The kinetic energy of a body is stated to increase by 300 percent. The corresponding increase in the momentum of the body will be ____%

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 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]

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

\(p = \sqrt {2m(KE)} \)

CALCULATION:

Let initial Kinetic energy = KE1 = E

Given that:

Final kinetic energy (K.E2) = K.E1 + 300 % of KE1 = E + 3E = 4E

The relation between the momentum and the kinetic energy is given by:

\(P = \sqrt {2m\;(KE)}\)

Final momentum (P') will be:

\(P' = \sqrt {2m\;(KE)_2} = \sqrt {2m\; × 4E} = 2 \sqrt {2m\;E} = 2 P\)

Increase in momentum (ΔP) = P' - P = 2P - P = P

% Increase = \(\frac{\Delta P}{P} \times 100=\frac{2P-P}{P} \times 100=100\)