Question

Plz give it fast..............

Answer 1

kinetic energy= mv^2/2

mass directly proportional to kinetic energy

square of velocity is directly proportional to kinetic energy.

Given

mass of A= mass of B

Suppose velocity be v

Velocity of A = v/2

velocity of B = 2v

Now

VA^2/Vb^2 = v^2/16v^2

So

ratio of kinetic energy of A:B = 1:16

Answer 2

$\huge\red\bigstar\huge\mathcal{\underline{ \underline{QUESTION}}}\red\bigstar$

for given bodies A and B having equal masses the velocity of a is half while the velocity of B is doubled find the ratio of their new kinetic energy.

$\huge\red\bigstar\huge\mathcal{\underline{ \underline{SOLUTION}}}\red\bigstar$

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let, mass Of A and B is m

therefore for the body B

mass=m

$(K_E)_{B}=\frac{1}{2}mv{}^{2}\\</p><p>now \: the\: velocity \:of \:the\: body\: A\: \\is\: doubled\: \\therefore.....</p><p>((K_E)_{B(new)}=\frac{1}{2}m(2v){}^{2}\\ \implies ((K_E)_{B(new)}=\frac{1}{\cancel{2}}m\times \cancel{4}v{}^{}\\\implies ((K_E)_{B(new)}=2mv{}^{2}$

therefore for the body A

mass=m

$(K_E)_A=\frac{1}{2}mv{}^{2}\\</p><p>now \: the\: velocity \:of \:the\: body\: A\: \\is\: halved\:\\therefore.....</p><p>((K_E)_{A(new)}=\frac{1}{2}m(\frac{v}{2}){}^{2}\\ \implies ((K_E)_{A(new)}=\frac{1}{8}mv{}^{2}$

therefore.....

$\frac{((K_E)_{B(new)}}{((K_E)_{A(new)}}=\frac{2mv{}^{2}}{\frac{1}{8}mv{}^{2}}\\\implies\frac{((K_E)_{A(new)}}{((K_E)_{B(new)}}=\frac{1}{16}\\ \implies \boxed{(K_E)_{A(new)}: (K_E)_{B(new)}=1:16}$

$\large\mathfrak{...hope\: this \:helps\: you.....}$