A man who is running has half the kinetic ene | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the mass of the man be $m$. Then the mass of the boy is $m_b = \frac{m}{2}$.Let the initial speed of the man be $v_m$ and the speed of the boy

Vedclass · Accepted Answer

$2(\sqrt{2}+1) \,ms^{-1}$

Vedclass · Answer

(B) Let the mass of the man be $m$. Then the mass of the boy is $m_b = \frac{m}{2}$.Let the initial speed of the man be $v_m$ and the speed of the boy be $v_b$.According to the problem,the kinetic energy of the man is half the kinetic energy of the boy:$\frac{1}{2} m v_m^2 = \frac{1}{2} (\frac{1}{2} m_b v_b^2) = \frac{1}{2} (\frac{1}{2} \cdot \frac{m}{2} v_b^2) = \frac{1}{8} m v_b^2$.Simplifying,$v_m^2 = \frac{1}{4} v_b^2$,which gives $v_m = \frac{v_b}{2}$.When the man speeds up by $1 \,ms^{-1}$,his new speed is $v_m' = v_m + 1 = \frac{v_b}{2} + 1$.Now,his kinetic energy equals the boy's kinetic energy:$\frac{1}{2} m (\frac{v_b}{2} + 1)^2 = \frac{1}{2} (\frac{m}{2}) v_b^2$.Dividing both sides by $\frac{m}{2}$,we get $(\frac{v_b}{2} + 1)^2 = \frac{v_b^2}{2}$.Taking the square root of both sides: $\frac{v_b}{2} + 1 = \frac{v_b}{\sqrt{2}}$.$1 = v_b (\frac{1}{\sqrt{2}} - \frac{1}{2}) = v_b (\frac{2 - \sqrt{2}}{2\sqrt{2}})$.$v_b = \frac{2\sqrt{2}}{2 - \sqrt{2}} = \frac{2\sqrt{2}(2 + \sqrt{2})}{4 - 2} = \frac{4\sqrt{2} + 4}{2} = 2\sqrt{2} + 2 = 2(\sqrt{2} + 1) \,ms^{-1}$.

$A$ 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

Similar Questions

What is the shape of the graph between the speed and kinetic energy of a body?

The kinetic energy acquired by a mass $m$ in travelling a certain distance $d$ starting from rest under the action of a constant force is directly proportional to

Give an important factor to control the speed of a vehicle in the area of a school or hospital.

If the kinetic energy of an object becomes four times its initial value,what will be its new momentum?

The kinetic energy of a body of mass $4 \ kg$ moving with a velocity of $(2 \hat{i}-4 \hat{j}-\hat{k}) \ ms^{-1}$ is (in $J$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module