Question

A man increases his speed by 2m/s, then his kinetic energy gets doubled. The original speed of the man is :
1. √2(√2 + 1) m/s
2. √2(√2 - 10)m/s
3. 2(√2 + 1)m\s
4. 2(√2 - 1) m/s

Answer 1

Correct Answer - Option 3 : 2(√2 + 1)m\s

CONCEPT:

Kinetic energy

• The energy possessed by a body due to the virtue of its motion is called kinetic energy.

\(\Rightarrow KE=\frac{1}{2}mv^{2}\)

Where KE = kinetic energy, m = mass and v = velocity

CALCULATION:

Given Δv = v₂ - v₁ = 2m/s and KE₂ = 2KE₁

Where Δv = increase in speed

• Initial kinetic energy,

\(\Rightarrow KE_{1}=\frac{1}{2}mv_{1}^{2}\)     -----(1)

• Final kinetic energy,

\(\Rightarrow KE_{2}=\frac{1}{2}mv_{2}^{2}\)     -----(2)

By equation 1 and equation 2,

\(\Rightarrow KE_{2}=2KE_{1}\)

\(\Rightarrow \frac{1}{2}mv_{2}^{2}=2\times\frac{1}{2}mv_{1}^{2}\)

\(\Rightarrow v_{2}^{2}=2v_{1}^{2}\)

\(\Rightarrow v_{2}=\pm\sqrt{2}v_{1}\)     -----(3)

Since,

\(\Rightarrow \Delta v = v_{2}-v_{1}=2m/s\)

By equation 3,

\(\Rightarrow \sqrt{2}v_{1}-v_{1}=2m/s\)

\(\Rightarrow v_{1}=\frac{2}{\sqrt{2}-1}\)     -----(4)

By dividing and multiplying \(\sqrt{2}+1\) in equation 4,

\(\Rightarrow v_{1}=\frac{2}{\sqrt{2}-1}\times\frac{\sqrt{2}+1}{\sqrt{2}+1}\)

\(\Rightarrow v_{1}=\frac{2(\sqrt{2}+1)}{\sqrt{2}^2-1^2}\)

\(\Rightarrow v_{1}=2(\sqrt{2}+1)\)m/s

• Hence, option 3 is correct.