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Projectile on incline

  1. Jul 20, 2014 #1
    1. The problem statement, all variables and given/known data
    A wedge of mass m and angle θ is placed on a smooth ground. As shown in the diagram A is a point on ground. A particle also having mass m is dropped from a height h at a horizontal distance x from A . It finally touches the ground at point B . Find √3AB in metres.

    Details and Assumptions

    Take m=5kg ,h=2√2 metre ,x=1metre.tanθ=1/√2

    The collision of the ball with the wedge is elastic and the wedge is free to move.
    THIS IS THE FIGURE https://d3pq38zxuosm5i.cloudfront.net/solvable/2676130fa9.e594e044b6.ZQmdPV.jpg
    2. Relevant equations
    AS the collision is elastic so momentum of system is conserved and kinetic energy is conserved.


    3. The attempt at a solution
    given tanθ=1/sq.root(2)
    using this we can find the the initial distance of ball above point A(v3)=3/√2.
    so velocity of ball when it collided wedge=√(3√2)g.
    since K.E. is conserved so 3√2g=v1^2+v2^2...............(1)(V1 is velocity of ball after collision and v2 is velocity of wedge after collision)
    now tanθ=1/sq.root(2) so θ=35.26
    applying law of conservation of momentum in Y-direction
    3/√2g=-v1 sin2θ+0...................(2)(considering upward direction -ve).
    Are my two equations correct.
    Answer of this question is 1.268.
     
  2. jcsd
  3. Jul 20, 2014 #2

    Nathanael

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    Isn't the initial distance above point A "h"? Did you mean just the height above the wedge? If so, wouldn't that be 2√2-√2 = √2?

    I don't think the Y-direction of conservation of momentum is important. The Earth is counteracting (and conserving) all of the [itex]ΔP_y[/itex] the entire time. (where P stands for momentum)

    Wouldn't you want to apply conservation of momentum in the x-direction along with conservation of energy to find the final velocity of the wedge?
     
  4. Jul 20, 2014 #3
    Let the particle fall on point E on the wedge
    tanA=1/sq.root(2)
    tanA=ED/AD but AD=1
    so [itex]\frac{1}{√2}[/itex]=ED/1
    SO ED= [itex]\frac{1}{√2}[/itex]
    height of particle above wedge=h-ED
    =2√2-[itex]\frac{1}{√2}[/itex]
    =[itex]\frac{4-1}{√2}[/itex]
    =[itex]\frac{3}{√2}[/itex]
     
  5. Jul 20, 2014 #4
    physics fig.png
    This is the figure.
     
  6. Jul 20, 2014 #5

    Nathanael

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    You're right, sorry, I don't know why I was thinking tanθ=x/"ED"

    Anyway, do you know what the final velocity of the wedge is? (and initial velocity of the ball after it bounces off the wedge?)

    Conservation of KE and Momentum should give you this answer
     
  7. Jul 20, 2014 #6
    Applying conservation of momentem in X-direction
    m√(3√2)g=mv1cos(90-2θ)+mv2(v1 and v2 are the velocities of particle and wedge after collision)
    √(3√2)g=v1sin2θ+v2....................(2)
    3√2g=v1^2+v2^2...............(1)(from conservation of kinetic energy)
    Could you please explain a bit more why was the application of conservation of momentum in Y-direction not effective?
     
  8. Jul 20, 2014 #7

    Nathanael

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    You mean √(3g√2) right? (The g is in the square root?)

    This is the speed of the particle right before it hits the wedge, correct?

    Isn't that speed completely vertical? So why would it be involved in "conservation of momentum in X-direction"?
     
  9. Jul 20, 2014 #8

    Nathanael

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    Sure.

    When you drop a ball, the momentum in the Y-direction changes, right?

    Well then how is Y-momentum conserved?
    It is conserved because the earth moves upwards with the same momentum as the ball moves down.

    The reason is of course that the ball pulls up on the Earth with the same force that the Earth pulls down on the ball.
    (The definition of "Force" is "change in momentum per change in time" ... So, if they always have the same force on them, then they always change momentum equally, (and oppositely,) and so momemtum is conserved)



    In the situation you have, the Earth is moving up with the same momentum the ball is moving down, so momentum in the Y direction is conserved.
    (If the momentum in the Y direction of the wedge changed, then either the Y-momentum of the ball or Earth would need to change accordingly, but it doesn't so they don't.)
     
  10. Jul 20, 2014 #9
    YES
    YES
    I just need two equation in terms of v1 and v2 to get the values of them.Why can't we get one of the equation by setting up conservation of momentum in X-direction?
    THANK YOU.
     
  11. Jul 20, 2014 #10

    TSny

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    Satvik. It looks like you are assuming that the velocity of the ball just after the bounce makes the same angle θ relative to the normal of the wedge as the velocity just before the bounce. This would be true if the wedge did not recoil, but it is not generally true if the wedge does recoil. (Think about a case where the mass of the wedge is very small compared to the mass of the ball. The velocity of the ball would hardly change when it hit the wedge in this case.)

