JEE MAIN - Physics (2021 - 20th July Evening Shift - No. 1)
If the Kinetic energy of a moving body becomes four times its initial Kinetic energy, then the percentage change in its momentum will be :
100%
200%
300%
400%
Explanation
K2 = 4K1
$${1 \over 2}$$mv$$_2^2$$ = 4$${1 \over 2}$$mv$$_1^2$$
v2 = 2v1
P = mv
P2 = mv2 = 2mv1
P1 = mv1
% change = $${{\Delta P} \over {{P_1}}} \times 100 = {{2m{v_1} - m{v_1}} \over {m{v_1}}} \times 100 = 100\% $$
$${1 \over 2}$$mv$$_2^2$$ = 4$${1 \over 2}$$mv$$_1^2$$
v2 = 2v1
P = mv
P2 = mv2 = 2mv1
P1 = mv1
% change = $${{\Delta P} \over {{P_1}}} \times 100 = {{2m{v_1} - m{v_1}} \over {m{v_1}}} \times 100 = 100\% $$
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