Question

The mass of a body is halved and velocity is doubled. Percentage increases in the K.E of the body is1)400%2)300%3)200%4)100%can u please help me with the question​

Answer 1

Kinetic energy is given by,

$\sf{\longrightarrow K=\dfrac{1}{2}\,mv^2}$

From this we get,

$\sf{\longrightarrow K\propto mv^2}$

Therefore,

$\sf{\longrightarrow \dfrac{K_1}{K_2}=\dfrac{m_1(v_1)^2}{m_2(v_2)^2}}$

Here,

$\sf{\longrightarrow m_2=\dfrac{m_1}{2}}$

$\sf{\longrightarrow \dfrac{m_1}{m_2}=2}$

And,

$\sf{\longrightarrow v_2=2v_1}$

$\sf{\longrightarrow \dfrac{v_1}{v_2}=\dfrac{1}{2}}$

Then,

$\sf{\longrightarrow \dfrac{K_1}{K_2}=2\times\dfrac{1^2}{2^2}}$

$\sf{\longrightarrow \dfrac{K_1}{K_2}=\dfrac{1}{2}}$

$\sf{\longrightarrow K_2=2K_1}$

Now, percentage increase in kinetic energy,

$\sf{\longrightarrow\delta K=\dfrac{K_2-K_1}{K_1}\times100}$

$\sf{\longrightarrow\delta K=\dfrac{2K_1-K_1}{K_1}\times100}$

$\sf{\longrightarrow\delta K=\dfrac{K_1}{K_1}\times100}$

$\sf{\longrightarrow\underline{\underline{\delta K=100\%}}}$

Hence (4) is the answer.