Newton S First Law Of Motion

Given:
$%5Cvec%7BP%7D(t)%20%3D%20%5Ccos(kt)%20%5C%2C%20%5Chat%7Bi%7D%20-%20%5Csin(kt)%20%5C%2C%20%5Chat%7Bj%7D%2C%20%5Cquad%20%7C%5Cvec%7BP%7D%7C%20%3D%201.$
The linear momentum $%5Cvec%7BP%7D$ is given by:

$%5Cvec%7BP%7D%20%3D%20m%5Cvec%7Bv%7D%20%5Cimplies%20%5Chat%7BP%7D%20%3D%20%5Cvec%7Bv%7D%2C$

where $%5Chat%7BP%7D$ is the unit vector in the direction of $%5Cvec%7BP%7D$ .

Step 1: Calculating the Velocity Vector
Differentiating $%5Cvec%7BP%7D(t)$ with respect to time to find the velocity vector:

$%5Cvec%7Bv%7D%20%3D%20%5Cfrac%7Bd%5Cvec%7BP%7D%7D%7Bdt%7D%20%3D%20-k%5Csin(kt)%20%5C%2C%20%5Chat%7Bi%7D%20-%20k%5Ccos(kt)%20%5C%2C%20%5Chat%7Bj%7D.$

The acceleration vector $%5Cvec%7Ba%7D$ is given by:

$%5Cvec%7Ba%7D%20%3D%20%5Cfrac%7Bd%5Cvec%7Bv%7D%7D%7Bdt%7D%20%3D%20-k%5E2%5Ccos(kt)%20%5C%2C%20%5Chat%7Bi%7D%20%2B%20k%5E2%5Csin(kt)%20%5C%2C%20%5Chat%7Bj%7D.$

Step 2: Calculating the Angle Between $%5Cvec%7BF%7D$ and $%5Cvec%7BP%7D$
The force $%5Cvec%7BF%7D$ is given by Newton’s second law:

$%5Cvec%7BF%7D%20%3D%20m%5Cvec%7Ba%7D.$

To find the angle $%5Ctheta$ between $%5Cvec%7BF%7D$ and $%5Cvec%7BP%7D$ , we use the dot product:

$%5Ccos%5Ctheta%20%3D%20%5Cfrac%7B%5Cvec%7BF%7D%20%5Ccdot%20%5Cvec%7BP%7D%7D%7B%7C%5Cvec%7BF%7D%7C%7C%5Cvec%7BP%7D%7C%7D.$

Substituting the expressions for $%5Cvec%7BF%7D$ and $%5Cvec%7BP%7D$ :

$%5Cvec%7BF%7D%20%5Ccdot%20%5Cvec%7BP%7D%20%3D%20(-k%5E2%5Ccos(kt))%5Ccos(kt)%20%2B%20(k%5E2%5Csin(kt))%5Csin(kt)%20%3D%20-k%5E2(%5Ccos%5E2(kt)%20%2B%20%5Csin%5E2(kt))%20%3D%20-k%5E2.$

Since $%7C%5Cvec%7BF%7D%7C%20%3D%20k%5E2$ and $%7C%5Cvec%7BP%7D%7C%20%3D%201$ , we have:

$%5Ccos%5Ctheta%20%3D%20%5Cfrac%7B-k%5E2%7D%7Bk%5E2%7D%20%3D%20-1%20%5Cimplies%20%5Ctheta%20%3D%20%5Cfrac%7B%5Cpi%7D%7B2%7D.$

Therefore, the angle between $%5Cvec%7BF%7D$ and $%5Cvec%7BP%7D$ is $%5Cfrac%7B%5Cpi%7D%7B2%7D$ .

Force Acting on the Block:
The force $F$ acting on the block increases linearly with time $t$ .

We can write:
$F%20%3D%20kt$ where $k$ is a constant of proportionality.

Using Newton's Second Law to Find Acceleration:
According to Newton's second law, the acceleration $a$ of the block is given by:
$F%20%3D%20ma$
Substituting $F%20%3D%20kt$ :
$ma%20%3D%20kt%20%5Cimplies%20a%20%3D%20%5Cfrac%7Bkt%7D%7Bm%7D$
where $m$ is the mass of the block.

Relation Between Acceleration and Time:
From the equation $a%20%3D%20%5Cfrac%7Bkt%7D%7Bm%7D$ , we see that the acceleration $a$ is directly proportional to time $t$ .

This means that as time $t$ increases, the acceleration $a$ also increases linearly.

Conclusion:
The correct graph representing the acceleration of the block with time is a straight line passing through the origin, as shown in Option (2).

High-Yield Trend

Chapter Questions
2 MCQs

Official Solution

Official Solution

Newton S First Law Of Motion

High-Yield Trend

Chapter Questions 2 MCQs

Official Solution

Official Solution

Chapter Questions
2 MCQs