Question

Consider an object moving of mass M moving with velocity V. calculate the ratio of momentum when :
(a) Velocity of an object is doubled
(b) Mass of an object is halved.

(c) both mass and velocity are increased by three times.

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Answer 1

To Find :

The Ratio of Momentum , when :

• Velocity of an object is doubled.

• Mass of an object is halved.

• Both Mass and velocity are increased by three times.

Given :

• Mass of the Body = M

• Velocity = V

We Know :

Momentum :

$\sf{\underline{\boxed{\overrightarrow{p} = m \times \overrightarrow{v}}}}$

Where ,

• p = Momentum.

• m = Mass of the body.

• v = Velocity of the Body.

Solution :

(1) When Velocity of the object is doubled :

Original Momentum :

Given :

• Mass = M
• Velocity = V

Thus , Momentum is :

$\sf{p = m \times v} \\ \\ \implies \sf{p = M \times V}$.

Hence , Orginal momentum is $\sf{p = M \times V}$

Momentum when velocity is halved .

Let the new momentum be p'.

Given :

• Mass = M

• Velocity = 2V

Substituting the values in the formula , we get :

$\sf{p' = m \times v} \\ \\ \implies \sf{p' = M \times 2V}$

Hence , the new Momentum is $\sf{p' = M \times 2V}$.

Ratio of the Momentum :

$\implies \sf{p' : p} \\ \\ \implies \sf{\dfrac{p'}{p}}$

Putting the value of p and p' in the Equation ,we get :

$\implies \sf{\dfrac{M \times 2V}{M \times V}} \\ \\ \\ \implies \sf{\dfrac{\not{M} \times 2\not{V}}{\not{M} \times \not{V}}} \\ \\ \\ \implies \sf{\dfrac{2}{1}} \\ \\ \implies \sf{2 : 1}$

Hence the ratio is 2 : 1.

(2) When Mass of the object is halved :

Orginal Momentum = $\sf{p = M \times V}$

Momentum When Mass is halved :

Let the new momentum be p'.

Given :

• Velocity = V

• Mass = $\dfrac{M}{2}$

Substituting the values in the formula , we get :

$\sf{p' = m \times v} \\ \\ \implies \sf{p' = \dfrac{M}{2} \times V}$

Hence , the new Momentum is $\sf{p' = M \times \dfrac{M}{2} \times V}$

Ratio of the Momentum :

$\implies \sf{p' : p} \\ \\ \implies \sf{\dfrac{p'}{p}}$

Putting the value of p and p' in the Equation ,we get :

$\implies \sf{\dfrac{\dfrac{M}{2} \times V}{M \times V}} \\ \\ \\ \implies \sf{\dfrac{\dfrac{\not{M}}{2} \times \not{V}}{\not{M} \times \not{V}}} \\ \\ \\ \implies \sf{\dfrac{1}{2}} \\ \\ \implies \sf{1 : 2}$

Hence the ratio is 1 : 2.

(3) When both Mass and velocity are increased by three times.

Orginal Momentum = $\sf{p = M \times V}$

Momentum when mass and Velocity is tripled :

Let the new momentum be p'.

• Mass = 3M

• Velocity = 3V

Substituting the values in the formula , we get :

$\sf{p' = m \times v} \\ \\ \implies \sf{p' = 3M \times 3V}$

Hence the new Momentum is $\sf{p' = 3M \times 3V}$

Ratio of the momentum :

$\implies \sf{p' : p} \\ \\ \implies \sf{\dfrac{p'}{p}}$

Putting the value of p and p' in the Equation ,we get :

$\implies \sf{\dfrac{3M \times 3V}{M \times V}} \\ \\ \\ \implies \sf{\dfrac{3\not{M} \times 3\not{V}}{\not{M} \times \not{V}}} \\ \\ \\ \implies \sf{\dfrac{9}{1}} \\ \\ \implies \sf{9 : 1}$

Hence the ratio is 9 : 1.

Answer 2

GIVEN :-

Mass of the body is M

Velocity of the body is V

TO FIND :-

Ratio of the Momentum when ,

Velocity of the object is doubled

Mass of the object is halved

Both Mass and velocity increased by three times

SOLUTION :-

Ratio = 1:9