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    A particle of mass 4.0 g is moving in a straight line at 1.0 km s−1. It experiences a collision of duration 3.0µs in which the direction of motion is deviated by 30◦ but the speed is unchanged. Calculate the magnitudes of (a) the change of momentum; (b) the force responsible.

    Another question that I need help. thanks
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    I'm not certain of this answer.

    p=(0.004)(0.001)

    cos(30)=x/(4x10^-6)
    x=(4x10^-6)cos(30)=3.46x10^-6

    change in momentum = (4x10^-6)-(3.46x10^-6)=5.36x10^-7

    force=change in momentum/time

    F=5.36x10^-7/3x10^-6=0.179N

    All numbers in standard units

    Hope this helps
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    (Original post by lallison1)
    I'm not certain of this answer.

    p=(0.004)(0.001)

    cos(30)=x/(4x10^-6)
    x=(4x10^-6)cos(30)=3.46x10^-6

    change in momentum = (4x10^-6)-(3.46x10^-6)=5.36x10^-7

    force=change in momentum/time

    F=5.36x10^-7/3x10^-6=0.179N

    All numbers in standard units

    Hope this helps
    I think you calculated the momentum wrong to begin with as it would be p = (0.004)(1000) = 4 kgms-1. so x = 3.46
    I then did the same as you for the next bit so change in p = 4 - 3.46 = 0.536 kgms-1. But the answer on the worksheet says it should be 2.1, so im unsure.
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    (Original post by lallison1)
    I'm not certain of this answer.

    p=(0.004)(0.001)

    cos(30)=x/(4x10^-6)
    x=(4x10^-6)cos(30)=3.46x10^-6

    change in momentum = (4x10^-6)-(3.46x10^-6)=5.36x10^-7

    force=change in momentum/time

    F=5.36x10^-7/3x10^-6=0.179N

    All numbers in standard units

    Hope this helps
    (Original post by Tuffyandtab)
    I think you calculated the momentum wrong to begin with as it would be p = (0.004)(1000) = 4 kgms-1. so x = 3.46
    I then did the same as you for the next bit so change in p = 4 - 3.46 = 0.536 kgms-1. But the answer on the worksheet says it should be 2.1, so im unsure.

    Hint: By definition, the rate at which the momentum changes is the force that causes the direction change. In other words, the change in momentum is the same vector as the force. (you guy's simply resolved the objects post event momentum in the x direction.)

    \text{F} = \frac{\Delta{\text{p}}}{\Delta{t  }}

    Because the pre and post event speed is unchanged, treat the problem as if we need to find the length of the chord for a segment of angle theta on a circle. i.e. graphically, the change in momentum vector is the same as the chord length:



    \Delta\text{p} = 2mv\text{ sin}(\frac{\theta}{2})

    Spoiler:
    Show







    \Delta\text{p} = 2 \text{ x } 0.004 \text{ x }1000 \text{sin}(\frac{30}{2})

    \Delta\text{p} = 8 \text{sin}(15) = 2.1 \text{kgms}^{-1}

    \text{F} = \frac{2.1}{3\text{ x }10^{-6}} = 0.70 \text{x}10^{6}\text{N}






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    (Original post by uberteknik)
    Hint: By definition, the rate at which the momentum changes is the force that causes the direction change. In other words, the change in momentum is the same vector as the force. (you guy's simply resolved the objects post event momentum in the x direction.)

    \text{F} = \frac{\Delta{\text{p}}}{\Delta{t  }}

    Because the pre and post event velocity is unchanged, treat the problem as if we need to find the length of the chord for a segment of angle theta on a circle. i.e. graphically, the change in momentum vector is the same as the chord length:



    \Delta\text{p} = 2mv\text{ sin}(\frac{\theta}{2})

    Spoiler:
    Show







    \Delta\text{p} = 2 \text{ x } 0.004 \text{ x }1000 \text{sin}(\frac{30}{2})

    \Delta\text{p} = 8 \text{sin}(15) = 2.1 \text{kgms}^{-1}

    \text{F} = \frac{2.1}{3\text{ x }10^{-6}} = 0.70 \text{x}10^{6}\text{N}






    Thanks for the help, but i don't quite understand where deltaP = 2mv sin(x/2) has come from. Could you explain some more please?
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    (Original post by Tuffyandtab)
    Thanks for the help, but i don't quite understand where deltaP = 2mv sin(x/2) has come from. Could you explain some more please?
    It's derived from a graphical momentum vector representation and the easiest calculation method.

    Referring to the diagram:



    Because the pre and post collision speed is the same, I drew an arc with internal angle theta = 30o representing the collision deviation angle. The pre and post momentum vectors are represented by the radius (r = mv) and become the hypotenuse of two right angled triangles:

    OB = pre collision momentum = mv
    OA = post collision momentum = mv

    Theta = deflection angle.

    The chord length BA, represents the change in momentum deltaP.

    AD = DB = r sin (theta/2)

    AB = 2r sin (theta/2).

    deltaP = 2mv sin (theta/2)

    This could also be solved following through with your original method:

    PX pre = mv

    deltaPX post = mv - mvcos(theta))

    PY pre = 0

    deltaPY post = mvsin(theta)

    Change in momentum = hypotenuse AB

    deltaP = SQRT{(mv - mvcos(30))2 + (mvsin(30))2}

    Same result, different method. One is less cumbersome than the other. It just relies on recognising the one that gets the answer quickest.
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    (Original post by uberteknik)
    Hint: By definition, the rate at which the momentum changes is the force that causes the direction change. In other words, the change in momentum is the same vector as the force. (you guy's simply resolved the objects post event momentum in the x direction.)

    \text{F} = \frac{\Delta{\text{p}}}{\Delta{t  }}

    Because the pre and post event velocity is unchanged, treat the problem as if we need to find the length of the chord for a segment of angle theta on a circle. i.e. graphically, the change in momentum vector is the same as the chord length:



    \Delta\text{p} = 2mv\text{ sin}(\frac{\theta}{2})

    Spoiler:
    Show








    \Delta\text{p} = 2 \times 0.004 \times 1000 \text{sin}(\frac{30}{2})

    \Delta\text{p} = 8 \text{sin}(15) = 2.1 \text{kgms}^{-1}

    \text{F} = \frac{2.1}{3 \times  10^{-6}} = 0.70 \text{x}10^{6}\text{N}







    Sorry, I might be a pain in the neck. I think you mean

    the pre and post event speed is unchanged

    for

    the pre and post event velocity is unchanged

    The question mentions there is no change in the speed.
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    (Original post by Eimmanuel)
    Sorry, I might be a pain in the neck. I think you mean

    the pre and post event speed is unchanged

    for

    the pre and post event velocity is unchanged

    The question mentions there is no change in the speed.
    Yes of course. Amended accordingly. Thanks you.
 
 
 
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