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A 9.8-g bullet is fired into a stationary block of wood having mass m = 5.04 kg. The bullet imbeds into the block. The speed of the bullet-plus-wood combination immediately after the collision is 0.609 m/s. What was the original speed of the bullet? (Express your answer with four significant figures.)

Respuesta :

Chenk13

Answer:

313.8 m/s

Explanation:

Momentum p must be conserved.

The momentum before the collision:

[tex]p=m_{bullet}v_{bullet} + m_{block}v_{block}[/tex]

The momentum after the collision:

[tex]p=(m_{bullet}+m_{block})v_{bullet+block}[/tex]

Solving both equations:

[tex]v_{bullet}=\frac{(m_{bullet}+m_{block})v_{bullet+block}}{m_{bullet}}[/tex]

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