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Standard VIII

Physics

Normal Force

A stationary ...

Question

A stationary particle explodes into two particles of masses $x$ and $y$ , which move in opposite direction with velocity $v_{1}$ and $v_{2}$ . The ratio of their kinetic energies $(E_{1} : E_{2})$ is

1

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\frac{x v_{2}}{y v_{1}}

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\frac{x}{y}

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\frac{y}{x}

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Solution

The correct option is D $\frac{y}{x}$
The momentum initially when mass exploded is zero, so
By conservation of momentum xv $_{1}$ = yv $_{2}$
The ratio of kinetic energies E $_{1}$ and $E_{2}$ is
$\frac{x v_{1}^{2}}{y v_{2}^{2}} = \frac{v_{1}}{v_{2}} = \frac{y}{x}$

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A stationary particle explodes into two particles of masses x and y, which move in opposite direction with velocity v1 and v2. The ratio of their kinetic energies (E1:E2) is

The correct option is D yxThe momentum initially when mass exploded is zero, soBy conservation of momentum xv1= yv2The ratio of kinetic energies E1 and E2 is xv21yv22=v1v2=yx

A stationary particle explodes into two particles of masses $x$ and $y$ , which move in opposite direction with velocity $v_{1}$ and $v_{2}$ . The ratio of their kinetic energies $(E_{1} : E_{2})$ is

The correct option is D $\frac{y}{x}$
The momentum initially when mass exploded is zero, so
By conservation of momentum xv $_{1}$ = yv $_{2}$
The ratio of kinetic energies E $_{1}$ and $E_{2}$ is
$\frac{x v_{1}^{2}}{y v_{2}^{2}} = \frac{v_{1}}{v_{2}} = \frac{y}{x}$