If momentum is increased by 20% , then K.E. i | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(a)$E = \frac{{{P^2}}}{{2m}}.$ If m is constant then $E \propto {P^2}$
$&rArr;$$\frac{{{E_2}}}{{{E_1}}} = {\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^2

Vedclass · Accepted Answer

$44$

Vedclass · Answer

(a)$E = \frac{{{P^2}}}{{2m}}.$ If m is constant then $E \propto {P^2}$
$&rArr;$$\frac{{{E_2}}}{{{E_1}}} = {\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^2} = {\left( {\frac{{1.2P}}{P}} \right)^2} = 1.44$
$&rArr;$ ${E_2} = 1.44{E_1} = {E_1} + 0.44{E_1}$
${E_2} = {E_1} + 44\% $ of ${E_1}$
i.e. the kinetic energy will increase by $44\%$

If momentum is increased by $20\%$ , then K.E. increases by ........... $\%$

Similar Questions

Which of the following statements is correct ?

If velocity of a body is twice of previous velocity, then kinetic energy will become

A particle of mass $m$ at rest is acted upon by a force $F$ for a time $t$. Its Kinetic energy after an interval $t$ is

A particle of mass ${m_1}$ is moving with a velocity ${v_1}$and another particle of mass ${m_2}$is moving with a velocity ${v_2}$. Both of them have the same momentum but their different kinetic energies are ${E_1}$and ${E_2}$respectively. If ${m_1} > {m_2}$ then

A $120\, g$ mass has a velocity $\vec v = 2\hat i + 5\hat j\,m{s^{ - 1}}$ at a certain instant. Its kinetic energy is .................. $\mathrm{J}$