6. A 145-g baseball moving 30.5 m/s strikes a stationary 5.75-kg brick resting on small rollers so it moves without significant friction. Af

Question

6. A 145-g baseball moving 30.5 m/s strikes a stationary 5.75-kg brick resting on small rollers so it moves without significant friction. After hitting the brick, the baseball bounces straight back, and the brick moves forward at 1.10 m/s. a. What is the baseball’s speed after the collision? (10pts) b. Find the total kinetic energy before the collision. (10pts) c. Find the total kinetic energy after the collision. (10pts)

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Thiên Ân 6 months 2021-07-15T10:04:13+00:00 1 Answers 4 views 0

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    2021-07-15T10:05:18+00:00

    Answer:

    Explanation:

    a )

    momentum of baseball before collision

    mass x velocity

    = .145 x 30.5

    = 4.4225 kg m /s

    momentum of brick after collision

    = 5.75 x 1.1

    = 6.325 kg m/s

    Applying conservation of momentum

    4.4225 + 0 = .145 x v + 6.325 , v is velocity of baseball after collision.

    v = – 13.12 m / s

    b )

    kinetic energy of baseball  before collision = 1/2 mv²

    = .5 x .145 x 30.5²

    = 67.44 J

    Total kinetic energy before collision = 67.44 J

    c )

    kinetic energy of baseball after collision = 1/2 x .145 x 13.12²

    = 12.48 J .

     kinetic energy of brick after collision

    = .5 x 5.75 x 1.1²

    = 3.48 J

    Total kinetic energy after collision

    = 15.96 J

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