    You have three unknowns: the recoil speed of the wedge, the x-component of velocity of the ball just after the bounce, and the y-component of the velocity of the ball just after the bounce.

    So far, you have only two equations: conservation of energy and conservation of the x-component of momentum of the ball-wedge system.

    [The y-component of momentum of the ball-wedge system is not conserved because there is a strong impulsive external force that acts on the ball-wedge system in the y direction during the collision. Can you identify this force?]

    You're going to need a third equation. What can you say about the component of the velocity of the ball that is parallel to the wedge during the collision?
     
  12. Jul 20, 2014 #11

    Nathanael

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    You can, it just wouldn't be the equation you wrote. Your equation involved the vertical momentum (right before the collision) which isn't relevant to the equation.
     
  13. Jul 21, 2014 #12
    I didn't think about that.You are right.



    Is it the reactionary force on wedge due to earth(or surface below)

    Let the particle be deflected by angle 'S' after collision relative to normal of the wedge with velocity v1---
    component of this velocity parallel to wedge= v1cos(90-S)
    =v1sinS
     
    Last edited: Jul 21, 2014
  14. Jul 21, 2014 #13

    TSny

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    Yes, that's right.


    OK. But rather than taking v1 and the angle S as two of the unknowns, it might be better to take v1x and v1y as unknowns. You're going to need them for handling the projectile motion later. So, your three unknowns are v1x, v1y, and v2 (the recoil speed of the wedge).

    (1) Can your write an expression for the component of v1 that is parallel to the surface of the wedge in terms of v1x, v1y, and the (known) angle θ?

    (2) Let vo be the velocity of the ball just before the collision. Can you write an expression for the component of vo that is parallel to the wedge surface just before the collision?

    What is the relation between (1) and (2)?
     
  15. Jul 21, 2014 #14
    v[itex]_{1x}[/itex]cosθ

    vo cos(90-θ)=vocosθ

    I thought of applying law of conservation of momentum in direction parallel to the wedge but in this direction a component of external force(reactionary force) on wedge is acting.So I can't do this.Is it right?
    I am not getting any way to establish relation between (1) and (2).
    THANK YOU.
     
    Last edited: Jul 21, 2014
  16. Jul 21, 2014 #15

    TSny

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    OK, but there will also be a contribution from v1y.

    Are you sure you have the right trig function here? For example, consider the special case where θ = 0.

    Right. Momentum of the ball-wedge system is not conserved in the direction parallel to the surface of the wedge. However, consider just the ball's change in velocity when it hits the wedge. What is the only impulsive force that acts on the ball during the collision? In what direction does it act? What can you say about the change in the component of the velocity of the ball that is perpendicular to this impulsive force?
     
  17. Jul 21, 2014 #16
    Yes I forgot that,it should be v1xcosθ-v1ysinθ.


    vocos(90-θ)=vosinθ.I wrote it vocosθ by mistake.



    I think the only Impulsive force that acts on the ball is reactionary force on the ball from the wedge.
    It acts in a direction normal to the wedge.
    change in the component of the velocity of the ball that is perpendicular to this impulsive force=final velociy in that direction -initial velocity in that direction.
    Since component of impulse in direction parallel to the plane is zero so
    v1xcosθ-v1ysinθ=vosinθ.
    Is it right?
     
  18. Jul 22, 2014 #17

    TSny

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    Yes, assuming +x is toward the left and +y is upward in your picture.
     
  19. Jul 23, 2014 #18
    Now I have three eq.
    v1xcosθ-v1ysinθ=vosinθ.............(3)
    √(3√2)g=v1sin2θ+vo....................(2)
    3√2g=v1^2+vo^2...............(1)
    But here is more than three variables.Do I need to make some more equations?
     
  20. Jul 23, 2014 #19

    TSny

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    I'm a bit confused with the notation. I thought we were using vo for the speed of the ball just before hitting the wedge.
    You found vo = √(3√2g).

    v2 is the symbol we were using for the recoil speed of the wedge.

    What does your equation (2) represent? Where did 2θ come from?

    Try to construct 3 equations in the three unknowns: v1x, v1y, v2. Your equation (3) looks good.
     
    Last edited: Jul 23, 2014
  21. Jul 23, 2014 #20
    From the conservation of kinetic energy ---
    vo^2=v1^2+v2^2............(2)
    From the conservation of momentum in X-direction----
    Vo=v1x+v2.......(1)
    Since v1=√(v1x^2+v2x^2)
    putting this value in eq ( 2)
    vo^2=v1x^2+v2x^2+v2^2

    or (3√2g)=v[itex]_{1x}[/itex][itex]^{2}[/itex]+v[itex]_{1y}[/itex][itex]^{2}[/itex]+v[itex]_{2}[/itex][itex]^{2}[/itex]........(2)
    and eq(1) becomes
    √(3√2g)=√(v[itex]_{1x}[/itex][itex]^{2}[/itex]+v[itex]_{1y}[/itex][itex]^{2}[/itex])+v[itex]_{2}[/itex]...........(1)



    Is it fine now?
     
    Last edited: Jul 23, 2014
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