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

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In summary: Summing it up, the problem presents a wooden block (mass M) hanging from a peg by a massless rope. A bullet (mass m, initial speed v0) collides with the block at time t = 0 and embeds in it. The system consists of the block and bullet, and the quantities being conserved between t = −10 s and t = +10 s are the total linear momentum of the system, the horizontal component of the linear momentum, the mechanical energy, and the angular momentum. However, since there are external forces acting on the system (such as gravity), none of the above quantities are conserved and therefore, the correct answer is E.
  • #1
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Homework Statement


https://aapt.org/physicsteam/2010/upload/2009_F-ma.pdf
See # 14


Homework 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

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?
 
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  • #2
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.

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
... and then what happens?

Usually it is quicker to take each option one at a time.
But you have a nice list ...
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
... "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.

Science is not about knowing answers, it is about asking uncomfortable questions.
Learning how to ask hard questions is basically the main point of science education.
 
  • #3
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
So the fact that there's an energy loss due to thermal energy negates everything?
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.
 
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  • #5
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 existence 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!
 
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  • #6
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
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
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.
 

Related to F = MA 2009 # 14 (Momentum Quick Conceptual Question)

What is F = MA?

F = MA is a mathematical formula that represents the relationship between force, mass, and acceleration. It states that the force (F) acting on an object is equal to its mass (M) multiplied by its acceleration (A).

What does momentum have to do with F = MA?

Momentum is a property of moving objects that is closely related to force and acceleration. The formula F = MA can be used to calculate the momentum of an object, as momentum is equal to the mass of an object multiplied by its velocity (which is a measure of its acceleration).

Why is F = MA important in physics?

F = MA is important in physics because it is one of the fundamental equations that helps us understand the behavior of objects in motion. It allows us to predict and calculate the forces acting on objects and how they will change their motion.

Can F = MA be applied to all types of motion?

Yes, F = MA can be applied to all types of motion, whether it is linear (straight-line) motion or rotational motion. This is because it is a universal equation that describes the relationship between force, mass, and acceleration.

How does F = MA relate to Newton's Laws of Motion?

F = MA is essentially a simplified version of Newton's Second Law of Motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. F = MA can also be used to understand and apply Newton's other laws of motion.

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