Question

A body of mass 5Kg is moving with velocity of $v=(2\hat{i}+6\hat{j})m/s$ at t = 0 sec. After time t=2 sec, velocity of body $v=(10\hat{i}+6\hat{j})m/s$ then change in momentum of body is
1.$40\hat{i}$kg m/s
2.$20\hat{i}$kg m/s
3.$30\hat{i}$kg m/s
4.$(50\hat{i}+30\hat{j})$kg m/s

Answer 1

This problem is based on two dimensional motion in x and y direction, to calculate change in momentum of body we need to find change in velocity in both direction and by multiplying with mass of the body so we can get change in momentum of body with direction.
Formula used:
Momentum can be defined as product of mass of a body and its velocity, it is a type of vector with both magnitude and direction
Change in momentum
$\Rightarrow \Delta p=M\times ({{\vec{V}}_{F}}-{{\vec{V}}_{I}})$

Complete answer:
Given, initial velocity${{\vec{V}}_{I}}=(2\hat{i}+6\hat{j})m/s$, which means the component of velocity on x axis is 2 m/s and on y axis is 6 m/s.
Final velocity${{\vec{V}}_{F}}=(10\hat{i}+6\hat{j})m/s$, similarly on x axis is 10 m/s and on y axis 6 m/s.
Mass of object is M=5 kg
Change in momentum is given by,
$\Rightarrow \Delta p=M\times ({{\vec{V}}_{F}}-{{\vec{V}}_{I}})$
$\Rightarrow \Delta p=5\times \left( (10\hat{i}+6\hat{j})-(2\hat{i}+6\hat{j}) \right)$
$\Rightarrow p=5\times \left( 8\hat{i} \right)=40\hat{i}$kg m/s
$\therefore$ the change in momentum of a body will be in x direction and equals to $40\hat{i}$kg m/s .

Therefore, option (1) is correct.

Additional information:
The rate of change of momentum with respect to time is proportional to the force (can be derived from the second law of motion) applied on a body and takes place in the direction of force itself. If the net force applied on the body is zero then change in momentum will eventually be zero.

Note:
In the above problem we need to consider the direction of velocities, so to calculate change in momentum we used a vector form in our required formula. Change in momentum is proportional to force so the unit can be given as Kg m/s or $[ML{{T}^{-1}}]$.