Question

A body of mass $2\,kg$ is projected from the ground with a velocity $20\,m{s^{ - 1}}$ at an angle ${30^ \circ }$ with the vertical. If ${t_1}$ is the time in seconds at which the body is projected and ${t_2}$ is the time in seconds at which it reaches the ground, the change in momentum in $kgm{s^{ - 1}}$ during the time ${t_2} - {t_1}$ is:
(A) $40$
(B) $40\sqrt 3$
(C) $50\sqrt 3$
(D) $60$

Answer 1

Hint The change in the momentum of the body can be determined by the sum of the momentum of the body of the mass when it moves vertically upwards with some angle and the momentum of the body of mass when it moves downwards.

Useful formula
The momentum of the object is given by,
$P = mv$
Where, $P$ is the momentum of the body, $m$ is the mass of the body and $v$ is the velocity of the body.

Complete step by step solution
Given that,
The mass of the body is given as, $m = 2\,kg$,
The velocity of the body is, $v = 20\,m{s^{ - 1}}$,
The body is projected at the angle with vertical is, $\theta = {30^ \circ }$.
Now, the initial momentum of the body of the mass is given as,
${P_1} = mv\cos \theta \,..................\left( 1 \right)$
By substituting the angle value in the above equation, then
${P_1} = mv\cos {30^ \circ }$
The value of the $\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$, then the above equation is written as,
${P_1} = \dfrac{{mv \times \sqrt 3 }}{2}\,...............\left( 2 \right)$
Now, the final momentum of the body of the mass is given as,
${P_2} = mv\cos \theta \,..................\left( 3 \right)$
By substituting the angle value in the above equation, then
${P_2} = mv\cos {30^ \circ }$
The value of the $\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$, then the above equation is written as,
${P_2} = \dfrac{{mv \times \sqrt 3 }}{2}\,...............\left( 4 \right)$
The change in the momentum is the sum of the two momentum,
$\Delta P = {P_2} + {P_1}$
Where, $\Delta P$ is the change in the momentum.
By substituting the equation (2) and equation (4) in the above equation, then
$\Delta P = \dfrac{{mv \times \sqrt 3 }}{2} + \dfrac{{mv \times \sqrt 3 }}{2}$
By adding the terms in the above equation, then the above equation is written as,
$\Delta P = \dfrac{{2\left( {mv \times \sqrt 3 } \right)}}{2}$
By cancelling the terms in the above equation, then the above equation is written as,
$\Delta P = mv \times \sqrt 3$
By substituting the mass of the body and the velocity of the object in the above equation, then
$\Delta P = 2 \times 20 \times \sqrt 3$
By multiplying the terms in the above equation, then
$\Delta P = 40\sqrt 3 \,kgm{s^{ - 1}}$

Hence, the option (B) is the correct answer.

Note The change in the momentum of the object is the subtraction of the two momentum but here the two momentum is added because, one momentum is moving upwards and the other momentum is moving downwards, so this direction change leads to the change in the sign of the momentum. So, both the momentums are added.