posted by esther on .
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.)
find the intial momentum of the block/bullet after impact, from that the velocity of the block. YOu now know that
solve for distance.
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.
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?
so the distance would be...... 2.96 m ?
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