JEE MAIN - Physics (2021 - 24th February Evening Shift - No. 23)
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
2
Explanation
Given that, $${{{M_1}} \over {{M_2}}} = {1 \over 2}$$
Also, p1 = p2 = p
$$ \Rightarrow $$ M1V1 = M2V2 = p
Also, we know that
$$K = {{{p^2}} \over {2M}} \Rightarrow {K_1} = {{{p^2}} \over {2{M_1}}}$$ & $${K_2} = {{{p^2}} \over {2{M_2}}}$$
$$ \Rightarrow {{{K_1}} \over {{K_2}}} = {{{p^2}} \over {2{M_1}}} \times {{2{M_2}} \over {{p^2}}} \Rightarrow {{{K_1}} \over {{K_2}}} = {{{M_2}} \over {{M_1}}} = {2 \over 1}$$
$$ \Rightarrow {A \over 1} = {2 \over 1}$$
$$ \Rightarrow $$ $$ \therefore $$ A = 2
Also, p1 = p2 = p
$$ \Rightarrow $$ M1V1 = M2V2 = p
Also, we know that
$$K = {{{p^2}} \over {2M}} \Rightarrow {K_1} = {{{p^2}} \over {2{M_1}}}$$ & $${K_2} = {{{p^2}} \over {2{M_2}}}$$
$$ \Rightarrow {{{K_1}} \over {{K_2}}} = {{{p^2}} \over {2{M_1}}} \times {{2{M_2}} \over {{p^2}}} \Rightarrow {{{K_1}} \over {{K_2}}} = {{{M_2}} \over {{M_1}}} = {2 \over 1}$$
$$ \Rightarrow {A \over 1} = {2 \over 1}$$
$$ \Rightarrow $$ $$ \therefore $$ A = 2
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