Two solids A and B of mass 1 kg and 2 kg respectively are moving with equal linear momentum. The ratio of their kinetic energies (K.E.)A : (K.E.)B will be ${{A \over 1}}$, so the value of A will be ________.
Answer (integer)
2
Solution
Given that, ${{{M_1}} \over {{M_2}}} = {1 \over 2}$<br><br>Also, p<sub>1</sub> = p<sub>2</sub> = p<br><br>$\Rightarrow$ M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> = p<br><br>Also, we know that<br><br>$K = {{{p^2}} \over {2M}} \Rightarrow {K_1} = {{{p^2}} \over {2{M_1}}}$ & ${K_2} = {{{p^2}} \over {2{M_2}}}$<br><br>$$ \Rightarrow {{{K_1}} \over {{K_2}}} = {{{p^2}} \over {2{M_1}}} \times {{2{M_2}} \over {{p^2}}} \Rightarrow {{{K_1}} \over {{K_2}}} = {{{M_2}} \over {{M_1}}} = {2 \over 1}$$<br><br>$\Rightarrow {A \over 1} = {2 \over 1}$
<br><br>$\Rightarrow$ $\therefore$ A = 2
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Linear Momentum and Kinetic Energy
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