A bullet is fired with a horizontal velocity of 1500 ft/s through a 6-lb block A and becomes embedded in a 4.95-lb block B. Knowing that blo

Question

A bullet is fired with a horizontal velocity of 1500 ft/s through a 6-lb block A and becomes embedded in a 4.95-lb block B. Knowing that blocks A and B start moving with velocities of 5 ft/s and 9 ft/s, respectively, determine (a) the weight of the bullet, (b) its velocity as it travels from block A to block B.

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Phúc Điền 4 years 2021-09-02T21:27:36+00:00 1 Answers 61 views 0

Answers ( )

  1. minhkhue
    0
    2021-09-02T21:29:08+00:00
    Reply

    Answer:

    weight of the bullet is 0.0500 lb

    velocity as it travels from block A to block B is 900 ft/s

    Explanation:

    given data

    horizontal velocity = 1500 ft/s

    mass block A = 6-lb

    mass block B = 4.95 lb

    blocks A velocity = 5 ft/s

    blocks B velocity = 9 ft/s

    solution

    we apply here law of conservation of momentum that is

    m × v(o) + m1 × (0) + m2 × (0) = m × v(2) + m1 × v1 + m2 × v2     …………….1

    put here value and we get m

    m = \frac{m1 \times v1 +m2 \times v2}{v(o) - v2}   …………..2

    m = \frac{6 \times 5 + 4.95 \times 9}{1500 - 9}  

    m = 0.0500 lb

    and

    when here bullet is pass through the A block then moment is conserve that is

    m × v(o) + m × v1 + m1 × v1   …………3

    v1 = \frac{m\times v(o)- m1\times v1}{m}    

    v1 = \frac{0.0500\times 1500 - 61\times 5}{0.05}  

    v1 = 900 ft/s

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