Question

If the mass of the body is changed to '9' times, then what should be the change in velocity such that its K.E.
remain same.​

Answer 1

Answer:

Three times of change in velocity

Explanation:

Check the image.

Answer 2

Answer:

The change in velocity such that its K.E. remains the same 2v₁/3.

Explanation:

The kinetic energy of a body is given as,

$K.E.=\frac{1}{2}mv^2$             (1)

K.E.=kinetic energy

m=mass of the body

v=velocity of the body with which it is moving

Now as per the question,

m₁=m

m₂=9m

The kinetic energy of the first body,

$K.E._1=\frac{1}{2}m_1v^2_1$            (2)

The kinetic energy of the second body,

$K.E._2=\frac{1}{2}m_2v^2_2$           (3)

∴  K.E. remains the same    (given in question)

Now,

$\frac{1}{2}m_1v^2_1=\frac{1}{2}m_2v^2_2$

$\frac{v^2_1}{v^2_2}=\frac{m_2}{m_1}$

$\frac{v_1}{v_2}=\sqrt{\frac{m_2}{m_1}}$                  (4)

By substituting the value of m₁ and m₂ in equation (4) we get;

$\frac{v_1}{v_2}=\sqrt{\frac{9m}{m}}=\frac{3}{1}$

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

Now change in velocity is given as,

$\Delta v=v_1-v_2=v_1-\frac{v_1}{3}=\frac{3v_1-v_1}{3}=\frac{2v_1}{3}$

Hence, the change in velocity such that its K.E. remains the same 2v₁/3.