Motion in a Plane

This is a Short Answer Type Question as classified in NCERT Exemplar

Explanation – the path of the ball observed by a boy standing on the footpath is parabolic. The horizontal speed of the ball is same as that of the car, therefore ball as well car travels equal horizontal distance, due to vertical speed, the ball follows parabolic path.

This is a Short Answer Type Question as classified in NCERT Exemplar

Explanation – the direction of (B $*C$ ) will be perpendicular the plane containing B and B by right hand rule. A $*(B*C)$ will lie in the plane of B and C and is perpendicular to vector A.

This is a Short Answer Type Question as classified in NCERT Exemplar

Explanation- if the football is kicked into the air vertically upwards. Acceleration of football will always vertically downwards and equal to acceleration due to gravity. But when it reaches the highest point its velocity is zero and acceleration is retarding.

This is a Short Answer Type Question as classified in NCERT Exemplar

Explanation- at point B it will gain the same speed u and after that speed increases and will be maximum just before reaching point c.

During journey from O to A speed decreases and will be maximum at point A and acceleration is always constant.

This is a Short Answer Type Question as classified in NCERT Exemplar

Explanation- vsin $\theta$ = vertical component

V_x= horizontal component of velocity =vcos $\theta$ = constant

V_y = vertical component of velocity =vsin $\theta$

Velocity will always be tangential to the curve in the direction of motion and acceleration is always vertically downward and is equal to g.

This is a Short Answer Type Question as classified in NCERT Exemplar

Explanation- the cyclist covers OPRQO path.

As we know whenever an object performing circular motion, acceleration is called centripetal acceleration and is always directed towards the centre.

so there will be centripetal acceleration a= v²/r

So a= 100/1km= 100/1000=0.1m/s² along RO.

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation – T_sand= $\frac{AP+QC}{1}+\frac{PQ}{v}=\frac{25\sqrt[]{2}+25\sqrt[]{2}}{1}+\frac{50\sqrt[]{2}}{v}$

 = 50 $\sqrt[]{2}+\frac{50\sqrt[]{2}}{v}=50\sqrt[]{2}\left(\frac{1}{v}+1\right)$

Time taken T_outside= $\frac{AR+RC}{1}s$

AR= $\sqrt[]{{75}^2+{25}^2}=25\sqrt[]{10}m$

RC= AR= $25\sqrt[]{10}m$

T_outside= 2AR= 50 $\sqrt[]{10}s$

T_sandoutside  . so After solving we get velocity is greater v= 0.81m/s

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation – r=cos $\theta ?+sin\theta$ ?…….1

$\theta =-sin\theta ?+cos\theta ?$ …….2

Multiplying eq1 by sin $\theta$  and 2 with cos $\theta$ and adding

Rsin $\theta +\theta cos\theta =sin\theta cos\theta ?+{sin}^2\theta j+{cos}^2\theta ?-sin\theta cos\theta ?$

= ?( ${cos}^2\theta +{sin}^2\theta$ )=j

= rsin $\theta +\theta cos\theta =j$

 n(rcos $\theta -\theta sin\theta$ )=i

b)r $\theta =cos\theta ?+sin\theta ?(-sin\theta ?+cos\theta ?)$

 = -cos $\theta sin\theta +sin\theta .cos\theta =0$ $\theta =90$

c)r=cos $\theta ?+sin\theta ?$

dr/dt=d/dt(cos $\theta ?+sin\theta ?$ )=w[-cos $\theta ?+sin\theta ?$ ]

d)L= M^oLT⁰

e)a=1unit , r= $\theta r=\theta [\mathrm{c}\mathrm{o}\mathrm{s}\theta ?+sin\theta ?]$

v= dr/dt= $\frac{d\theta }{dt}r+\theta \frac{d}{dt}[\mathrm{c}\mathrm{o}\mathrm{s}\theta ?+sin\theta ?]$

v= $\frac{d\theta }{dr}r+\theta [-\mathrm{c}\mathrm{o}\mathrm{s}\theta ?+sin\theta ?]\frac{d\theta }{dt}$

= $\frac{d\theta }{dt}r+\theta w\theta =wr+$ w $\theta \theta$

a= $\frac{d}{dt}\left[wr+w\theta \theta \right]=\frac{d}{dt}\left[\frac{d\theta }{dt}r+\frac{d\theta }{dt}\left(\theta \theta \right)\right]$

a= $\frac{d^2\theta }{d{t}^2}r+\frac{d\theta }{dt}$ $\frac{dr}{dt}+\frac{d^2\theta }{d{t}^2}\theta \theta +\frac{d\theta }{dt}\frac{d}{dt}\left(\theta \theta \right)$

= $\frac{d^2\theta }{d{t}^2}r+w^2\theta +\frac{d^2\theta }{d{t}^2}\theta \theta +w^2\theta +w^2\theta (-r)$

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation- a) for x direction u_x= u+v_ocos $\theta$

u_y=velocity in y direction= v₀sin $\theta$

now tan $\theta =\frac{u_y}{u_x}=\frac{u_{o}sin\theta }{u+u_{o}cos\theta }$

$\theta ={tan}^{-1}\frac{u_{o}sin\theta }{u+u_{o}cos\theta }$

b) let t be the time flight y =0 u_y=v_osin $\theta ,a_y=-g,t=T$

y= u_yt+1/2 a_yt²

0= v_osin $\theta T$ + $\frac{1}{2}\left(-g{T}^2\right)$

So T = $\frac{2u_{o}sin\theta }{g}$

c) horizontal range R, = (u+v_ocos $\theta$ T= (u+v_ocos $\theta$ ) $\frac{2u_{o}sin\theta }{g}$

d) for range to be maximum dR/d $\theta =0$

$\frac{v_o}{g}\left[2ucos\theta +v_{o}cos2\theta *2\right]=0$

$2ucos\theta +2v_o\left[2{cos}^2\theta -1\right]=0$

4v_ocos^{2 $\theta +2ucos\theta -2v_o=0$}

So cos $\theta$ = $\frac{-u?\sqrt[]{u^2+8v_o^2}}{4v_o}$

$\theta ={cos}^{-1}\left[\frac{-u\pm \sqrt[]{u^2+8v_o^2}}{4v_o}\right]$

e) cos $\theta$ = $\frac{-v_o?\sqrt[]{u^2+8v_o^2}}{4v_o}=\frac{-1+3}{4}=\frac{1}{2}$

so $\theta =60$

f) if u=0 ${\theta }_{max}={cos}^{-1}\left[\frac{-0\pm \sqrt[]{u^2+8v_o^2}}{4v_o}\right]={cos}^{-1}\left(\frac{1}{\sqrt[]{2}}\right)=45$ ⁰

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation – speed of river V_r= 3m/s

Speed of swimmer V_s= 4m/s

(a) when swimmer starts swimming due north then its resultant velocity

V= $\sqrt[]{v_r^2+v_s^2}=\sqrt[]{3^2+4^2}=5m/s$

tan so 'N

(b) to reach at point B resultant velocity will be

V= $\sqrt[]{v_s^2-v_r^2}=\sqrt[]{4^2-3^2}=\sqrt[]{7}m/s$

tan $\theta =\frac{v_r}{v}=\frac{3}{\sqrt[]{7}}$

(c) time taken by swimmer t =d/v= d/4s

in case b time taken by swimmer to cross the river

t₁=d/v=d/ $\sqrt[]{7}$

so t1