Question

The change in momentum of a body is 50kg m/s in time 4 sec. what is the force acting on it?

Answer 1

In the above Question , the following information is given -

The change in momentum of a body is 50kg m/s in time 4 sec.

To find -

What is the force acting on the body ?

Solution -

[ ∆ Momentum ] / [ t ] = 50 kg m / s .

=> [ ∆ P ] / [ t ] = 50 kg m / s

=>[ P_Final - P_Initial ] / t = 50 kg m / s

=> [ m× v_final - m×v_initial ] / [ t ] = 50 kg m / s

=> [ m ] × { v_final - v_initial } / [ t ] = 50 kg m / s

=> m × a = 50 kg m / s [ Note that a = { v - u } / t ]

Now we know that -

Force acting on body -

=> Mass of the body × It's acceleration

=> 50 Newton

This is the required answer . .

_________________

Shortcut -

Notice carefully that 1 Newton = 1 kg m / s .

So , the change in Momentum is equal to the force applied on the body .

___________________

$\rule{150}{1.5}$

Answer 2

$\huge\sf\pink{Answer}$

☞ Force = 50 N

$\rule{110}1$

$\huge\sf\blue{Given}$

✭ Change in momentum of a body is 50 kg m/s

✭ Time = 4 seconds

$\rule{110}1$

$\huge\sf\gray{To \:Find}$

◈ The force acting

$\rule{110}1$

$\huge\sf\purple{Steps}$

➝ $\sf\bigg\lgroup\dfrac {\triangle Momentum}{ t}\bigg\rgroup= 50 kg m/s$

➝ $\sf\bigg\lgroup \dfrac{\triangle P}{ t} \bigg\rgroup= 50 kg m/s$

➝ $\sf\bigg\lgroup\dfrac{ P_{Final} - P_{Initial}}{t} \bigg\rgroup= 50 kg m/s$

➝ $\sf\bigg\lgroup \dfrac{m\times V_{Final} - m\times V_{Initial}}{t}\bigg\rgroup = 50 kg m/s$

➝ $\sf\dfrac{m(V_{Final} - V_{Initial})}{t} = 50 kg m/s$

➝ $\sf\red{m \times a = 50 kg m / s}$

$\bullet\underline{\textsf{As Per the Question}}$

$\underline{\boxed{\sf {Force = Mass\times Acceleration} }}$

≫ $\sf\orange{Force = 50 N}$

$\rule{170}3$