A 0.050 kg bullet strikes a 5.0 kg wooden block with a velocity of 909 m/s and embeds itself in the block which fies off its stand. what was

Question

A 0.050 kg bullet strikes a 5.0 kg wooden block with a velocity of 909 m/s and embeds itself in the block which fies off its stand. what was the final velocity of the bullet?

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Verity 4 years 2021-09-04T04:51:49+00:00 1 Answers 77 views 0

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  1. binhan
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    2021-09-04T04:52:58+00:00
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    Answer:

    The final velocity of the bullet is 9 m/s.

    Explanation:

    We have,

    Mass of a bullet is, m = 0.05 kg

    Mass of wooden block is, M = 5 kg

    Initial speed of bullet, v = 909 m/s

    The bullet embeds itself in the block which flies off its stand. Let V is the final velocity of the bullet. The this case, momentum of the system remains conserved. So,

    mv=(m+M)V\\\\V=\dfrac{mv}{m+M}\\\\V=\dfrac{0.05\times 909}{0.050+5}\\\\V=9\ m/s

    So, the final velocity of the bullet is 9 m/s.

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