A 12.0 g bullet was fired horizontally into a 1 kg block of wood. The bullet initially had a speed of 250 m/s. The block of wood was hanging

Question

A 12.0 g bullet was fired horizontally into a 1 kg block of wood. The bullet initially had a speed of 250 m/s. The block of wood was hanging from a 2 m long piece of (massless) string. After the collision the block/bullet combined object swings upward on the string. Find the height the block/bullet combined object rises.

a. 0.66m
b. 0.45m
c. 0.12 m
d. 0.35 m
e. 0.27m

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Farah 2 weeks 2021-07-19T23:30:56+00:00 1 Answers 3 views 0

Answers ( )

    0
    2021-07-19T23:32:24+00:00

    Answer:

    So height though which combination of block bullet rises is 0.45 m

    Explanation:

    We have given  mass of the bullet  m_1=12gram=0.012kg

    Velocity of the bullet v_1=250m/sec

    Mass of block of wood m_2=1kg

    Block is at rest so v_2=0m/sec

    From conservation of momentum.

    m_1v_1+m_2v_2=(m_1+m_2)v

    0.012\times 250+0\times 0=(1+0.012)v

    v=2.964m/sec

    From third equation of motion

    h=\frac{v^2}{2g}=\frac{2.964^2}{2\times .8}=0.45m

    So option (b) will be the correct answer.

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