# Numerical values of the momentum and kinetic energy of a particle are equal. Velocity of the particle is:

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Numerical values of the momentum and kinetic energy of a particle are equal. Velocity of the particle is:

1. 1 ms-1
2. 2 ms-1
3. 4 ms-1
4.

Momentum cannot be equal to the kinetic energy

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Correct Answer - Option 2 : 2 ms-1

Concept:

• Kinetic energy (K.E): The energy possessed by a body by the virtue of its motion is called kinetic energy.

The expression for kinetic energy is given by:

$KE = \frac{1}{2}m{v^2}$

Where m = mass of the body and v = velocity of the body

• Momentum (p): The product of mass and velocity is called momentum.

Momentum (p) = mass (m) × velocity (v)

The relationship between the kinetic energy (KE or k) and Linear momentum is given by:

As we know,

$KE = \frac{1}{2}m{v^2}$

Divide numerator and denominator by m, we get

$KE = \frac{1}{2}\frac{{{m^2}{v^2}}}{m} = \frac{1}{2}\frac{{\;{{\left( {mv} \right)}^2}}}{m} = \frac{1}{2}\frac{{{p^2}}}{m}\;$ [p = mv]

$\therefore KE = \frac{1}{2}\frac{{{p^2}}}{m}\;$

$p = \sqrt {2mKE}$

Calculation:

Let, m be the mass of the body and v be the velocity of the body.
Now, momentum of the body is

p = mv

Kinetic energy is

$KE = \frac{1}{2}m{v^2}$

As per given condition,

Say,

$Momentum = Kinetic\;energy = x$

$\Rightarrow mv= {1\over2}mv^2=x$

$\Rightarrow {1\over2}mv^2=x$

$\Rightarrow {1\over2m}m^2v^2=x$

we know (mv)2 = x2

$\Rightarrow {1\over2m}x^2=x$

$\Rightarrow x=2m$

$\Rightarrow mv=2m \Rightarrow v=2$

Hence the correct answer is 2 m/s.