
A running man has half the kinetic energy of that of a boy of half of his mass. The man speeds up by 1 $m/s$ . So as to have the same kinetic energy as that of the boy. The original speed of the man is:
$\begin{align}
& A.\sqrt{2}m/s\\
& B.(\sqrt{2}-1)m/s\\
& C.\dfrac{1}{\sqrt{2}}m/s\\
& D.\dfrac{1}{1\sqrt{2}}m/s\\
\end{align}$
Answer
585.9k+ views
Hint: Apply the formula of the kinetic energy which is $KE = \dfrac{1}{2} mv^2$ and find the relation between the velocities of the man and the boy. Then find the value of the original speed of the man using the given condition for the equal kinetic energies of the man and the boy.
Step by step solution:
The value of the mass of the man and the velocity of the man is unknown and hence let us consider the mass of the man as m
The value of the velocity the man be v
So we get the kinetic energy of the man using the formula given as :
The kinetic energy of the man is used using the formula as :
$KE = \dfrac{1}{2} mv^2$
The value of the kinetic energy of the man becomes: $KE_m = \dfrac{1}{2} mv^2$
Now the mass of a boy can be written as: $\dfrac{m}{2}$
The velocity of the boy is assumed as $v_b$
Now the given condition is $\dfrac{1}{2} m v^2 = \dfrac{1}{4} \times \dfrac{m}{2} v_b ^2$
We get $v = \dfrac{v_b}{2}$
Now the value of the kinetic energy is the same as that of the boy when the value of the velocity of the man is increased by 1 meter per second
So we can write the condition as : $\dfrac{1}{2} m (v+1)^2 = \dfrac{1}{2} \times \dfrac{m}{2} (v_b)^2$
Substituting the value of the $v_b$ we get the value of the initial velocity of the man as
$V = \dfrac{\sqrt2 -1 }{1} m/s $
The value of the speed of the man initially becomes: ($\sqrt2 -1$) m/s
So the correct option is option B
Note: One of the main possible mistakes that we can make in this kind of problem is finding the condition to apply for the given problem. The given is that when the velocity of the man's speed is increased by one meter per second the kinetic energy is the same. We need to carefully apply the condition to get the value of the original speed of the person.
Step by step solution:
The value of the mass of the man and the velocity of the man is unknown and hence let us consider the mass of the man as m
The value of the velocity the man be v
So we get the kinetic energy of the man using the formula given as :
The kinetic energy of the man is used using the formula as :
$KE = \dfrac{1}{2} mv^2$
The value of the kinetic energy of the man becomes: $KE_m = \dfrac{1}{2} mv^2$
Now the mass of a boy can be written as: $\dfrac{m}{2}$
The velocity of the boy is assumed as $v_b$
Now the given condition is $\dfrac{1}{2} m v^2 = \dfrac{1}{4} \times \dfrac{m}{2} v_b ^2$
We get $v = \dfrac{v_b}{2}$
Now the value of the kinetic energy is the same as that of the boy when the value of the velocity of the man is increased by 1 meter per second
So we can write the condition as : $\dfrac{1}{2} m (v+1)^2 = \dfrac{1}{2} \times \dfrac{m}{2} (v_b)^2$
Substituting the value of the $v_b$ we get the value of the initial velocity of the man as
$V = \dfrac{\sqrt2 -1 }{1} m/s $
The value of the speed of the man initially becomes: ($\sqrt2 -1$) m/s
So the correct option is option B
Note: One of the main possible mistakes that we can make in this kind of problem is finding the condition to apply for the given problem. The given is that when the velocity of the man's speed is increased by one meter per second the kinetic energy is the same. We need to carefully apply the condition to get the value of the original speed of the person.
Recently Updated Pages
Locomotion occurs in Earthworm with the help of aSetae class 11 biology CBSE

A Velocitytime graph for a body of mass 10 kg is shown class 11 physics CBSE

The IUPAC name of the given structure is A3 methyl class 11 chemistry CBSE

Following are the runs scored by two batsmen in 5 cricket class 11 statistics CBSE

A group consists of 4 girls and 7 boys In how many-class-11-maths-CBSE

Calculate the number of electrons constituting one class 11 physics CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

