AP EAMCET201922 Apr 2019Morning ShiftPhysicsWork, Power and EnergyActual
A man who is running has half the kinetic energy of a boy of half his mass. The man speeds up by (1 ~ms ⁻¹ ) and then has the same kinetic energy as the boy. The initial speed of the boy is
Options
- A( 2 +1 ~ms ⁻¹ )
- B(2( 2 +1) ms ⁻¹ )
- C( 2 ~ms ⁻¹ )
- D(2 ~ms ⁻¹ )
Correct answer
B. (2( 2 +1) ms ⁻¹ )
Step-by-step solution
According to the question, let the mass of man be (m ). So, mass of boy is ( m 2 ). Let speed of man be (v₁ ) and that of boy be (v₂ ). ( ) Kinetic energy of man (= 1 2 ) kinetic energy of boy or ( KE _ man = 1 2 KE _ boy ) ( 1 2 m v₁^2= 1 2 1 2 m 2 v₂^2 or v₁^2 v₂^2 = 1 4 ) ( v₁ v₂ = 1 2 ) ( v₁= 1 2 v₂ )...(i) If the velocity of man increased by (1 ~m / s ) then, new velocity of man, (v₁^ = ( v₂ 2 +1 )[ ) From Eq. (i)] Now, kinetic energy of man (KE) = kinetic energy of Boy ( ( KE _ B ) ) ( aligned 1 2 m ( v₂ 2 +1