Motion in a Plane

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- b 

Explanation – u= a? + b? as u in the first quadrant, hence both components a and b will be positive. For v= p? + q? , as it is positive x direction and located downward hence x component p will be positive and y component q will be negative.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- d

Explanation- a scalar quantity is independent of direction hence has the same value for observers with different orientations of the axes.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- b

Explanation-A = i+j

 B = i-j

 A.B=|A|B|cos $\theta$

(? +? ). (? -? ) = $\left(\sqrt[]{1+1}\right)\left(\sqrt[]{1+1}\right)$ cos $\theta$

Where $\theta$  is the angle between A and B

Cos $\theta$ = $\frac{1-0+0-1}{\sqrt[]{2\sqrt[]{2}}}$ = $\frac{0}{2}=0$

$\theta$  = 90^o

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

Explanation- given |A|=2 and |B|=4

a)|A $*B$ |=AB cos $\theta$ = 0

so 2 $*4$ cos $\theta$ =0

so $\theta =90$  so it matches with ii

B) |A $*$ B|= ABcos $\theta$ =8

2 $*4$ cos $\theta$ =8

So $\theta$ = 0 so it matches with option i

c) |A $*$ B|= ABcos $\theta$ =4

so $\theta$ =60 so it matches with option iv

d) |A $*$ B|= ABcos $\theta$ =-8

so $\theta$ =180 so it matches with option iii

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

Explanation- given |A|=2 and |B|=4

a)|A $*B$ |=AB sin $\theta$ = 0

so 2 $*4$ sin $\theta$ =0

so $\theta =0$ so it matches with iv

B) |A $*$ B|= ABsin $\theta$ =8

2 $*4$ sin $\theta$ =8

So $\theta$ = 90 so it matches with option iii

c) |A $*$ B|= ABsin $\theta$ =4

so $\theta$ =30 so it matches with option i

d) |A $*$ B|= ABsin $\theta$ =4 $\sqrt[]{2}$

so $\theta$ =45 so it matches with option ii

Explanation – here A and B vectors are joint by head and tail. So C= A+B

(a) from fig iv it is clear that c=a+b

(a) from fig iii it is clear that c+b=a so a-c=b

(b) from fig I it is clear that b=a+c so b-a =c

(c) from ii it is clear that -c= a+b so a+b+c=0

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

Explanation – a) radius of earth =6400km= 6.4 $*{10}^{6}m$

Time period = 1 day = 24 $*60*60$ = 86400s

Centripetal acceleration a= w²r= R(2 $\pi /T$ )²=4 $\pi$ ²R/T

= $\frac{4*{\left(\frac{22}{7}\right)}^2*6.4*{10}^6}{({24*60*60)}^2}$ = 0.034m/s²

b) time = 1yr=365 $*24*60*60$ days= 365=3.15 $*{10}^{7}s$

centripetal acceleration = Rw²= $\frac{4{\pi }^{2}R}{T^2}$

^{$=\frac{4*{\frac{22}{7}}^2*1.5*{10}^{11}}{({3.15*{10}^7)}^2}=5.97*{10}^{-3}m/s$ 2}

$\frac{a_c}{g}=\frac{5.97*{10}^{-3}}{9.8}=\frac{1}{1642}$

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

Explanation- when it be at position P, drops a bomb to hit a target T

Let $\theta$

Speed of the plane =720km/h = 720 $*5/18$ = 200m/s

Altitude of the plane P'T = 1.5km= 1500m

If bomb hits the target after time t then horizontal distance travelled by the bomb PP'=u $*t$ =200t

Vertical distance travelled by the bomb P'T=1/2gt²

1500 = ½ (9.8)t²

So t²= 1500/4.9, t = $\sqrt[]{\frac{1500}{4.9}}=17.49s$

PP'=200 (17.49)m=

tan $\theta$ =P'T/P'P=1500/200 (17.49)=0.49287= tan23⁰12'

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

Explanation – due to air resistance particle energy as well as horizontal component of velocity keep on decreasing making the fall steeper then rise. When we are neglecting air resistance path is parabola when we consider air resistance then path is asymmetric parabola.

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

Explanation – the boy throws the ball at an angle of 60.

Horizontal component of velocity 4cos $\theta$ = 10cos60

=10 (1/2)

=5m/s.

so horizontal speed of the car is same, hence relative velocity of car and ball in the horizontal direction will be zero.