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Homework Help: Elastic collison at an angle

  1. Jun 11, 2015 #1
    1. The problem statement, all variables and given/known data
    A racquet ball with mass m = 0.224 kg is moving toward the wall at v = 15.1 m/s and at an angle of θ = 28° with respect to the horizontal. The ball makes a perfectly elastic collision with the solid, frictionless wall and rebounds at the same angle with respect to the horizontal. The ball is in contact with the wall for t = 0.078 s.
    I need to find:

    What is the magnitude of the change in momentum of the racquet ball?

    What is the magnitude of the average force the wall exerts on the racquet ball?

    2. Relevant equations
    F=.(5*M*V^2)/ Delta(X)

    3. The attempt at a solution
    I was able to find momentum before the collision, which came out to be 3.3824, but I'm unsure how to add the vectors to get the correct momentum vectors to find the momentum after it hits the wall which will let me find the change in momentum.
  2. jcsd
  3. Jun 12, 2015 #2


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    The problem is improperly stated, and unsolvable as stated, because it only gives the angle 'to the horizontal', whereas what is needed is the angle to the normal to the wall (a line perpendicular to the wall). Assuming the wall is vertical (a reasonable assumption), the ball could be travelling at 28 degrees to the horizontal, almost parallel to the wall, and suffer a negligible change in momentum when it bounces. Or it could be travelling at 28 degrees to the normal, which gives the biggest change in momentum. Or anything in between.

    The author probably meant to say that the direction was 28 degrees to the normal, so let's assume that. Then draw a diagram showing the wall as a straight line, the line of the ball coming in to hit the wall and the line at which it bounces away. Both those lines are at 28 degrees to the normal coming out from the wall at the collision point. Now use those lines to work out the component of the ball's velocity towards the wall.
  4. Jun 12, 2015 #3
    I don't think it's unsolvable as stated. Wouldn't it look something like this?
    I don't think I see how the depiction could be misconstrued.

    With regard to the OP's problem, as AndrewKirk noted, try adding the momentum vectors. Remember that velocity and momentum are vectors, not scalars. Their quantities have associated directions. Do you remember how to add vectors? You'll need a coordinate system, which is just an economic way of stating that you need to establish perpendicular axes that define positive and negative directions of change in distance.
  5. Jun 12, 2015 #4


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    It's not that the diagram can be misconstrued, it's that it's incomplete. It doesn't show the magnitude of the velocity vector. That magnitude will vary with the angle of the plane in which the diagram is drawn to the plane of the wall - the azimuth angle. If that angle is very small then the velocity will be small. The velocity will be maximised when the angle is a right angle.

    Think about the following two situations in a squash court:

    1. you are in the front RH corner and hit the ball almost parallel to the wall, at an angle 28 degrees to the horizontal, so that it just grazes the front wall

    2. you are at the rear of the court and hit the ball in a direction whose horizontal component is directly towards (at right angles to) the wall, and at 28 degrees to the horizontal

    These are both at 28 deg to the horizontal. But the second will hit the wall much harder than the first.
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