Question

A bullet, mass 10 g, is fired into a block of wood, mass 390 g, lying on a smooth surface. Then both (bullet & wood) move together with a velocity of 10 m/s. a) What was the speed of the bullet before collision? b) What is the K.E before and after the collision?

Answer 1

Mass of bullet m1 = 10g 
Mass of block m2 = 200g 
Velocity of bullet before impact u1 = 500 m/s 
Velocity of block before impact u2 = 0 
Velocity of block+bullet after impact v = ? 
Momentum before impact M1 = m1u1+m2u2 
Or, M1 = m1u1 ------(1) 
(since u2=0) 
Momentum after impact M2 = m1v + m2v (since bullet gets embedded in block and both move with speed v) 
Or, M2 = (m1+m2)v-----(2) 
By conservation of momentum 
M2 = M1-----(3) 
From (1), (2) and (3) 
(m1+m2)v = m1u1 
Or, v = m1u1/(m1+m2) 
= 10*500/(10+200) m/s 
= 10*500/210 m/s 
= 23.8 m/s 
Ans: 23.8 m/s

Answer 2

Answer:

hyyy buddy

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