Answers

2015-05-13T17:40:41+05:30
we know that [ m1.u1+ m2 u2 = m1.v1 + m2.v2 ]

here initial velocity of the rifle (u1) =0
its mass (m1) = 4kg

mass of the bullet (m2)  = 50g=.05kg
its velocity before firing  (u2) = 0

hence we get, m1u1 + m2u2 =0

the momentum of the bullet and the rifle after firing = m1v1 + m2v2 =0 
i.e. 4kg.v1 +0.05*35 = 0
     4 v1 +1.75 = 0
     4.v1 = -1.75
     v1 = -1.75/ 4 = 0.4375 = 0.44 (approx.)




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you may ask ur doubts if any..
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:P
thanks for that
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  • Brainly User
2015-05-13T18:26:04+05:30
M1 = 4 kg 
m2 = 0.05 kg 
v2=35m/s
v2 ?
seedha   potential ko conserve kr do 
m1u1 + m2u2 = m1v1+m2v2
initial velocity har ek me zero h to 
m1v1+m2v2 = 0 
4×v1 + 0.05 ×35 = 0 
4v1 =1.75 
v1 =1.75/4
v1 = 0.43 m/s



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