A 4.25-g bullet traveling horizontally with a velocity of magnitude 375 m/s is fired into a wooden block with mass 1.12 kg, initially at res

Question

A 4.25-g bullet traveling horizontally with a velocity of magnitude 375 m/s is fired into a wooden block with mass 1.12 kg, initially at rest on a level frictionless surface. The bullet passes through the block and emerges with its speed reduced to 114 m/s. Part A How fast is the block moving just after the bullet emerges from it

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Kiệt Gia 6 months 2021-08-28T11:19:14+00:00 2 Answers 2 views 0

Answers ( )

    0
    2021-08-28T11:20:35+00:00

    Answer:

    0.9904 m/s

    Explanation:

    To solve this problem we need to use the conservation of momentum:

    m1v1 + m2v2 = m1’v1′ + m2’v2′

    Using this equation, we have:

    m_bullet*v_bullet = m_bullet_after*v_bullet_after + m_block*v_block

    0.00425 * 375 = 0.00425 * 114 + 1.12 * v_block

    1.12 * v_block = 1.5938 – 0.4845

    1.12 * v_block = 1.1093

    v_block = 1.1093 / 1.12

    v_block = 0.9904 m/s

    0
    2021-08-28T11:20:55+00:00

    Answer:

    0.99 m/s

    Explanation:

    From the question,

    Note: The collision between the bullet the the wooden block is elastic.

    Total momentum before collision = Total momentum after collision.

    mu+Mu’ = mv+Mv’……………. Equation 1

    Where m = mass of the bullet, u = initial velocity of the bullet, M= mass of the wooden block, u’ = initial velocity of the wooden block, v = final velocity of the bullet, v’ = final velocity of the wooden block

    Since, u’ = 0 m/s

    Therefore,

    mu = mv+Mv’

    make v’ the subject of the equation

    v’ = (mu-mv)/M……………….. Equation 2

    Given: m = 4.25 g = 0.00425 kg, u = 375 m/s, M = 1.12 kg, v = 114 m/s.

    Substitute into equation 2

    v’ = [(0.00425×375)-(0.00425×114)]/1.12

    v’ = (1.59375-0.4845)/1.12

    v’ = 0.99 m/s

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