# physics

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A heavy wooden block rests on a flat table and a high-speed bullet is fired horizontally into the block, the bullet stopping in it. How far will the block slide before coming to a stop? The mass of the bullet is 10.5 g, the mass of the block is 10.5 kg the bullet’s impact speed is 750 m/s, and the coefficient of kinetic friction between the block and the table is 0.220. (Assume that the bullet does not cause the block to spin.)

• physics -

find the intial momentum of the block/bullet after impact, from that the velocity of the block. YOu now know that

KEblockAndBullet=mu*totalmass*g*distance
solve for distance.

• physics -

Once sliding begins, there is a friction force equal to
f = (M+m)*g*0.22 = 22.7 N

The initial speed V' of the block-with-bullet, after impact is given by the law of conservation of momentum:
m*750 = (M+m)V
V' = (.0105/10.51)*750 = 0.75 m/s

As a last step, equate the kinetic energy of the sliding block-plus-bullet to the work done against friction.

(1/2)(M+m)V'^2 = f X

and calculate the sliding distance X.

m and M are bullet and block mass, respectively.

• physics -

1/2 (10.5 kg) (750 m/s) + 1/2 (.0105 kg) (750 m/s) = .220 N * 10.5105 kg (9.81) distance

is this what i'm solving for?

• physics -

so the distance would be...... 2.96 m ?

• physics -

No. You use the velocity V' AFTER collision (0.75 m/s), and you square it, as in the formula I wrote.

Your friction force f also looks wrong

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