A bullet of mass 20g strilkes a block of mass 980g and is embedded in it. The block is in contact with a spring whose force constant = 100N/m. After the collision the spring is compressed upto 10cm. Find- 1. Velocity of block after collision. 2. Magnitude of velocity of bullet. 3. Loss in kinetic energy due to collision
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m=20/1000=0.02 kg.
M=0.98kg.
Also, k=100N/m.
it is given that, x=0.1m.
a)velocity of the block system,
(m+M)v²=kx²=>v²=[100×(0.1)²]/0.98+0.02)=>v=1m/s.
since total momentum is conserved, so
(m+M)v=mv'+MV'=>1=0.98×V'=>V'=1.02m/s. which is the velocity of the block .
b)but the velocity of the bullet is 0 m/s after the collision.
c)loss in K.E is also 0 J as we have considered that the bullet loses its velocity as it is embedded in the block and it loses all of its velocity at a very small time.
M=0.98kg.
Also, k=100N/m.
it is given that, x=0.1m.
a)velocity of the block system,
(m+M)v²=kx²=>v²=[100×(0.1)²]/0.98+0.02)=>v=1m/s.
since total momentum is conserved, so
(m+M)v=mv'+MV'=>1=0.98×V'=>V'=1.02m/s. which is the velocity of the block .
b)but the velocity of the bullet is 0 m/s after the collision.
c)loss in K.E is also 0 J as we have considered that the bullet loses its velocity as it is embedded in the block and it loses all of its velocity at a very small time.
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