A 9.71 g bullet, moving horizontally with an initial speed v0, embeds itself in

Question

A 9.71 g bullet, moving horizontally with an initial speed v0, embeds itself in a 1.45 kg pendulum bob that is initially at rest. The length of the pendulum is L = 0.815 m. After the collision, the pendulum swings to one side and comes to rest when it has gained a vertical height of 12.4 cm.

Is the kinetic energy of the bullet-bob system immediately after the collision greater than, less than, or the same as the kinetic energy of the system just before the collision?

Find the initial speed of the bullet.

How long does it take for the bullet-bob system to come to rest for the first time?

Explanation / Answer

Given

mass of the bullet    m1 = 9.71 g

mass of pendulum bob   m2 = 1.45 kg

vertical height   h = 12.4 cm

length of the pendulum L = 0.815 m

initiallly bob is at rest so vo ' = 0 m/s

   Using law of conservation of momentum

                m1 vo + m2 v o' = ( m1 + m2 ) v

                 m1 vo = ( m1 + m2 ) v ........(1)

Both potential energy at the bottom and kinetic energy at the top are zero.

                So  

            ( 1/2) ( m1+ m2) v2 = ( m1 + m2 ) gh

               (1 /2 )( 1.4597)v2 = ( 1.4597)(9.8)(0.124 m )

       there fore v = 1.5589m/s  

substitute v value in equation (1)

            9.71*10-3 vo = (1.4597)(1.5589m/s )

   initial speed of the bullet vo = 234 m/s

Time period of pendulum

          T = 2 ( L / g ) 1/2

             = 2 ( 0.815 m / 9.8 ) 1/2

              = 1.81 s

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