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

Question

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 (A) √2 m / s (B) (√2-1) m / s (C) 2 m / s (D) (√2+1) m / s

Tardigrade Admin · Accepted Answer

According to problem KB=2 Km (1/2) mB vB2=2((1/2) mm vm2) (1/2) mB vB2=2((1/2) mm(vm+1)2) Solving vm=√2+1 m / s

Q. 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

$2 m / s$

$(2 - 1) m / s$

$2 m / s$

$(2 + 1) m / s$

According to problem
$K_{B} = 2 K_{m}$
$\frac{1}{2} m_{B} v_{B}^{2} = 2 (\frac{1}{2} m_{m} v_{m}^{2})$
$\frac{1}{2} m_{B} v_{B}^{2} = 2 (\frac{1}{2} m_{m} (v_{m} + 1)^{2})$
Solving
$v_{m} = 2 + 1 m / s$

Q. 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 1m/s, then his KE becomes equal to KE of the boy, the original speed of the man is

2​m/s

(2​−1)m/s

2m/s

(2​+1)m/s

According to problem KB​=2Km​ 21​mB​vB2​=2(21​mm​vm2​) 21​mB​vB2​=2(21​mm​(vm​+1)2) Solving vm​=2​+1m/s

Q. 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

$2 m / s$

$(2 - 1) m / s$

$2 m / s$

$(2 + 1) m / s$

According to problem
$K_{B} = 2 K_{m}$
$\frac{1}{2} m_{B} v_{B}^{2} = 2 (\frac{1}{2} m_{m} v_{m}^{2})$
$\frac{1}{2} m_{B} v_{B}^{2} = 2 (\frac{1}{2} m_{m} (v_{m} + 1)^{2})$
Solving
$v_{m} = 2 + 1 m / s$