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  • A body falling on the ground from height of , rebounds to a fight <img alt="2.5 \mathrm{~m

A body falling on the ground from height of 10 \mathrm{~m}, rebounds to a fight 2.5 \mathrm{~m}, the percentage loss in K.E. will be

 

Option: 1

45 \%
 


Option: 2

60 \%
 


Option: 3

75 \%
 


Option: 4

40 \%


Answers (1)

Let \mathrm{v_1\: and \: v_2} be the velocity of the body just before and after the collision

 \mathrm{k_1 =\frac{1}{2} m v_1^2=m g h_1 }       .........(1)

\mathrm{\&\: k_2 =\frac{1}{2} m v_2^2=m g h_2 }    ..........(2)

Dividing we get-

\mathrm{\frac{v_{1}^{2}}{v_{2}^{2}}=\frac{h_{1}}{h_{2}}=\frac{10}{2.5}=4}

\mathrm{ \frac{v_1}{v_2}=\sqrt{4} }

\mathrm{ \frac{v_1}{v_2}=2 }
Now, percentage loss in kinetic energy-

\mathrm{ =\frac{m g\left(h_1-h_2\right)}{m g h_1} \times 100 }

\mathrm{ =\frac{10-2.5}{10} \times 100=75 \% }

Hence option 3 is correct.



 

 






 

Posted by

Kshitij

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