A 0.140-kg baseball is dropped and reaches a speed of 1.20 m/s just before it hits the ground. It rebounds with a speed of 1.00 m/s. What is

by

A 0.140-kg baseball is dropped and reaches a speed of 1.20 m/s just before it hits the ground. It rebounds with a speed of 1.00 m/s. What is the change of the ball’s momentum

0 thoughts on “A 0.140-kg baseball is dropped and reaches a speed of 1.20 m/s just before it hits the ground. It rebounds with a speed of 1.00 m/s. What is”

  1. Answer:

    The change in momentum is Δp= 0.028 kg m/s

    Explanation:

    An impulse describes a change in momentum. The change in momentum of an object is its mass times the change in its velocity.

    The change in moment is given by the expression below

    Δp=m⋅(Δv)=m⋅(vf−vi) .

    Given data

    mass m= 0.140-kg

    initial velocity vi= 120 m/s

    final velocity vf= 1 m/s

    substituting we have

    Δp=m⋅(Δv)=0.14⋅(1−1.2)

    Δp=m⋅(Δv)=0.14⋅(-0.2)

    Δp= 0.028 kg m/s

    The change in momentum was found to be Δp= 0.028 kg m/s

    Reply

Leave a Comment