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  • What percentage of kinetic energy of a moving particle is transferred to a stationary particle when it strikes the stationary particle of <img alt="\mathrm{5}" src="https://entrancecorner.oncodecog

What percentage of kinetic energy of a moving particle is transferred to a stationary particle when it strikes the stationary particle of \mathrm{5} times its mass ?

( Assume the collision to be head-on elastic collison )

Option: 1

\mathrm{50.0%}


Option: 2

\mathrm{66.6%}


Option: 3

\mathrm{55.6%}


Option: 4

\mathrm{33.3%}


Answers (1)

best_answer

By linear momentum conservation

\mathrm{m(u)+ 5m (0)=m(v_1)+ 5m (v_2)}

\mathrm{u=v_{1}+5v_{2}}                          ..............(1)

Coefficient of restitution equation

\mathrm{e=\frac{v_{2}-v_{1}}{u-0}=1} \\                         

\mathrm{v_{2}-v_{1}=u}                           ...............(2)

Add (1) and (2)

\mathrm{6v_{2}=2u} \\

\mathrm{v_{2}=\frac{u}{3}} \\

Percentage energy of the bigger block   \mathrm{=\frac{\frac{1}{2} \times 5 m \times \frac{u^{2}}{9}}{\frac{1}{2} m u^{2}} \times 100=55.5 \%} \\

Hence the correct answer is option 3.

Posted by

vishal kumar

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