# Another Klassical Mechanics Dilemma!

Hey guys, I'm a university freshman @ NTU, Singapore, Computer Engineering course . Yet, I'm really into physics, especially klassical ones . Ok, a few days ago, a hiskool student asked me a question, I must say that I'm really stuck in this dilemma! . Give it a try if you guys can, Thanks!

Question: Two wedges of the same mass M, same shape and height h are laid movable on the floor. A mass m is put on the top of wedge 1; it is let to slide down wedge 1 and go up wedge 2. Calculate the maximum height mass m can reach wedge 2 in terms of m, M and h, suppose that the surfaces are perfectly frictionless.

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HallsofIvy
Homework Helper
The floor the wedges are sitting on is also frictionless? If not, and the wedges don't move, the problem is easy, no matter what the masses are, in order to conserve energy, the mass m must rise back to height h.
If the floor is frictionless, the wedges will also move. The point is that horizontal momentum must be conserved (not vertical since there is a force downward. Determine how the mass m and first wedge must move in order to conserve horizontal momentum and energy. After the mass moves onto the second wedge, remember that the first wedge now continues at that velocity. Now determine how the second wedge and mass m must move so that, along with the motion of the first wedge, horizontal momentum and energy are conserved. Finally, the maximum height will occur when the vertical velocity of mass m is 0 (so its horizontal velocity is the same as the second wedge) and total horizontal momentum and energy are still conserved.

yup. the principle is correct as mentioned above. the wedges are movable.
from the first wedge the ball slides:
momentum conservation:
$$mv=-MV_1$$
energy is conserved:
$$\frac{1}{2}mv^2+\frac{1}{2}MV^2=mgh$$
from these two we can get:
$$v^2=\frac{2gh}{1+\frac{m}{M}}$$

then this ball hits the second wedge, and climbs up.
when it reaches max height, the speed of ball and wedge are the same.
momentum conservation:
$$mv=(m+M)V$$
(this big V is different from above, since this is the final V of the wedge and ball)
energy:
$$\frac{1}{2}mv^2=mgh^' + \frac{1}{2}(m+M)V^2$$
we can then eliminate big V. rearranging,

i got:
$$h^'=h \frac{M^2+Mm-m^2}{(M+m)^2}$$

where $h^'$ is the final height.

you can check it yourself.

sorry, a bit mistyping above. from the first wedge the ball slides:
momentum conservation:
$$mv=-MV_1$$
energy is conserved:
$$\frac{1}{2}mv^2+\frac{1}{2}MV^2=mgh$$
from these two we can get:
$$v^2=\frac{2gh}{1+\frac{m}{M}}$$

then this ball hits the second wedge, and climbs up.
when it reaches max height, the speed of ball and wedge are the same.
momentum conservation:
$$mv=(m+M)V$$
(this big V is different from above, since this is the final V of the wedge and ball)
energy:
$$\frac{1}{2}mv^2=mgh^{'} + \frac{1}{2}(m+M)V^2$$
we can then eliminate big V. rearranging,

i got:
$$h^{'}=h \frac{M^2+Mm-m^2}{(M+m)^2}$$

where $h^'$ is the final height.

you can check it yourself.

hi UglieDuckie, is my answer correct or not?

have you checked it with your friend?

sniffer said:
hi UglieDuckie, is my answer correct or not?

have you checked it with your friend?
Man, your calculations are brilliant, thanks for paying attention!
After doing just the same steps, I came to the solution:

h' = h
____________
(1+m/M)^2

Check the ans in this attachment! #### Attachments

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i made an algebraic mistake. yours is correct.

thanks.