Question

two boys having masses in the ratio 3:2 run on a level road with equal linear momentum. if the kinetic energy of heavier boy is 700 joule then kinetic energy of lighter boy is?

Answer 1

Answer:

Kinetic energy of the lighter boy will be 1050 J

Explanation:

We have given the ratio of the mass of the r=two boys is 3 :2

Let the mass of the first boy is $m_1\ and\ velocity\ is\  v_1$

And mass of second boy is $m_2\ and\ velocity\ is\ v_2$

So $\frac{m_1}{m_2}=\frac{3}{2}$

We have given that they that they have same momentum

We know that momentum is given by P = m v

So $\frac{P_1}{P_2}=\frac{m_1v_1}{m_2v_2}$

As $P_1=P_2$ and  $\frac{m_1}{m_2}=\frac{3}{2}$

So $\frac{1}{1}=\frac{3v_1}{2v_2}$

$\frac{v_1}{v_2}=\frac{2}{3}$

Now we know that kinetic energy is given by $Ke=\frac{1}{2}mv^2$

So $\frac{Ke_1}{Ke_2}=\frac{\frac{1}{2}m_1v_1^2}{\frac{1}{2}m_2v_2^2}$

$\frac{700}{Ke_2}=\frac{\frac{1}{2}\times 3\times 2^2}{\frac{1}{2}\times 2\times 3^2}$

$\frac{700}{Ke_2}=\frac{12}{18}$

$Ke_2=1050J$