Arifle 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?

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

grav force prop 1/r squared 2r all squared is 4rsquared 1/4rsquared quartered

friction

explanation:

they us to move forward and backwards