How Is Speed Calculated in Elastic Collisions?

[chickens]

Do help me in the question below. Thanks in advance.

http://forum.lowyat.net/uploads/post-68-1089800451.jpg

 

[arildno]

Welcome to PF!
If you've read the guidelines, you would see that no one will bother to do your homework, but are willing to help you out provided you show what you've done so far.
So, to begin with:
How would you define work and energy ? (Question a)

 

[chickens]

Oh, let's see,

Q a )Work - The work done by a constant force F when the displacement of its point of application in S is given by the scalar products of F and S.

Energy - Energy enables a body to do work.

I also did b (i) (ii), c (i) ...after that I'm stuck on the following questions...

for c (ii) i assume mu^2 = (m1+m2)V (conservation of momentum principle) but i wasn't able to get the answer :/

Please help me .

 

[arildno]

Much better!
But, what on Earth do you mean by:
"for c (ii) i assume mu^2 = (m1+m2)V (conservation of momentum principle)"
Are we talking momentum or energy here??

Now to help yo along a bit, let $$E_{0}$$ be the original energy of the ball just as it strikes the square.
We let P stand for the potential energy increase the square gains by changing from position (a) to position (b) (you have already calculated P, right?).

What does it mean that, by transferring 30% of $$E_{0}$$ to the square, the square is "just able to reach position (b)"?
Surely, this must mean that the square falls to rest at (b), that is, has no velocity!
Hence, the initial kinetic energy of the square is $$\frac{30}{100}E_{0}$$
whereas its final potential energy (measured relative to the initial position) is given by P.
By conservation of mechanical energy, we should have:
$$\frac{30}{100}E_{0}=P$$

 

[chickens]

Yes I do understand that formula. But in c(ii), it wants the speed at which the ball strikes the block. During a very short period, the ball will touch the block eventhough its a elastic collision...so that's why i assume $$mu^2 = (m_{1}+m_{2})v$$ but i really don't know if that's correct.

 

[Doc Al]

Yes I do understand that formula. But in c(ii), it wants the speed at which the ball strikes the block. During a very short period, the ball will touch the block eventhough its a elastic collision...so that's why i assume $$mu^2 = (m_{1}+m_{2})v$$ but i really don't know if that's correct.

It's not only not correct, it's nonsense. Energy and momentum are two different things--they don't even have the same units! So... don't mix them up.

There is no question that mechanical energy is conserved: it's stated in the problem. So use that fact.

arildno explained how to approach this problem. But let me ask a few questions that may lead you in the same direction: (1) How much energy is needed to tip the block? (you already calculated that before you got to question c) (2) If that energy is 30% of the ball's original KE, how much is the ball's original KE? (3) If you know the KE, what's the speed?

You say you got the answer for c (i)? How did you get it? If so, the answer for c (ii) follows from the definition of kinetic energy: $KE = 1/2 mv^2$.