# Homework Help: F = MA 2009 # 14 (Momentum Quick Conceptual Question)

1. Jan 29, 2013

### SignaturePF

1. The problem statement, all variables and given/known data
See # 14

2. Relevant equations
None really, perhaps keep in mind that
1) p is conserved in the absence of external forces
2) L is conserved in the absence of external torque
3) Mech E is conserved if no energy is lost due to external forces

3. The attempt at a solution
I'm not sure how to attack this problem. My reasoning was that the bullet, as it is embedded into the block, has most of its velocity absorbed by the interior of the block (not rigorous at all). This means that the block does not move at the velocity it should and thus all A, B, C, D are false.

How can I do this rigorously?

2. Jan 29, 2013

### Simon Bridge

14. A wooden block (mass M) is hung from a peg by a massless rope. A speeding bullet (with mass m and initial speed v0) collides with the block at time t = 0 and embeds in it. Let S be the system consisting of the block and bullet. Which quantities are conserved between t = −10 s and t = +10 s?

(A) The total linear momentum of S.
(B) The horizontal component of the linear momentum of S.
(C) The mechanical energy of S.
(D) The angular momentum of S as measured about a perpendicular axis through the peg.
(E) None of the above are conserved.

... and then what happens?

Usually it is quicker to take each option one at a time.
But you have a nice list ...
... "external" to what? Anyway - rephrase your list as questions:

1. are their any external forces to the system being brought to bear?
2. are their any external torques to the rotating part?
3. is their any energy loss (note: does not have to be due to "external forces")?
... from those answers you have characterized the system so the options will make sense.

Learning how to ask hard questions is basically the main point of science education.

3. Jan 29, 2013

### SignaturePF

I thought
1) no external forces
2) no external torque
3) energy loss due to heat/friction inside the block

So the fact that there's an energy loss due to thermal energy negates everything?

4. Jan 30, 2013

### Simon Bridge

Well... no: that's not what your list is telling you. You can conserve momentum and not (mech) energy ...

You want to take another look at #2, at t<0 what is the angular momentum? What is it at t>0? Are these numbers the same?

Now compare your answers to the three questions with each option provided.

Last edited: Jan 30, 2013
5. Jan 30, 2013

### fgb

UPDATE: I had misread the problem, nevermind what I said before :P

UPDATE2: I read it right, I erased my comment for nothing ¬¬ Will re-write. Sorry.

Something like that: In these "simple" problems, forces essentially come from gravity or contact. An external force will come from contact to the external world (or existance of gravity with an external object), and an external torque will have to be produced by that external force.

Is any part of the system (bullet + block) in contact to the external world (rope, Earth, etc.)? Does if feel a force from those parts? Does this force produce a torque about an axis through the peg? :)

Also, notice that the problem considers a "long" time interval before and after the collision. Try to visualize how the system will behave when the block is hit by the bullet, it really helps!

Last edited: Jan 30, 2013
6. Jan 30, 2013

### Simon Bridge

Case in point - let the bullet be mass m and speed v
block is mass M, initial speed 0
if the final speed of m+M is u, then:

conserving momentum: $mv = (m+M)u$

this means that $u=mv/(m+M)$
... so the final speed is slower than the initial speed like you intuited - but the overall momentum still stays the same.

now look at the kinetic energy:

initially: $K_i = \frac{1}{2}mv^2$
finally: $K_f = \frac{1}{2}(m+M)u^2 = \frac{1}{2}(m+M)[mv/(m+M)]^2 =\frac{1}{2}(mv)^2/(m+M)$

comparing them: $$\frac{K_f}{K_i} = \frac{m}{m+M}$$ i.e. $K_i > K_f$ and kinetic energy is not conserved even though momentum is.

What happened to it?
Well it got lost in heat, sound and so on... which was your original thought.

Note: fgb has a valid point about how gravity (etc) applies here though - modifying the above.

7. Jan 30, 2013

### fgb

At first I had thought that the correct answer was D. Since as the block moves it feels a force due to the rope (which has a horizontal component after the rope tilts a little) there are external forces and no conservation of linear momentum. Tricky question since you are told to consider a fairly large time interval - momentum is still conserved during and short after the collision, but not a little later as the block moves.

Mechanical energy is lost in the collision due to deformation, as the bullet sticks inside the block.

The force due to the rope (the external force), also, always passes through the axis, so it does not produce an external torque and at first I thought angular momentum is conserved. However, taking gravity into account, gravity indeed produces an external torque and therefore no quantity is conserved, so I would say the correct answer is E.

8. Jan 30, 2013

### Simon Bridge

Actually the question kinda defines "before" as at the instant t=-10s, and "after" as the instant t=+10s. Anyhow: if we include gravity - then there is an external force messing with the system. Anything with momentum being conserved will get messed up. That leaves energy - but the collision is inelastic - and "none of the above".

I had misread the question - I thought it referred to the collision at t=0, which is where it usually appears.

To be super-rigorous, you'd have to construct the equations for each case.
However, most of the questions in this paper can be tackled by just going through each of the alternatives one at a time after stopping to think of the processes involved. There's usually two options that look likely, which is where you look harder.