Motion in a Plane

 $x=4sin\left(\frac{\pi }{2}-\omega t\right)$ - (i)

$y=4sin\left(\omega t\right)$ - (ii)

From (i) and (ii) cos2 $\omega t+si{n}^2\omega t={\left(\frac{x}{4}\right)}^2+{\left(\frac{y}{4}\right)}^2=1$

$\Rightarrow x^2+y^2={\left(4\right)}^2$

 $E=\frac{1}{2}m{u}^2$

$E_{Highestpoint}=\frac{1}{2}m{\left(\frac{u}{2}\right)}^2=\frac{1}{8}m{u}^2=\frac{E}{4}$

Linear motion is one-dimensional motion. It refers to motion in a single direction or in a straight line. In linear motion, the object either moves forward or backward along one axis, i.e. x-axis. For example - a ball dropped from a height vertically downward or a car moving straight on a road. Motion in a Plane refers to an object moving in two dimensions, usually along x and y axes. For example, a football kicked at an angle.

The angle of projection is used to find the trajectory, horizontal range of a projectile, maximum height, and time of flight. For example, the maximum range on level ground is given by the 45-degree angle.

The vectors like velocity, displacement, and acceleration act along different directions in the two-dimensional motion. Resolving the vectors into the vertical and horizontal components allows the application of one-dimensional kinematic equations in each direction separately. It helps solve the problems more accurately and also simplifies the analysis.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Explanation- |A+B|=|A-B|

$\sqrt[]{\left|{A|}^2+\left|{B|}^2+2\right|A\right|\left|B\right|cos\theta }=\sqrt[]{\left|{A|}^2+\left|{B|}^2-2\right|A\right|\left|B\right|cos\theta }$

$\left|{A|}^2+\left|{B|}^2+2\right|A\right|\left|B\right|cos\theta$  = $\left|{A|}^2+\left|{B|}^2-2\right|A\right|\left|B\right|cos\theta$

4|A|B|cos $\theta$ =0

|A|²+|B|²cos $\theta$ =0

A=0 or B=0 so $\theta =90$ . so A perpendicular B

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a, b, c

Explanation- (i) speed will constant throughout

(ii) velocity will be tangential in the direction of motion

(iii) centripetal acceleration will be a= v²/r, will always be towards centre of the circular path.

(iv) angular momentum is constant in magnitude and direction out of the plane perpendicularly as well.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a, c

Explanation – as we know average acceleration is a_av= $\frac{?v}{?t}=\frac{v_2-v_1}{t_2-t_1}$

But when acceleration is not uniform V_av is not equal to v₁+v₂/2

So we can write $?v=\frac{?r}{?t}$

$?r=r_2-r_1$ = v₂-v₁ (t₂-t₁)

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- c

Explanation– as the given track y=x² is a frictionless track thus total energy will be same throughout the journey.

Hence total energy at A = total energy at P . at B the particle is having only Ke but at P some KE is converted to P

Hence (KE)_B = (KE)_P

Total energy at A = PE= total energy at B = KE= total energy at P

= PE+KE

Potential energy at A is converted to KE and PE at P hence

(PE)P< (PE)A

Hence (height)P= (height)A

As height of p < height of A

Hence path length AB > path length BP