A bullet of mass 7g is fired into a block of metal of weight 7kg.The block is free to move.Calculate the velocity of bullet if the velocity of block with bullet is 0.7m/s.

Is the answer 350.35m/s?If not then provide the correct one

Dear nitin,

By considering the conservation of momentum

m1*V1 = (m1+m2)*v

finding V1,after substuting the values

m1= 7g

m2=7000g

v=0.7 m/s

we get V1=700.7 m/s

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Suryakanth –IITB

applying conservation of momentam

     let intial velocity of bullet is Vb

 intial momentam=MbVb               Mb is the mass of bullet

 finally the whole system moves with velocity.7m/s

  final momentam=(M+Mb).7

                          =4.9049

intitial momentam = final momettam

so  Vb =4.9049/.007

         =700.7m/s

Mass of bullet = 7/1000 kg = 0.007 kg

Mass of block = 7 kg

Momentum is conserved in this case. So:

m1*u1 + m2*u2 = m1*v1 + m2*v2

m - mass [1 - bullet 2- block]

u - Initial velocity

v - final velocity

0.007*u1 = 4.9

u1 = 49/0.007

u1 = 7000 metre per sec.

Mass of bullet(m1) = 7 g = 0.007 kgMass of block(m2) = 7 kgSince the block is free to move and it is stationary, initial velocity (u2) = 0Final velocity of the bullet and block = 0.7 m/sInitial velocity of the bullet(u1) = ?According to the law of conservation of momentum,m1u1 + m2u2 =(m1 + m2)v(Since the bullet and the block are combining to form a single mass)Putting the values of m1, m2, u1, u2 and v, we get0.007u1 + 7 × 0 =(0.007 + 7) × 0.7u1 = 7.007 × 0.7/0.007u1 = 700.7 m/sTherefore, the velocity of the bullet is 700.7 m/s.