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A rifle bullet with mass 8.00 g strikes and embeds itself in a block with mass 0.992 kg that rests on a frictionless, horizontal surface and is attached to a coil spring. The impact compresses the spring 15.0 cm. Calibration of the spring shows that a force of 0.750 N is required to compress the spring 0.250 cma) Find the magnitude of the block's velocity just after impact. b) What was the initial speed of the bullet?

Respuesta :

krissydudley549

Answer:

ok so here we go

Explanation:

(ma)(va,1)+(mb)(vb,1)=(ma+mb)(v2)(ma)(va,1)+(mb)(vb,1)=(ma+mb)(v2).

Plugging in the givens results in (.008)(va,1)=v2(.008)(va,1)=v2.

Then, I considered the energies of the system as a whole. Spring force is conservative.

I reason that at the point at which the spring is at maximum compression of .15 m, the final velocity of the block is zero.

So, (1/2)(ma)(va,1)2=(1/2)(300)(.15)2(1/2)(ma)(va,1)2=(1/2)(300)(.15)2

I get 300 as the spring constant because F=kxF=kx.

Solving for va,1va,1, I got 29.0 m/s. I then multiplied 29.0 by .008 to get the velocity of the system, v2

(1/2)(ma+mb)(v2)2

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