A body of mass 1kg, moving with a velocity of 4m/s collides with a body of mass 0.5kg moving with a velocity of 2m/s. If tge velocity of the first body after collision is 8/3m/s, find the velocity of the second body after collision?
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Net initial momentum=(40*4)+(60*2)=280kgm/s.
As collision is perfectly inelastic, final mass of both solids(both stuck together)=100 kg
Final vel. of combined solids=Tot. Mom./Tot. Mass=280/100=2.8m/s
As K.E.=0.5*m*v^2
Initial K.E.=(.5*40*4^2)+(0.5*100*2^2)=320+120=440J
Final K.E.=0.5*100*2.8^2=392J
Loss in K.E.=440–392=48 J
As collision is perfectly inelastic, final mass of both solids(both stuck together)=100 kg
Final vel. of combined solids=Tot. Mom./Tot. Mass=280/100=2.8m/s
As K.E.=0.5*m*v^2
Initial K.E.=(.5*40*4^2)+(0.5*100*2^2)=320+120=440J
Final K.E.=0.5*100*2.8^2=392J
Loss in K.E.=440–392=48 J
Anonymous:
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Answered by
4
Given :
mass of first object = 1 kg
initial velocity = 4 m/s
final velocity = 8/3 m/s
mass of second object = 0.5 kg
initial velocity = 2 m/s
final velocity = ?
By Law of conservation of energy :
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
= > 1 × 4 + 0.5 × 2 = 1 × 8/3 + 0.5 v
= > 4 + 1 = 8/3 + 0.5 v
= > 0.5 v + 8/3 = 5
= > 0.5 v = 5 - 8/3
= > 0.5 v = ( 15 - 8 ) /3
= > 0.5 v = 7/3
= > 1/2 v = 7/3
= > v = 14/3 m/s
= > v = 4.666 m/s
ANSWER:
The velocity would be 4.67 m/s
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