Question

A 3.45-g bullet embeds itself in a 1.37-kg block, which is attached to a spring of force constant 785 N/m.

If the maximum compression of the spring is 5.48 cm, find the initial speed of the bullet in m/s.

Find the time for the bullet-block system to come to rest in seconds.

Answer 1

suppose bullet and box just after collision moves with velocity =v

so system KE = 0.5* ( 3.45*10^-3 +1.37)*v^2

when system compresses the spring ,all its KE goes into spring PE so by energy conservation

system KE = Spring PE

0.5* ( 3.45*10^-3 +1.37)*v^2 = 1/2*k*Xm^2

0.5* ( 3.45*10^-3 +1.37)*v^2 = 0.5*785* ( 0.0548)^2

from here v = 1.3 m/s

now apply momentum conservation  

Pi = Pf

P_bullet = P( bullet +box)

3.45*10^-3* V_bullet =  ( 3.45*10^-3 +1.37)*1.3

V_bullet   =   ( 3.45*10^-3 +1.37)*1.3 / (3.45*10^-3 )

V_bullet = 517.53 m/s answer

now time is taken to come to rest

when bullet hits the box the system has maximum speed after that it start to decrese to zero it means it was at equilibrium position initial and then goes to one extreme position​

so time taken = 1/4* Time period = 1/4 * 2pi *sqrt( m/k) =

= 0.25*2pi*sqrt(  ( 3.45*10^-3 +1.37) /785)

=0.0657039 sec answer