The KE of a body is increased by 44 %. What i | vedclass

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

Question

(B) The relationship between kinetic energy $(K)$ and momentum $(P)$ is given by $K = \frac{P^2}{2m}$,which implies $P = \sqrt{2mK}$.Let the initial k

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(B) The relationship between kinetic energy $(K)$ and momentum $(P)$ is given by $K = \frac{P^2}{2m}$,which implies $P = \sqrt{2mK}$.Let the initial kinetic energy be $K_1$ and the final kinetic energy be $K_2 = K_1 + 0.44 K_1 = 1.44 K_1$.Let the initial momentum be $P_1$ and the final momentum be $P_2$.Since $P \propto \sqrt{K}$,we have $\frac{P_2}{P_1} = \sqrt{\frac{K_2}{K_1}} = \sqrt{1.44} = 1.2$.Therefore,$P_2 = 1.2 P_1$.The percentage increase in momentum is $\frac{P_2 - P_1}{P_1} 	imes 100 = (1.2 - 1) 	imes 100 = 20 \%$.

The $KE$ of a body is increased by $44 \%$. What is the percentage increase in its momentum?

Similar Questions

The kinetic energy of a body of mass $4 \,kg$ and momentum $6 \,Ns$ will be (in $\,J$)

Two bodies of masses $1 \,g$ and $4 \,g$ are moving with equal kinetic energies. The ratio of the magnitudes of their linear momenta is

$A$ graph between kinetic energy $K$ and momentum $P$ of a particle is plotted as shown in the figure. The mass of the moving particle is ........ $kg$.

Two bodies of masses $1 \ g$ and $9 \ g$ have equal kinetic energy. The ratio of their momenta is:

Two bodies of masses $m_1$ and $m_2$ are moving with the same kinetic energy. If $P_1$ and $P_2$ are their respective momenta,the ratio $\frac{P_1}{P_2}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module