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Bullet blocks problem

  1. Apr 21, 2014 #1
    1. The problem statement, all variables and given/known data

    A block A of mass MA = 1 kg is kept on a smooth horizontal surface and attached by a light thread to another block B of mass MB=2 kg .Block B is resting on ground and thread and pulley are massless and frictionless.A bullet of mass m 0.25 kg moving horizontally with velocity of u=200m/s penetrates through the block and comes out with a velocity of 100m/s.Find the height through which the block B will rise.

    Ans :5.21m

    2. Relevant equations



    3. The attempt at a solution

    When bullet penetrates through block A ,an impulsive force P acts on it.

    ∫Pdt = 0.25(200-100) = 25Ns

    Considering block A,let the impulsive tension be T

    ∫Pdt - ∫Tdt = 1(vA)

    Considering block B ,

    ∫Tdt = 2vB

    I guess I am missing something ,is it vA = vB because of inextensibility of the string connecting the masses? But that doesn't give correct answer .

    I would be grateful if somebody could help me with the problem.
     

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    Last edited: Apr 21, 2014
  2. jcsd
  3. Apr 21, 2014 #2

    ehild

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    The speed of both blocks must be equal. You can assume that the bullet penetrates through the block so fast that you can ignore any displacements of the blocks during that time. But you know their speed from conservation of momentum. And then it is a decelerating motion...

    ehild
     
  4. Apr 21, 2014 #3
    Thanks ehild !

    Does that mean first an impulsive tension acts which gives the blocks an initial velocity and afterwards a constant tension exists in the string?

    Due to impulsive tension why can't the initial velocity of block B greater than that of block A i.e vB > vA ? In that case block B will be under free fall .

    Conservation of momentum or Impulse momentum theorem ?
     
    Last edited: Apr 21, 2014
  5. Apr 21, 2014 #4

    ehild

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    There can be different approximations. I assumed that both blocks make a system that gets common speed from the bullet. I do not understand your impulsive tensions.

    ehild
     
  6. Apr 21, 2014 #5
    I am not sure but I tried this on my first look at the problem and this approach didn't give me the right answer.

    From conservation of momentum, I got speed of ##M_A## equal to ##25\,m/s##. Hence, the speed of ##M_B## is ##25\,m/s##.

    I got the magnitude of deacceleration equal to ##20g/3##. Hence, max height achieved by B is:
    $$h=\frac{v^2}{2a}=\frac{625\times 3}{40g} \approx 4.6875\,m$$
     
  7. Apr 21, 2014 #6
    How does the bullet give the blocks a common speed ? The bullet passes through the block A in a very short period ,say dt .A large contact force arises between the bullet and the block A .This in turn gives rise to large tension in the string for a short duration .This tension is responsible for the initial velocity of block B .

    If it were not for impulsive tension , how else would block B get initial velocity ?
     
  8. Apr 21, 2014 #7
    Momentum is not conserved in this problem . Please refer my attempt in the OP .From that you will get v = 25/3 m/s .
     
  9. Apr 21, 2014 #8

    SammyS

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    It must be assumed that the thread is inextensible . It doesn't stretch !
     
  10. Apr 21, 2014 #9
    Okay...But what about the possibility of block B getting a higher initial velocity then block A due to the impulse ?
     
    Last edited: Apr 21, 2014
  11. Apr 21, 2014 #10
    Could conservation of energy play a role in the problem?
     
  12. Apr 21, 2014 #11

    ehild

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    How? that would require a varying tension in the string.




    ehild
     
    Last edited: Apr 21, 2014
  13. Apr 21, 2014 #12
    That would require block B accelerating relative to block A which implies a force. Other than gravity and the string, is there another force present?
     
  14. Apr 21, 2014 #13

    haruspex

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    That could indeed happen if there were any elasticity in the string. But how to prove it doesn't when there's no elasticity is an interesting question. I don't find the answers so far convincing.
    No, only that the impulsive tension is greater than it needs to be to get the blocks moving at the same speed.
    It doesn't require an ongoing force, merely a sufficiently large initial impulsive tension.
     
