Question

A running man has half the kinetic energy of that of a boy of half of his mass. The man speeds up by 2 m/s so as to have the same kinetic energy as that of the boy. The initial speed of the man was:
1. \(\sqrt2 \,m/s\)
2. \(\sqrt2-1\,m/s\)
3. \(\frac{2}{\sqrt2-1}\,m/s\)
4. \(\frac{1}{\sqrt2}\,m/s\)

Answer 1

Correct Answer - Option 3 : \(\frac{2}{\sqrt2-1}\,m/s\)

CONCEPT:

Kinetic energy

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

\(⇒ KE=\frac{1}{2}mv^{2}\)

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

CALCULATION:

Given m_M = m, m_B = \(\frac{m}{2}\), KE_M1 = \(\frac{KE}{2}\), and KE_M2 = KE_B = KE

• The kinetic energy of the boy is given as,

\(⇒ KE_B=\frac{1}{2}m_Bv_B^{2}\)

\(⇒ KE=\frac{1}{2}\times\frac{m}{2}v_B^{2}\)

\(⇒ KE=\frac{1}{4}mv_B^{2}\)     ----(1)

• The initial kinetic energy of the man is given as,

\(⇒ KE_{M1}=\frac{1}{2}m_Mv_{M1}^{2}\)

\(⇒ \frac{KE}{2}=\frac{1}{2}mv_{M1}^{2}\)

\(⇒ KE=mv_{M1}^{2}\)     ----(2)

• The final velocity of the man is given as,

⇒ v_M2 = V_M1 + 2

• The final kinetic energy of the man is given as,

\(⇒ KE_{M2}=\frac{1}{2}m_Mv_{M2}^{2}\)

\(⇒ KE=\frac{1}{2}m(v_{M1}+2)^{2}\)     ----(3)

By equation 1 and equation 2,

\(⇒ mv_{M1}^{2}=\frac{1}{4}mv_B^{2}\)

\(⇒ v_{M1}^{2}=\frac{1}{4}v_B^{2}\)     ----(4)

By equation 1 and equation 3,

\(⇒ \frac{1}{2}m(v_{M1}+2)^{2}=\frac{1}{4}mv_B^{2}\)

\(⇒ (v_{M1}+2)^{2}=\frac{1}{2}v_B^{2}\)     ----(5)

By equation 4 and equation 5,

\(⇒ (v_{M1}+2)^{2}=2v_{M1}^{2}\)

\(⇒ v_{M1}=\frac{2}{\sqrt2-1}\,m/s\)

• Hence, option 3 is correct.