Some Momentum and KE physics

[Justin218]

Any of you guys help me out with any of these problems? Thanks

7. Two objects move on a level frictionless surface. Object A moves east with a momentum of 24 kg*m/s. Object B moves north with momentum 10 kg*m/s. They make perfectly inelastic collision. What is the magnitude of their combined momentum after the collision?
a. 14
b. 26
c. 24
d. cannot be determined without knowing masses and velocities.

11. In an inelastic collision, if the momentum is conserved, then which of the following statements is true about kinetic energy?
a. KE is also conserved.
b. KE is gained.
c. KE is lost.
d. None of the above.

17. A small object collides with a large object and sticks. Which object experiences the larger magnitude of momentum change?
a. large object
b. small object.
c. both experience same magnitude of momentum changes.
d. Cannot be determined.

18. Which of the following is an accurate statement?
a. The momentum of a projectile is constant.
b. the momentum of a moving object is constant.
c. if an object is acted on by a non-zero net external force, its momentum will not remain constant.
d. If the KE of an object is doubled, its momentum will also double.

24. A sailboat of mass M is moving with a momentum p. Which of the following represents its KE?
a. (p^2)/(2m)
b. .5mp^2
c. mp
d. mp/2

 

[jaydee]

Law of momentum: Total momentum before = total momentum after.
Momentum= mass*velocity.
Inelastic collision means that the objects repel each other upon impact (as opposed to elastic collision on question 17).
Law of conservation of energy: Energy can not be created nor destroyed.

From this you should be able to figure it out.

 

[MereMortal]

<< Inelastic collision means that the objects repel each other upon impact (as opposed to elastic collision on question 17). >>



This statement is incorrect. An elastic collision is one in which kinetic energy is conserved; the kinetic energy is not conserved in an inelastic collsion. When the colliding bodies stick together after the collision, the collision is said to be completely or perfectly inelastic.

 

[goodoptics]

7.-a vector type of problem. since momentum(p) is conserved, magnitude of p(after) = magnitude p(initial) = sqrt(24^2+10^2) = 26 kg*m/s. Ans:c

11. In an inelastic collision, part of the total KE is lost to the collision.

17. Cannot be determined unless density or mass of the individual objects are given. Momentum is about mass and velocity, not size.

18. Ans: c Use F_ext = dp/dt (definition). If F_ext != 0, it's equivalent to say dp/dt != 0. Hence, p is not constant. The other three are obviously inaccurate.

24. use p = mv and KE = .5mv^2...it's pretty straight forward.

 

[PowerEngineer]

Well, it's been a long time since I took physics, but here are my guesses:

7. Since the collision is elastic, I'd expect that the total momentum following the collision should be equal to the momentum before the collision. So what's the momentum before the collision. Note that the 10 kg-m/s and the 24 kg-m/s are at 90 degrees to each other -- a dead giveaway that we're dealing with a right triangle solution. The total momemntum is (10*10 + 24*24)^0.5 which comes out to 26 kg-m/s

11. In an inelastic collision, some of the kinetic energy is used to deform the masses which means that KE is lost. As an obvious example, think of silly putty hitting the floor.

17. This strikes me as a bit of a trick question. I think they want you to say that the smaller object does, because it has the largest change in velocity. I'd bet that the right answer is both experience the same magnitude of change, equal but opposite (but I haven't time right now to check this).

18. The first two choices don't rule out acceleration, in which case their momentum can't be constant. The third choice is the flip side of the coin and is correct. The last choice is obviously not true (except in rare instances).

24. if p=mv and ke=0.5*mv^2, then ke=0.5*(mv)^2/m which is the same as p^2/2m.


Looks like Optics is quicker than I am....and we agree on four out of five at least.

 

[goodoptics]

<< Well, it's been a long time since I took physics, but here are my guesses:

7. Since the collision is elastic, I'd expect that the total momentum following the collision should be equal to the momentum before the collision. So what's the momentum before the collision. Note that the 10 kg-m/s and the 24 kg-m/s are at 90 degrees to each other -- a dead giveaway that we're dealing with a right triangle solution. The total momemntum is (10*10 + 24*24)^0.5 which comes out to 26 kg-m/s

11. In an inelastic collision, some of the kinetic energy is used to deform the masses which means that KE is lost. As an obvious example, think of silly putty hitting the floor.

17. This strikes me as a bit of a trick question. I think they want you to say that the smaller object does, because it has the largest change in velocity. I'd bet that the right answer is both experience the same magnitude of change, equal but opposite (but I haven't time right now to check this).

18. The first two choices don't rule out acceleration, in which case their momentum can't be constant. The third choice is the flip side of the coin and is correct. The last choice is obviously not true (except in rare instances).

24. if p=mv and ke=0.5*mv^2, then ke=0.5*(mv)^2/m which is the same as p^2/2m.


Looks like Optics is quicker than I am....and we agree on four out of five at least.

>>



I agree that 17 is quite tricky for one, we don't know the directions the initial velocities. Also, please note that smaller != lighter, same for larger != heavier. Hence, we really can't comment on the magnitude of the momentum change.

 

[jaydee]

Is it? I'm not too good on exact terminology, he's right, I got it backwards.

 

[PowerEngineer]

<< I agree that 17 is quite tricky for one, we don't know the directions the initial velocities. Also, please note that smaller != lighter, same for larger != heavier. Hence, we really can't comment on the magnitude of the momentum change. >>



You may be right, Optics....but I'm sticking with my gut response here. I realize that size and mass are not necessarily linked. Assuming the total momentum of the two bodies is conserved, then the change in momentum experienced by the first body would have to be exactly offset by the change in momentum experienced by the second body. This must mean that the magnitudes of the change are equal (but their directions are opposite).

Even before thinking this through, however, the wording of the questions raised warning flags in my mind. It feels like an invitation to see a difference where there actually isn't one. All my test-taking instincts shout out to me that the answer is C!!!

 

[Justin218]

I agree 17 is a tricky one but I'm going to agree with Power. I'm thinking both have to compensate eachother. Either way, I'm going to argue it if I get it wrong . Thanks for helping me out guys.

 

[goodoptics]

<< I agree 17 is a tricky one but I'm going to agree with Power. I'm thinking both have to compensate eachother. Either way, I'm going to argue it if I get it wrong . Thanks for helping me out guys. >>



After some rethinking, I have to agree with you guys. c is the correct answer.

 

[Justin218]

but don't feel bad Optic, I had D down before I even posted this