If the kinetic energy of a particle increases by 44%, then the increase in its momentum 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:

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

Or, 

\(\Rightarrow p = \sqrt {2mKE} \)

CALCULATION:

Given - KE₂/KE₁ = 144/100 and let p₁ = 100

• The kinetic energy of a particle in 1^st case is given by:

\(\Rightarrow KE_1 = \frac{1}{2}\frac{{{p^2_1}}}{m}\;\)      -------- (1)

• The kinetic energy of a particle in 2nd case is given by:

\(\Rightarrow KE_2 = \frac{1}{2}\frac{{{p^2_2}}}{m}\;\)       ------ (2)

On dividing equation 2 by 1, we get

\(\Rightarrow \frac{KE_2}{KE_1}=(\frac{p_2}{p_1})^2\)

\(\Rightarrow \frac{p_2}{p_1}=\sqrt{\frac{KE_2}{KE_1}}\)

\(\Rightarrow \frac{p_2}{p_1}-1=\sqrt{\frac{KE_2}{KE_1}}-1\)

\(\Rightarrow \frac{\Delta p}{p_1}-1=\sqrt{\frac{KE_2}{KE_1}}-1\)

\(\Rightarrow \Delta p=100\times \sqrt{\frac{144}{100}}-1=20\, \%\)