1.

In the above question the percentage loss of kinetic energy of the body during its collision with the ground is :

Answer»

0.25
0.5
0.75
0.99

Solution :Here the percentage Loss of energy is given by:
`Loss=("K.E. before COLLISION-K.E. sfter collision")/("K.E. before collision")XX100`
=`(1/2mv_1^2-1/2mv_2^2)/(1/2mv_1^2)xx100``(because v_1=2v_2)`
=`((2v_2)^2-v_2^2)/(4v_2^2)xx10`
=`(3v_2^2)/(4v_2^2)xx100=3/4xx100=75%`


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