Name: Variable Mass System
Uploaded: 2019-07-19
Duration: 12 min 25 s
Description: In this video, we will look at how the laws of motion and their mathematical formulation apply to a variable mass system like a rocket. Don't miss it.

Question

A particle of mass

2 m

is projected at an angle of

45 º

with horizontal with a velocity of

20 \sqrt{2} m / s .

After

1 s

explosion takes place and the particle is broken into two equal pieces. As a result of explosion one part comes to rest. Find the maximum height attained from the ground by the other part. Take

g = 10 m / s^{2} .

Accepted Answer

Given,
Mass of particle, $2 m$
Angle of projection is, $θ = 45 °$
Projection velocity is, $v = 20 \sqrt{2} m s^{- 1}$
Acceleration due to gravity, $g = 10 m s^{- 2}$
Time after which explosion takes place is, $t = 1 s$

In vector form
Projection velocity is, $\vec{v} = v \cos (45 °) \hat{i} + v \sin (45 °) \hat{j} = 20 \hat{i} + 20 \hat{j}$
Acceleration is, $\vec{a} = - g \hat{j} = - 10 \hat{j}$

Using first equation of motion in vector form, we have $\vec{v} = \vec{u} + \vec{a} t$
At $t = 1 s$ , velocity will be,
$45 º$
${\vec{v}}_{1} = (20 \hat{i} + 10 \hat{j})$

Height achieved at this height is calculated by second equation of motion, $\vec{s} = \vec{u} t + \frac{1}{2} \vec{a} t^{2}$
$\Rightarrow \vec{s} = (20 \hat{i} + 20 \hat{j}) t + \frac{1}{2} (- 10 \hat{j}) t^{2}$
$\Rightarrow \vec{s} = (20 \times 1 \hat{i} + (20 - 5) \hat{j})$
$\Rightarrow \vec{s} = 20 \hat{i} + 15 \hat{j}$

At this place explosion takes place, so linear momentum must be conserved as during explosion net force on the particle will be zero.
$\Rightarrow 2 m {\vec{v}}_{1} = m {\vec{v}}_{2} + m \times \vec{0}$ , where ${\vec{v}}_{2}$ represents velocity of one half part also other half comes to rest as per given question.
$\Rightarrow {\vec{v}}_{2} = 2 (20 \hat{i} + 10 \hat{j}) = 40 \hat{i} + 20 \hat{j}$

Now for maximum height, we have to work on $y$ component only, so, on applying third equation of motion, $v^{2} = u^{2} + 2 a s$
$\Rightarrow 0 = {v_{2 y}}^{2} + 2 (- g) h_{2}$
$\Rightarrow 0 = 20^{2} - 20 h_{2}$
$\Rightarrow h_{2} = 20 m$

So maximum height from ground is, $H_{m a x} = h_{1} + h_{2} = 15 + 20 = 35 m$
Case is shown in below figure,

Important Questions on Centre of Mass, Momentum and Collisions

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