Question

9.
14.
A man is running on horizontal road has half the
kinetic energy of a boy of half of his mass. When man
speeds up by 1 m/s, then his KE becomes equal to
KE of the boy, the original speed of the man is
(1) JZ ms (2) (12-1) mis
(3) 2 m/s
147 (12+1) mis
Aakash Educational Services Pvt. Ltd.-Regd. Office. Aakash​

Answer 1

Answer:

2.4 m/s

Explanation:

Original speed of man = v

Speed of boy = V

Mass of man = m

Mass of boy = m/2

At original speed of man

Kinetic energy of man = 0.5 × Kinetic energy of boy

0.5mv² = 0.5 × 0.5(m/2)V²

v² = V²/4

v = V/2

V = 2v

When man speeds up by 1 m/s

Kinetic energy of man = Kinetic energy of boy

0.5m(v + 1)² = 0.5(m/2)V²

(v + 1)² = V²/2

(v + 1)² = (2v)²/2

(v + 1)² = 4v²/2

(v + 1)² = 2v²

v² + 2v + 1 = 2v²

v² - 2v - 1 = 0

v = 2.4 m/s

Answer 2

Answer:

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