Fasiha A.

asked • 04/01/20

Need help with physics question!

I attempted this problem but I don't where to get the k constant from to complete the equation.

1/2(1.2+0.01)V^2=1/2kx^2


A rifle bullet with mass 10.00g strikes and

embeds itself in a block with mass 1.2

kg that rests on a friction less, horizontal surface

and is attached to a coil spring. The impact

compresses the spring 15.0cm. Calibration

of the spring shows that a force of 0.850N is required to compress the spring 0.250cm.

What is the velocity of the bullet?

1 Expert Answer

By:

Fasiha A.

Hi, thank you for replying. I just wanted to confirm the answer I got after putting it into the equation is 2.5m/s. Is that correct?
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04/01/20

Richard P.

tutor
Not quite. 2.5 m/s is the speed of the combined block+bullet just after the collision. Since momentum is conserved (to a very good approximation) in the collision, the velocity of the bullet just before the collision is (1.21/ .01) 2.5 = 302.5 m/s. The approximation is that the duration of the collision itself is so small that the block+bullet moves very little during the collision itself. The 15 cm of travel takes place almost entirely after the collision itself is over.
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04/01/20

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