# momentum

Discussion in 'MCAT Study Question Q&A' started by Sonyfan08, Jul 23, 2011.

1. ### Sonyfan08 7+ Year Member

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TBR # 4 pg. 192: The situation is a bullet colliding into a block. If the bullet mass keeps increasing, why does the resulting system have a maximum final speed?

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2. ### TieuBachHo 7+ Year Member

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One thing that I notice here that ppl usually get quick responses. You might want to provide a general picture / description (using words or images) for others to understand. That way, they can help you even if they don't have your references.

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### Sonyfan08 7+ Year Member

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Thanks for the suggestion!

4. ### TieuBachHo 7+ Year Member

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...

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5. ### zwander 7+ Year Member

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Imagine this scenario. The block which the bullet is striking is 1kg. How say you start the experiment with a bullet that is .001kg. The speed will drastically change for the bullet when is strikes the block because it is so much smaller. But say the bullet is not 100000000kg. That is huge. Thus when such a large object strikes another object, the final speed changes very little, so at very large masses the speed of the bullet plateaus.

6. ### TieuBachHo 7+ Year Member

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You are right? The final speed right after the collision. I misinterpret the question. But the bullet mass increases? This is a bit confusing too.

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8. ### TieuBachHo 7+ Year Member

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EDIT from above (take multiple times) LOL:

The maximum speed of the block, not the system (right after the collision), so it increases to a maximum. The bullet mass doesn't increase but the bullet mass before the collision (=0) when the block rests at 0. The mass of the bullet entering as it trikes the block increases. Big differences, but the wordings are confusing.

Edit: If anything, the bullet mass should decrease due to burnt-off in collision. But I doubt they mean that.

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Thanks!

10. ### zwander 7+ Year Member

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m=mass bullet
M=mass block

So if mv=(m+M)v and m>>M, then v final is essentially the same as v initial for extremely large bullets.

Edit: and that final v is the same for m as it is for M

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### Sonyfan08 7+ Year Member

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So I have another question, lol:

If there was a pole on top of a fulcrum with two objects hanging on either end was in rotational and translational equilibrium, and one of the object was moved to disturb this.. the forces will no longer be in translational equilibrium & rotational equilibrium. Now the forces no longer act at the fulcrum, but now all of the forces act at the center of mass of the pole. So, I guess by moving one of the object, it moved the center of mass?

12. ### TieuBachHo 7+ Year Member

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You are asking about torque causing the pole to be displaced out of equilibrium? The center of mass of the system is not the same as the center of mass of the pole. By moving the object(s), you displace its mean location due to relative distribution of masses of the whole. Make sense? "Mean location" is just another word for a relative location.

13. ### zwander 7+ Year Member

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Yea, you definitely move the center of mass when you move objects that were previously in trans and rotational equilibrium. In terms of rotational equilibrium, you are increasing the distance through which the force is acting if you increase the lever arm, so you lose rotational equilibrium, and then as the pole begins to tilt, you have work being done of the pole by gravity (a component of the force becomes parallel to the length of the rod), so you'll lose translational equilibrium.

So yea, I think at equilibrium, the fulcrum was at the center of mass to balance the forces, but if you move the weights and not the fulcrum, then it can no longer be in equilibrium.

I'm pretty sure that the certain of mass of an object only changes when a net force acts on the system. And yea in this case, the center of mass changes when a force moves the weights unbalancing the pole.

14. ### dmplz707

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So what was the answer to the original question, is it C?

15. ### salim271 Patience is tough. :/ 2+ Year Member

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Its A, with 0 or close to zero mass, the final speed is very low, and with a increasing mass the final speed maxes out at some point and doesnt change anymore.

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### Sonyfan08 7+ Year Member

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One more question:

BR pg. 202, #5. (regarding momentum)

situation I: Khoi states toward a stationary Jen at a constant speed of 2 m/s. They collide and move together following collision.

Situation II: Jen skates toward a stationary Khoi at a constant speed of 2 m/s. They collide and move together following collision.

Khoi is 2 X the mass of Jen.

The question asks, in which situation does Jen feel the largest magnitude of impulse.

The answer was that Jen felt the same magnitude of impulse in either situation since her change in V is same in each situation. Khoi was heavier = greater momentum & greater force, and since impulse is = F*t, I thought situation I had greater impulse. Where is my logic incorrect?

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### Sonyfan08 7+ Year Member

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If a person sits on a stool and holds a 5 kg mass in each hand, and stretches out his hand, he will spin slower. But why is that? Torque is F*r. If the lever arm is the distance btw his shoulder and hand, wouldn't the torque be greater, hence greater acceleration? I'm assuming it's because inertia is greater, but I'm having trouble visualizing/understanding exactly what is inertia.

18. ### zwander 7+ Year Member

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So first of all, this is conservation of angular momentum problem, so really there is no torque involved in the system. That is one of the conditions for conservation of momentum, no net forces acting on the system. To figure out why he spins slow you need to know angular momentum (L)= angular velocity x moment of inertia. Moment of inertia is analogous to mass if that it measures an objects resistance to forces and it is proportional to m and the r^2 of the distribution of mass. thus as you extend your arms, the inertia in the system increases, and because angular momentum is conserved, angular velocity must decrease.

19. ### zwander 7+ Year Member

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You are correct though is saying the greater the lever arm the greater the torque thus the larger the angular momentum you have in a system if you apply a same force 2r away as compared to just r.

20. ### plzNOCarribbean

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How do you bump a thread? I dont want to start a new one but I just asked a question in the old one and it hasn't come up on the main screen

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### Sonyfan08 7+ Year Member

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@plznocarribean: hmm. I'm not sure why it didnt bump. If you responded to the post, it should bump up the thread..

Another question: What is the difference between angular momentum and linear momentum?

22. ### zwander 7+ Year Member

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Conceptually, they are essentially the same. They both essentially say in a closed system absent of any external factors (such as force or torque) this "momentum" of the system will not change, but only transfer between objects in the system. The equations to work with each one are different however, although the variables are analogous (angular velocity to velocity, Moment of inertia to mass for example). If you understand the concepts of linear momentum (cars colliding), probably the most important being the requirement for the absence of outside forces, you will be good to apply in to rotational momentum.

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### Sonyfan08 7+ Year Member

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Thanks for all of the responses so far. It's really helping me. =)

How does angular momentum exist if the tangential velocity of a rotating body is constantly changing?

24. ### indianjatt 7+ Year Member

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impulse = mdeltaV. In the first case her velocity goes from 0 m/s to 1.33 secs. Delta V = 1.333

In the second case she goes from 2 m/s to .67 m/s, so she also has Delta V = 1.33.

Since her mass is the same, impulse I = impulse II

Regarding impulse = F*t, think of the force on her. F = ma. a=dv/dt. For both cases dv/dt = 1.333 so her Force is the same in both cases. Therefore, she experiences equal impulse. Also, you're thinking that Khoi has a higher Force, but he doesn't. Newton's Third law states that Fa=-Fb. Khoi weighs 2x (2m). dv/dt = .67. F = 2m(.67) = 1.333 m. Jen weighs x (m). dv/dt = 1.33. F = m1.33 = 1.333 m . Same force

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