A superball with a mass m = 61.6 g is dropped from a height h = [02]____________________ m. It hits the floor and then rebounds to a new hei

Question

A superball with a mass m = 61.6 g is dropped from a height h = [02]____________________ m. It hits the floor and then rebounds to a new height that is 88.5% of the initial height. Use upward as the positive direction. a. What is the momentum change (kg m/s) of the ball during its collision with the floor? b. Suppose you drop a ball of putty of the same mass from the same height. The putty sticks to the floor. What then would be the putty-ball’s momentum change (kg m/s) during its collision with the floor?

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Doris 4 years 2021-08-19T09:44:35+00:00 1 Answers 31 views 0

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  1. kimcuc
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    2021-08-19T09:45:41+00:00
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    There is not enough data to solve the problem, but I’m assuming the initial height as h = 10 m for the question to have a valid answer and the student can have a reference to solve their own problem

    Answer:

    (a) \Delta P=1.67 \ kg.m/s

    (b) \Delta P=0.86\ m/s

    Explanation:

    Change of Momentum

    The momentum of a given particle of mass m traveling at a speed v is given by

    P=m.v

    When this particle changes its speed to a value v’, the new momentum is

    P'=m.v'

    The change of momentum is

    \Delta P=m.v'-m.v

    \Delta P=m.(v'-v)

    Defining upward as the positive direction, we’ll compute the change of momentum in two separate cases.

    (a) The initial height of the superball of m=61.6 gr = 0.0616 Kg is set to h= 10 m. This information leads us to have the initial potential energy of the ball just after it’s dropped to the floor:

    U=m.g.h=0.0616\cdot 9.8\cdot 10 =6.0368\ J

    This potential energy is transformed into kinetic energy just before the collision occurs, thus

    \displaystyle \frac{1}{2}mv^2=6.0368

    Solving for v

    \displaystyle v=\sqrt{\frac{6.0368\cdot 2}{0.0616}}

    v=-14\ m/s

    This is the speed of the ball just before the collision with the floor. It’s negative because it goes downward. Now we’ll compute the speed it has after the collision. We’ll use the new height and proceed similarly as above. The new height is

    h'=88.5\% (10)=8.85\ m

    The potential energy reached by the ball at its rebound is

    U'=m.g.h'=0.0616\cdot 9.8\cdot 8.85 =5.342568\ J

    Thus the speed after the collision is

    \displaystyle v'=\sqrt{\frac{5.342568\cdot 2}{0.0616}}

    v'=13.17\ m/s

    The change of momentum is

    \Delta P=0.0616\cdot (13.17+14)

    \Delta P=1.67 \ kg.m/s

    (b) If the putty sticks to the floor, then v’=0

    \Delta P=0.0616\cdot (0+14)

    \Delta P=0.86\ m/s

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