I gave.

Closet Monkey

finality wrote: »

But for the A1 you need a little more detail than I gave.

Thank you for the answer. I never would have guessed you could do it that way. I don't know what else you could have added, your answer seems perfect.

Also I'm curious to know why the i-components have a negative value? I may have missed something in answering these kind of questions.

aarond280

Closet Monkey wrote: »

Thank you for the answer. I never would have guessed you could do it that way. I don't know what else you could have added, your answer seems perfect.

Also I'm curious to know why the i-components have a negative value? I may have missed something in answering these kind of questions.

It's negative because it's going in a negative direction. The i direction is negative when its traveling to the left or the negative i axis, the j is negative when it is travelling downwards or when its going towards the negatve j-axis

finality

Closet Monkey wrote: »

Thank you for the answer. I never would have guessed you could do it that way. I don't know what else you could have added, your answer seems perfect.

Also I'm curious to know why the i-components have a negative value? I may have missed something in answering these kind of questions.

They're going left, that's usually the negative direction. It doesn't actually matter though, you could also assign left as the positive direction, as long as you remember that anything going right would then be negative.

Closet Monkey

Oh, I see. I keep mixing up the i-plane and j-plane.

Well thank you anyhow for answering my question.

Closet Monkey

Anyone willing to answer another question. I am almost finished Impacts but there is just one part of a question I'm stuck on. (Question below!)


Here is my answer for part (a) and part (b). I can't get part (c) at all.






Also I don't know if part (b) is right or wrong. If I didn't say s = s +1 and added 1 instead of take 1 away I would get a different answer. Can you clarify for me why do you take 1m away from the distance? Is it because they are both travelling in the same direction or opposite directions. i.e like this:

Closet Monkey

Any answers please?

Osric

Your solution in (b) is almost correct, except it should be 1 - (1-e)/(1+e).
(1-e)/(1+e) is the distance the 1st (left) sphere has travelled in the time it takes the second one to hit the wall, so you take that from the 1m (original distance from wall and sphere2) to get the new distance

For (c).
1st you get the velocity of sphere 2 after it hits the wall, which is just e * u(1+e)
Now, sphere1 is still moving at u(1-e), and sphere2 is now moving at eu(1+e). When they collide, they will together cover a distance of 2e/(1+e) - the distance between the two from (b). So, distance sphere1 travels + distance sphere2 travels = 2e/1+e. From that, where t is time until collision,
u(1-e)t + eu(1+e)t = 2e/1+e. Factorise and multiply out to get t(u + ue^2) = 2e/1+e. from that, you get t = 2e/(1+e)(u)(1+e^2). Then, the distance of the collision from the wall = distance travelled by sphere2 in that time (from initial position at wall => collision point). So d = 2e/(1+e)(u)(1+e^2) * e(u)(1+e). Stuff cancels and you get 2e^2/1+e^2.

Closet Monkey

Thank you for answering my question. I've been waiting for 2 weeks for an answer.

I never found the velocity of the ball after it hit the wall. That's probably the main reason I couldn't find the distance.

Also is the velocity of sphere 2 not -e*u(1+e)?

Osric

Closet Monkey wrote: »

Thank you for answering my question. I've been waiting for 2 weeks for an answer.

I never found the velocity of the ball after it hit the wall. That's probably the main reason I couldn't find the distance.

Also is the velocity of sphere 2 not -e*u(1+e)?

Technically, yes. But since it's colliding with the wall (a stationary particle), you can ignore the minus for the sake of the question. It'll just mean that the velocity of sphere2 is negative relative to sphere1, but you're dealing with the distance they travel, and since you don't need proper velocities for another collision (momentum equation), you just take the absolute value of the velocity

Closet Monkey

Ok, thank you. I think that was the reason I didn't get the right answer after you showed me how to do it. I should've just left it with a + than.

bpb101

there the answer for the second question asked

Closet Monkey

Wow, thank you! I should've know you could use Speed = Distance/Time since there is no acceleration. Thank you for the detailed answer.