# 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

105 Points
12 years ago

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

All the best.

Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

Suryakanth –IITB

509 Points
12 years ago

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

Karthik Eyan
45 Points
12 years ago

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.

Stephen Curry
13 Points
5 years ago
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.