A 1.00 kg block is attached to a horizontal spring with spring constant 2500 N/m. The block is at rest on a frictionless surface. A 20 g bul

Question

A 1.00 kg block is attached to a horizontal spring with spring constant 2500 N/m. The block is at rest on a frictionless surface. A 20 g bullet is fired into the block, in the face opposite to the spring and sticks.

What is the bullet’s speed if the subsequent oscillations have an amplitude of 10.0 cm?

(Express your answer in m/s as a pure number).

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Vân Khánh 6 months 2021-07-24T02:08:40+00:00 1 Answers 116 views 0

Answers ( )

    -1
    2021-07-24T02:10:24+00:00

    Answer:

    The speed of the bullet is = 27.36 \frac{m}{s}

    Explanation:

    Mass of the block (M) = 1 kg

    Spring constant = 2500 N per meter

    Mass of the bullet (m_{b}) = 20 gm = 0.2 kg

    Amplitude A = 0.1 m

    Now the final velocity of block when bullet is entered into the block is given by

    V_f = \sqrt{\frac{k}{M + m_b} } A

    V_f = \sqrt{\frac{2500}{1 + 0.2} } 0.1

    V_f = 4.56 \frac{m}{sec}

    From conservation of momentum principal

    P_{after} = P_ {before}

    (M+m_{b} )V_f = m_{b} v_b + M v_B

    Since v_B = 0

    1.2 × 4.56 = 0.2 v_b

    v_b = 27.36 \frac{m}{s}

    Therefore the speed of the bullet is = 27.36 \frac{m}{s}

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