A 2.40-g bullet embeds itself in a 1.10-kg block, which is attached to a spring of force constant 760 N/m .
fistly we find the velocity of after collision
velocity V1=x[k/m+M]^1/2
=0.054[760/1.1+0.0024]^1/2
=1.42 m/s
now we find the initial velocity of bullet
initial velocity Vo=[m+M/m]V1
=[0.0024+1.1/0.0024]*1.42
=652.3 m/s
now we find the time block-bullet comes to rest
time =V1/g=1.42/9.8=0.145 sec
A 2.40-g bullet embeds itself in a 1.10-kg block, which is attached to a spring of...
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