Question

two bodies of masses 2 gram and 4 gram having same kinetic energy and having their ratio of Linear Momentum as​

Answer 1

Answer: Ratio of their Linear Linear Momentum = √2/2

Explanation:

Let the mass and velocity of 1st body be m1 and v1

Let the mass and velocity of 2nd body be m2 and v2.

Given, Kinetic energy is same:

Hence,

(1/2) * m1 * (v1) ^2 = (1/2) * m2 * (v2)^2

Given,m1 = 2 grams and m2 = 4 grams

Hence, (v1 )^2 = 2(v2) ^2

=) (v1) ^2 / (v2)^2 = 2/1

=) v1 / v2 = √2/1 - Equation (1)

Now, ratio of their Linear Momentum is :

=) (m1 * v1) / (m2 * v2) = 2 * V1 / 4* v2

=) v1 /(2*v2) = ( 1/2)* v1 / v2

=) (1/2) * √2 (Substitution value of v1 / v2

from equation 1)

=) √2/2

Answer 2

Concept:

The kinetic energy of both the masses are equal.

Let v1 and v2 be the velocity of masses 2kg and 4kg respectively.

Formulas used:

$K. E. = \frac{1}{2} m {v}^{2}$

$p = m v$

Calculation:

As the kinetic energy of both the bodies are equal.

$\frac{1}{2} m { |v1| }^{2} = \frac{1}{2} m { |v2| }^{2}$

$\frac{1}{2} \times 2 \times { |v1| }^{2} = \frac{1}{2} \times 4 \times { |v2| }^{2}$

${ (\frac{v1}{v2} })^{2} = 2$

$\frac{v1}{v2} = \sqrt{2}$

Therefore, the ratio of the magnitude of their momentum is

$\frac{m \times v1}{m \times v2}$

$\frac{2 \times v1}{4 \times v2} = \sqrt{2} \times \frac{2}{4}$

Final Answer:

Therefore, the ratio of the magnitude of the momentum is √2:2.