A stationary particle explodes into two particles of masses x and y , which move in opposite directi

English

Gujarati

Hindi

5.Work, Energy, Power and Collision

medium

A stationary particle explodes into two particles of masses $x$ and $y$, which move in opposite directions with velocity $v_1$ and $v_2$. The ratio of their kinetic energies $\left(E_1: E_2\right)$ is

A

$1$

B

$\frac{x v_2}{y v_1}$

C

$\frac{x}{y}$

D

$\frac{y}{x}$

Solution

(d)

Momentum will be conserved

$0=x v_1+y v_2$

$-x v_1=y v_2 \ldots (1)$

$\frac{k_1}{k_2}=\frac{\frac{1}{2} x v_1^2}{\frac{1}{2} y v_2^2}$

Using $(1)$ $\left(\frac{v_1}{v_2}\right)^2=\frac{y^2}{x^2}$

$\frac{k_1}{k_2}=\frac{x}{y} \cdot\left(\frac{y^2}{x^2}\right)$

$=\frac{y}{x}$

Standard 11

Physics

Share

Similar Questions

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

medium

(AIPMT-1989)

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

medium

(AIPMT-2004)

If the K.E. of a particle is doubled, then its momentum will

medium

Two particles of the same mass are moving in circular orbits because of force, given by $F(r) = \frac{{ – 16}}{r}\, – \,{r^3}$ The first particle is at a distance $r = 1,$ and the second, at $r = 4.$ The best estimate for the ratio of kinetic energies of the first and the second particle is closest to

hard

(JEE MAIN-2018)

A particle is dropped from a height $h$. A constant horizontal velocity is given to the particle. Taking $g$ to be constant every where, kinetic energy $E$ of the particle w. r. t. time t is correctly shown in

medium