4. A 26.0-g arrow leaves a bowstring at a velocity of 46 m/s. a. What is the change in momentum of the arrow?

Question

4. A 26.0-g arrow leaves a bowstring at a velocity of 46 m/s.

a. What is the change in momentum of the arrow?

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Thu Nguyệt 4 years 2021-08-10T10:11:08+00:00 1 Answers 15 views 0

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  1. ngocdiep
    0
    2021-08-10T10:12:34+00:00
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    Answer:

    1.196kgm/s

    Explanation:

    The change in momentum \delta\rho of the arrow is given by;

    \delta\rho=mv-mu...............(1)

    where m is the mass of the arrow, u is the initial velocity and v is the final velocity.

    Given;

    m = 26.0g = 0.026kg

    u = 0 since the arrow was initially at rest before being released.

    v = 46m/s.

    Since the intial velocity is zero, equation (1) reduces to

    \delta\rho=mv...............(2)

    Hence the change in momentum is calculated thus;

    \delta\rho=0.026kg*46m/s\\\delta\rho=1.196kgm/s

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