```                   A block of mass 0.50 kg is moving with a speed of 2.00 m/s on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is
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3 years ago

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```                    Dear vipin
doing conservation of momentum
new  v = 0.5*2/1+0.5
= 1/1.5 =2/3
energy loss = k.e final-k.e initila
= 0.5(1.5*(0.66*0.66) - 0.5*2*2)
= -0.6666
therefore energyloss = 0.6666n-m

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3 years ago
```                    Dear Vipin,
We know that momentum should be conserved
gives 0.5*2 = (1+0.5)*v
where v i the velocity of entire mass system moving after collision...
v=(1/1.5) m/s
energy lost = kinetic energy of the system before collision - kinetic energy of the system aftr collision
= 1/2*0.5*22 -1/2* 1.5 *(1/1.5)2
=0.6667 Joules
All the best for IITJEE preparation,

With Regards,
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3 years ago