A running man has the same kinetic energy as that of a boy of half his mass. The man speed up by 2m/s and the boy changes his speed by ‘x’ m/s so that the kinetic energies of the boy and the man are equal. Then x=

The value of x is 2√2 m/s or 2.83 m/s.

Explanation:

Let the mass of the man (mm) = 2m.

Therefore, the mass of the boy (mb) = ½ mass of man = ½ × 2m = m

Let velocity of the man (vm) be v1.

Let the velocity of the boy (vb) be v2.

Let the kinetic energy of the man be Km.

Let the kinetic energy of the boy be Kb.

But the kinetic energy of the man is the same as that of the boy i.e

Km = Kb

½•mm•vm² = ½•mb•vb²

Mass of the man (mm) = 2m.

velocity of the man (vm) = v1.

Mass of the boy (mb) = m.

velocity of the boy (vb) = v2

½ × 2m × v1² = ½ × m × v2²

2v1² = v2²

Take the square root of both side

√2v1 = v2 (1)

Now, the speed of the man increased by 2 and that of the boy by x i.e

Mass of the man (mm) = 2m.

velocity of the man (vm) = v1 + 2

Mass of the boy (mb) = m.

velocity of the boy (vb) = v2 + x

Km = Kb

½•mm•vm² = ½•mb•vb²

½ × 2m × (v1 + 2)² = ½ × m × (v2 + x)²

2 (v1 + 2)² = (v2 + x)²

Take the square root of both side

√2 (v1 + 2) = (v2 + x)

Clear bracket

√2v1 + 2√2 = v2 + x (2)

From equation 1 above,

v2 = √2v1

Substituting the value of v2 into equation 2.

√2v1 + 2√2 = v2 + x

√2v1 + 2√2 = √2v1 + x

Collect like terms

√2v1 – √2v1 + 2√2 = x

2√2 = x

x = 2√2 m/s = 2.83 m/s

Therefore, the value of x is 2√2 m/s or 2.83 m/s.

the duration of a photographic flash is related to an rc time constant, which is 0.100 μs for a certain camera.