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A block of mass 0.50 kg is moving with speed of 2.00 ms^(-1) 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 collisions is |
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Answer» 0.16 J `KE_i=1/2"mu"^2+1/2M(0)^2=1/2xx0.5xx2xx2+0=1J` For collision applying CONSERVATION of linear momentum `mxxu=(m+M)xxv` `:. 0.5xx2=(0.5+1)xxvimplies v=2/3 m//s` FINAL kinetic energy of the system is `KE_f=1/2(m+M)v^2=1/2(0.5+1)xx2/3xx2/3=1/3J` `:.` Energy LOSS during collision=`(1-1/3)J=0.67J` |
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