Question #228208
A horizontal force of 0.80N is required to pull a 5kg block across a table top at a constant speed. With the block initially at rest, a 20g bullet fired horizontally into the block causes the block to slide 1.5m before coming to rest again. Determine the speed v of the bullet , where the bullet is assumed to be embedded in the block.
1
Expert's answer
2021-08-23T12:40:43-0400

Given:

M=5.0kgM=5.0\:\rm kg

m=0.02kgm=0.02\:\rm kg

F=0.80NF=0.80\:\rm N

d=1.5md=1.5\:\rm m


The law of conservation of momentum gives

mu=(M+m)vmu=(M+m)v

The energy-work theorem gives

(M+m)v22=Fd\frac{(M+m)v^2}{2}=Fd

Finally, the initial speed of a bullet

u=M+mm2FdM+mu=\frac{M+m}{m}\sqrt{\frac{2Fd}{M+m}}

u=5.0+0.020.0220.81.55.0+0.02=174m/su=\frac{5.0+0.02}{0.02}\sqrt{\frac{2*0.8*1.5}{5.0+0.02}}=174\:\rm m/s


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