  15. Apr 21, 2014 #14

    SammyS

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    I treated it pretty much like an inelastic (not perfectly inelastic) collision of a 0.25 kg bullet (That's one MASSIVE bullet) with a 3 kg object, then used conservation of energy with PE based on the 2 kg block.

    I came up with 5.31 m .
     
  16. Apr 21, 2014 #15
    They assume that the bullet collides with the system of the two blocks, as a whole.
    I mean, this way you get their result.

    If you look at a more realistic scenarios, it is not even necessary that the bloc B rise from the ground. If block A is long and "soft", so that the bullet looses momentum slowly, the tension in the spring may be always less than the weight of B. At least in principle.

    Maybe we can solve a scenario assuming a constant force between bullet and bloc A and an elastic string, to study more interesting/realistic versions of the problem.

    Edit. Sammy already posted along the lines of my first paragraph. I've got 5.21 m though.
     
  17. Apr 21, 2014 #16
    Since the string is inelastic is it safe to assume that vB≤vA ?

    This is exactly the point that makes me unsure that the two blocks move together with the same speed .

    Do you agree that initially an impulsive tension acts and thereafter a constant tension exists in the string ?

    If you look at my attempt in the OP ,if value of ∫Tdt is quite large than the block B might have a velocity larger than that of block A.

    I still do not understand how can we assume that the two blocks move with the same speed.

    This concept of impulsive tension comes up quite often in the problems ,so I would like to be sure about it.
     
    Last edited: Apr 21, 2014
  18. Apr 21, 2014 #17
    How did you assume that the two blocks move together ? If the initial velocity of block A is more than that block B ,the inextensibility of the string makes sure that the two move together .But the inextensibility of the string doesn't rule out that the velocity of B can't be more than that of A.
     
  19. Apr 21, 2014 #18

    SammyS

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    What mechanism is present in this problem which can make the speed of Block B be more than that of Block A ?
     
  20. Apr 21, 2014 #19
    Impulsive tension
     
  21. Apr 21, 2014 #20

    ehild

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    What do you call "impulsive tension"? I learnt about momentum, I know that the force is equal to the time derivative of momentum, and the momentum of a closed system is conserved. I also know that conservation of momentum does not apply if there is some "impulsive" external force, from a constraint, that can become indefinitely large during the interaction. Is there any constraint for the motion of the bodies along the string? Both are free to move.

    There is no such thing that infinite force in real life. And inextensible strings do not exist. All strings are extensible a bit. If you think that the connected bodies do not move together, you should take the elasticity of the string into consideration. It is clear that the bullet interacts with block A and it starts to move. That motion causes stress in the string, it stretches a bit. That disturbance travels along the string and reaches block B, and causes to move it, when the tension at that place overcomes gravity. .The acceleration of B is the effect, and the initial motion of the first block is the cause. It can happen, that B moves faster than A when the bullet leaves A, but you need to derive it from the properties of the string. The problem says that the string is inextensible, that means the blocks start to move together, as parts of a rigid body.


    There is also the possibility, that you push block B upward, and this way it moves faster than block A. That makes the string slack. No force opposes gravity at B, and no retarding tension acts to block A. While it interacts with the bullet, it will accelerate faster than in case of a taut string. After the bullet leaves, it will move with a constant speed, while block B will decelerate because of gravity. Sooner or later the string becomes taut and the blocks get to move together.

    As the bullet and the blocks and the string make a system, the component of the momentum along the string is conserved as the only external force is gravity and it is constant, so its effect can be ignored during the interaction with the bullet. You do not know what happens exactly during the interaction with the bullet and after a short time, but the result of the collision process is a taut string and two blocks moving together.

    ehild
     
    Last edited: Apr 21, 2014
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