Question

A running man has half the kinetic energy that a boy of half his mass has. The man speeds up by 1 m/sec and then has the same kinetic energy as the boy. What were the original speeds of man and boy
respectively?
(1) 2.4, 4.8 m/sec
(2) 2.4, 3.4 m/sec
(3) 3.4, 4.8 m/sec
(4) 3.4, 6.8 m/sec


Answer with explanation or else you'll be reported. ​

Answer 1

Solution :-

• Mass of man = M

• Speed of man = Vm

• Mass of boy = M/ 2

• Speed of boy = Vb

Now as per the given condition ,

➜ KE m = KE b / 2

➜ M x Vm²/2 = M x Vb² / 4 x 2

➜ Vm = Vb / 2 ... (1)

Now ,

• New speed of the man = Vm + 1

Everything else remains the same .

➜ KE m ' = KE b '

➜ M x  (Vm + 1 )² / 2 = M x Vb² /2 x  2

➜ (Vm + 1 )² = Vb² / 2

➜ (Vm + 1 )²= 4 Vm/ 2 ....( From 1 )

➜ Vm + 1 = √2 Vm

➜ √2Vm -  Vm = 1

➜ Vm = 1 / √2 - 1

➜ Vm = 1 / 1.414 - 1

➜ Vm = 1 / 0.414

➜ Vm = 2.4 m /s

Now let's substitute the value of Vm in (1),

➜ Vb = Vm x 2

➜ Vb = 2.4 x 2

➜ Vb = 4.8 m/s

The original speed of the man and the boy are 2.4 m/s and 4.8 m/s respectively .