Posted by esther on Wednesday, March 10, 2010 at 2:27pm.
A heavy wooden block rests on a flat table and a highspeed 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 bulletâ€™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  bobpursley, Wednesday, March 10, 2010 at 2:30pm
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  drwls, Wednesday, March 10, 2010 at 2:38pm
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 blockwithbullet, 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 blockplusbullet 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  esther, Wednesday, March 10, 2010 at 2:53pm
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  esther, Wednesday, March 10, 2010 at 2:56pm
so the distance would be...... 2.96 m ?

physics  drwls, Wednesday, March 10, 2010 at 5:24pm
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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