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A level Physics Elastic Collision Problem watch

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Size:  504.9 KBThe question appears to only include the vertical height and I am struggling to figure out how to answer the questions with the given information, any guidance would be much appreciated
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    (Original post by Physicist Matt)
    The question appears to only include the vertical height and I am struggling to figure out how to answer the questions with the given information, any guidance would be much appreciated
    You can use the formula of
    [Loss in GPE = Gain in KE]
    mgh=(1/2)mv^2
    since it comes to rest and the collision is elastic, all of particle A's Kinetic Energy is transferred to particle B.
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    (Original post by Physicist Matt)
    Name:  IMG_2702.jpg
Views: 53
Size:  504.9 KBThe question appears to only include the vertical height and I am struggling to figure out how to answer the questions with the given information, any guidance would be much appreciated
    It's a lot simpler than at first seems.

    The key is to recognise that the first ball has gravitational potential energy and that all of this must be converted to kinetic energy (wholly horizontal velocity becaue the first ball follows an arc) when it reaches the bottom of it's swing.

    Ball A

    mgh = \frac{1}{2}mv^{2}

    2gh = v^{2}

    v = \sqrt{2gh}

    h = 0.1m

    a = 9.81 ms-2

    v = 1.40 m/s (2.s.f.) in a horizontal direction at the bottom of it's swing.

    Since the balls are equal mass, once again ignoring friction, energy conservation means the horizontal velocity of the first ball A is wholly transferred to the resting ball B in the horizontal direction.

    The problem now becomes straightforward SUVAT resolved into vertical and horizontal components.

    Ball B must now fall a vertical distance h in time t.

    Ball B

    s = ut + \frac{1}{2}at^{2}

    u = 0 (initial vertical velocity)

    s = 0.1m (vertical distance)

    s = \frac{1}{2}at^{2}

    \frac{2s}{a} = t^2

    t = \sqrt{\frac{2s}{a}}

    t = 0.1427843 seconds

    Finally, again using SUVAT, the second ball travels a horizontal distance d in time t:

    v = 1.40 m/s

    s = vt

    s = 1.40 x 0.1427843 = 0.20 m

    Hence ball B travels a horizontal distance of 0.2m (20cm keeping units consistent with the question).

    Spoiler:
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    If one wanted to be clever (not recommended at this stage), the calculation part can be simplified and left until right at the end.

    d = v_{h}t

    d = \sqrt{2gh}\sqrt{\frac{2s}{a}}

    since h = s = 0.1m

    simplifying:

    d^{2} = 2gh\frac{2s}{a} = 4h^{2}

    d = \sqrt{4h^{2}}

    d = 2h

    d = 2 x 0.1m = 0.2m


 
 
 
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