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

3 years ago


Answers : (2)


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,

Adapa Bharath

3 years ago

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