Question

Two bodies with same mass moving with velocities with the ratio 1: 3. Their kinetic energy is in the ratio

a) 1 : 3

b) 1 : 6

c) 1 : 9

d) none of these

I don't want the process I want only the solution.


please tell me this answer it's my request , u can take ur time but please don't skip and tell me the answer of this question.​

Answer 1

Answer:

The ratio of their kinetic energies is 1 : 9.

Explanation:

Given that,

• Two bodies with same mass.
• It's velocities with the ratio 1:3.

Let's first body be A and second body be B.

• Velocity of body A is 1 m/s.
• Velocity of body B is 3 m/s.

As we know that,

$\green \bigstar \: \boxed{\sf \red{ K.E = \frac{1}{2} mv {}^{2} }}$

According to Question,

$\longrightarrow\sf \: \frac{K.E_A }{K.E_B } = \frac{ \frac{1}{2}m_A(v_A) {}^{2} }{ \frac{1}{2} m_B(v_B) {}^{2} } \\ \\ \longrightarrow\sf \: \frac{K.E_A }{K.E_B } = \frac{ \cancel{\frac{1}{2}}m_A(v_A) {}^{2} }{ \cancel{\frac{1}{2}} m_B(v_B) {}^{2} } \\ \\ \longrightarrow\sf \: \frac{K.E_A }{K.E_B } = \frac{ m_A(v_A) {}^{2} }{m_B(v_B) {}^{2} } \\ \\ \longrightarrow\sf \: \frac{K.E_A }{K.E_B } = \frac{ \cancel{m_A } \times (1) {}^{2} }{ \cancel{m_A} \times (3) {}^{2} } \\ \\ \longrightarrow\sf \: \frac{K.E_A }{K.E_B } = \frac{ 1 }{ 9 }.$

${ \boxed{ \sf{ \pink{K.E_A : K.E_B = 1 : 9}}}} \: \green\bigstar$

Answer 2

Two bodies are having same mass . Let it be ' m '

Ratio of their velocities = 1 : 3

Kinetic Energy : It is given by half of product of mass and square of velocity

$\pink{\bigstar}\ \; \sf K.E=\dfrac{1}{2}mv^2$

____________________

$\sf \dfrac{K.E_1}{K.E_2}=\dfrac{\frac{1}{2}m_1v_1^2}{\frac{1}{2}m_2v_2^2}\\\\\to \sf \dfrac{K.E_1}{K.E_2}=\dfrac{mv_1^2}{mv_2^2}\\\\\to \sf \dfrac{K.E_1}{K.E_2}=\dfrac{v_1^2}{v_2^2}\\\\\to \sf \dfrac{K.E_1}{K.E_2}=\bigg(\dfrac{v_1}{v_2}\bigg)^2\\\\\to \sf \dfrac{K.E_1}{K.E_2}=\bigg(\dfrac{1}{3}\bigg)^2\\\\\to \sf \dfrac{K.E_1}{K.E_2}=\dfrac{1}{9}\ \; \orange{\bigstar}$

Option c

More Info :

Mechanical Energy :

It is given by the sum of both Potential energy and kinetic energy .

$\sf \bigstar\ \; M.E=P.E+K.E$