A 0.086 kg bullet is fired at 266 m/s and collides with a 6.3 kg wooden block that is initially stationary. The bullet embeds into the wood

Question

A 0.086 kg bullet is fired at 266 m/s and collides with a 6.3 kg wooden block that is initially stationary. The bullet embeds into the wood block such that the collision is perfectly inelastic. Use conservation of momentum to find the speed of the wooden block (with bullet embedded) at the instant after the collision.

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Kim Chi 4 years 2021-07-21T07:38:52+00:00 1 Answers 3 views 0

Answers ( )

  1. Bo
    0
    2021-07-21T07:40:03+00:00
    Reply

    Answer:

    The value of speed of wooden block at the instant after the collision   V_f = 262.41  \frac{m}{s}

    Explanation:

    Given data

    Mass of block = 6.3 kg

    Mass of bullet = 0.086 kg

    Velocity of bullet V_0 = 266 \frac{m}{s}

    Final velocity of whole system is

    V_f = M_{b}\frac{ V_0}{M_b + (M_{block})}

    Put all the values in above formula we get

    V_f = \frac{(6.3)(266)}{(6.3 + 0.086)}

    V_f = 262.41  \frac{m}{s}

    This is the value of speed of wooden block at the instant after the collision.

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