# Conservation laws (tricky, conceptual,not homework)

1. Nov 24, 2008

### f(x)

The problem statement, all variables and given/known data
Consider a small body of mass m placed over a larger body of mass M whose surface is horizontal near the smaller mass and gradually curves to become vertical at height h.

The smaller mass is pushed on the longer one at speed v and the system is left to itself.
Assuming all surfaces to be frictionless, find the speed of the smaller mass when it breaks off the larger mass.

My attempt
This problem has been frustrating me for quite some time now. :uhh:
I start by assuming the bigger block is moving at speed V to the left when the smaller block is on the vertical and just about to break off.
Then by conserving momentum in horizontal direction : $$V=\frac{mv}{M+m}$$
since the small block is also having a velocity component V to the left wrt ground

Now, i am going to use conservation of energy, and this is where I have doubts.
I am not sure whether to choose only the small block & ground as the system , or to choose both blocks & ground as the system. (I am getting answer in neither)

I think that in choosing only the small block and ground as system, an external force by the large block is acting (normal reaction) and thus i cannot use conservation of energy directly. Is this hypothesis correct ?

So, choosing the both blocks and ground as system :
The small block is having a velocity component upwards and velocity component left wards which is V. Let the required speed of the small block at the time of breaking off be x
Since $$F_{ext}=0$$
$$\frac{1}{2}mv^2 = \frac{1}{2}MV^2 + mgh + \frac{1}{2}mx^2$$
Plz verify whether this equation is correct

Solving, i get $$x^2=\frac{(v^2-2gh)(m+M)^2-mMv^2}{(m+M)^2}$$

while the answer is printed as $$x^2=\frac{(M^2+Mm+m^2)}{(M+m)^2}v^2-2gh$$

Sorry for the length, but i desperately need some input
Thx a lot !

2. Nov 24, 2008

### Chi Meson

Energy is conserved, as you noted. The normal force never causes motion in the direction of that force, therfore, the normal force does not do any work and does not need to be considered in the CoE formulae

But you need another equation, since you have two unkowns.

What else is always conserved?
Edit: oops, you already noted this.

Edit again: TinyTim caught it first. Dang.

Last edited: Nov 24, 2008
3. Nov 24, 2008

### tiny-tim

Hi f(x)!

The length is fine … you've explained the problem admirably clearly.

ah, I expect you're thinking of am ordinary frictionless roller-coaster or pendulum-type situation, where energy is conserved even though the mass is constrained to move along a curve …

that works because the work done is zero … since there's no friction, the external force is normal to the surface, while the motion is by definition parallel to it …

but here, the force is still normal to the surface, but the motion (relative to the Earth) isn't parallel to it, because the whole system is moving.

So: general principle: energy is conserved only if work done is zero: and in this case you're right (but for the wrong reason … a normal force usually is ok): you must use the whole system.
erm … they're the same!

(M2 + mM + m2) = (M + m)2 - mM

4. Nov 25, 2008

### f(x)

Thx a lot tinytim and Chi Meson for your help :)

Ah ok thanks

Oh :surprised
I thought the answer meant $$\frac{(M^2+Mm+m^2)(v^2-2gh)}{M+m)^2}$$ but i was unsure about the reasoning, still wanted to confirm
Thx a lot for those corrections tinytim :)

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