Question

A particle having a mass 0.5 kg is projected under gravity with a speed of 98m/s at an angle of 60$^{0}$. The magnitude of the change in momentum in kgm/s of the particle after 10s is.
A. 490
B. 98
C. 49
D. 0.5

Answer 1

Resolve the velocity of the particle into its horizontal and vertical components. Therefore, the particle will have some vertical and horizontal momentums. The force of gravity acts downwards. Therefore, only the vertical momentum will change. Calculate the change in momentum by using p=mv and kinematic equation $v-u=+at$.

Formula used:
p=mv
$v-u=+at$
$\Delta p=F\Delta t$

Complete step by step answer:
It is given that a particle is projected with a speed of u= 98$m{{s}^{-1}}$ at an angle of 60$^{0}$.
To make the things easy, resolve the initial velocity of the particle into its horizontal and vertical components. Let the horizontal component be ${{u}_{x}}$ and the vertical component be ${{u}_{y}}$.
Now, once the particle is released, it will be under the force of gravity acting in the downward direction. Due to the gravitational force, the particle will have an acceleration equal to the acceleration due gravity g. The direction of the acceleration is downwards.

Since, the particle is accelerating downwards only, the vertical velocity of the particle will change with time and the horizontal velocity will remain constant with time.

The particle will have some momentum in both directions. However, the vertical momentum of the particle will change with time whereas the horizontal momentum will remain constant since no force is acting in this direction.

Therefore, there will be a change in momentum only along the vertical direction.

Momentum of the particle is given as p=mv.
Let the particle’s initial vertical momentum be ${{p}_{y}}$ and its vertical velocity and momentum after 10s be ${{v}_{y}}$ and $p{{'}_{y}}$.
Therefore, the change in momentum of the particle is $\Delta p=p{{'}_{y}}-{{p}_{y}}=m{{v}_{y}}-m{{u}_{y}}$
$\Rightarrow \Delta p=m\left( {{v}_{y}}-{{u}_{y}} \right)$ … (i).

It is given that m=0.5kg . So we have to calculate ${{v}_{y}}-{{u}_{y}}$. For this we will use the kinematic equation $v-u=+at$.
Here, v=${{v}_{y}}$, u=${{u}_{y}}$, a=-g and t=10s
Therefore,

$\Rightarrow {{v}_{y}}-{{u}_{y}}=(-g)(10)$
Let us consider g=9.8$m{{s}^{-2}}$.
$\Rightarrow {{v}_{y}}-{{u}_{y}}=(-9.8)(10)=-98m{{s}^{-1}}$.
Substitute the values of m and ${{v}_{y}}-{{u}_{y}}$ in equation (i).
$\Rightarrow \Delta p=0.5\left( -98 \right)=-49kgm{{s}^{-1}}$
Therefore, the magnitude of change of momentum of the particle is $49kgm{{s}^{-1}}$.
Hence, the correct answer is option C.

Note:
Alternative method to solve the question.
We know that when a force acts on a particle, the momentum of the particle along the direction of the force changes with time.

If the force is constant, then the change in momentum is given as $\Delta p=F\Delta t$.
Since in the given case the gravitational force (${{F}_{g}}$) is acting downwards, only the vertical momentum of the particle will change.
We know ${{F}_{g}}=-mg$ and $\Delta t$ is given as 10s.
Therefore, $\Delta p=-mg(10)=-0.5\times 9.8\times 10=-49kgm{{s}^{-1}}$.
Hence, the magnitude of change in the momentum of the particle is $49kgm{{s}^{-1}}$.