Motion in a Plane

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New answer posted•a year ago

Motion in a Plane Physics NCERT Exemplar Solutions Class 11th Chapter Four

A, B and C are three non-collinear, non co-planar vectors. What can you say about direction of A × (B × C)?

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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.

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New answer posted•a year ago

Motion in a Plane Physics NCERT Exemplar Solutions Class 11th Chapter Four

A football is kicked into the air vertically upwards. What is its (a) acceleration, and (b) velocity at the highest point?

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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 a

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New answer posted•a year ago

Motion in a Plane Physics NCERT Exemplar Solutions Class 11th Chapter Four

A ball is thrown from a roof top at an angle of 45° above the horizontal. It hits the ground a few seconds later. At what point during its motion, does the ball have

(a) Greatest speed.

(b) Smallest speed.

(c) Greatest acceleration? Explain

9 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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 alwa

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New answer posted•a year ago

Motion in a Plane Physics NCERT Exemplar Solutions Class 11th Chapter Four

A particle is projected in air at some angle to the horizontal, moves along parabola as shown in Fig., where x and y indicate horizontal and vertical directions, respectively. Show in the diagram, direction of velocity and acceleration at points A, B and C.

7 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

Explanation- vsin $θ$ = vertical component

V_x= horizontal component of velocity =vcos $θ$ = constant

V_y = vertical component of velocity =vsin $θ$

Velocity will always be tangential to the curve in the direction of motion and acceleration is a

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New answer posted•a year ago

Motion in a Plane Physics NCERT Exemplar Solutions Class 11th Chapter Four

A cyclist starts from centre O of a circular park of radius 1km and moves along the path OPRQO as shown Fig. 4.3. If he maintains constant speed of 10ms–1, what is his acceleration at point R in magnitude and direction?

6 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

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New answer posted•a year ago

Motion in a Plane Physics NCERT Exemplar Solutions Class 11th Chapter Four

A man wants to reach from A to the opposite corner of the square C (Fig. 4.10). The sides of the square are 100 m. A central square of 50m × 50m is filled with sand. Outside this square, he can walk at a speed 1 m/s. In the central square, he can walk only at a speed of v m/s (v < 1). What is smallest value of v for which he can reach faster via a straight path through the sand than any path in the square outside the sand?

5 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

Explanation – T_sand= $\frac{A P + Q C}{1} + \frac{P Q}{v} = \frac{25 \sqrt[]{2} + 25 \sqrt[]{2}}{1} + \frac{50 \sqrt[]{2}}{v}$

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

Time taken T_outside= $\frac{A R + R C}{1} s$

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

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New answer posted•a year ago

Motion in a Plane Physics NCERT Exemplar Solutions Class 11th Chapter Four

Motion in two dimensions, in a plane can be studied by expressing position, velocity and acceleration as vectors in Cartesian co-ordinates where are unit vector along x and y directions, respectively and Ax and Ay are corresponding components of (Fig). Motion can also be studied by expressing vectors in circular polar co-ordinates as A=A_rr + $A_{θ} θ$ where ? =r/r=cos $θ i + s i n θ j$ and $θ = - s i n θ i + c o s θ j$ are unit vectors along direction in which ‘r’ and ‘q ’ are increasing.

(a) Express in terms of q.

(b) Show that both q are unit vectors and are perpendicular to each other.

(c) Show that d(r)/dt , where w =dq/dt q and dq/dt = -wr

(d) For a particle moving along a spiral given by r=aqr , where a = 1 (unit), find dimensions of ‘a’.

(e) Find velocity and acceleration in polar vector representation for particle moving along spiral described in (d) above.

12 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

Explanation – r=cos $θ ? + s i n θ$ ?…….1

$θ = - s i n θ ? + c o s θ ?$ …….2

Multiplying eq1 by sin $θ$ and 2 with cos $θ$ and adding

Rsin $θ + θ c o s θ = s i n θ c o s θ ? + {s i n}^{2} θ j + {c o s}^{2} θ ? - s i n θ c o s θ ?$

= ?( ${c o s}^{2} θ + {s i n}^{2} θ$ )=j

= rsin $θ + θ c o s θ = j$

n(rcos $θ - θ s i n θ$ )=i

b)r $θ = c o s θ ? + s i n θ ? (- s i n θ ? + c o s θ ?)$

= -cos $θ s i n θ + s i n θ . c o s θ = 0$ $θ = 90$

c)r=cos $θ ? + s i n θ ?$

dr/dt=d/dt(cos $θ ? + s i n θ ?$ )=w[-cos $θ ? + s i n θ ?$ ]

d)L= M^oLT⁰

e)a=1unit , r= $θ r = θ [c o s θ ? + s i n θ ?]$

v= dr/dt= $\frac{d θ}{d t} r + θ \frac{d}{d t} [c o s θ ? + s i n θ ?]$

v= $\frac{d θ}{d r} r + θ [- c o s θ ? + s i n θ ?] \frac{d θ}{d t}$

= $\frac{d θ}{d t} r + θ w θ = w r +$ w $θ θ$

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New answer posted•a year ago

Motion in a Plane Physics NCERT Exemplar Solutions Class 11th Chapter Four

A cricket fielder can throw the cricket ball with a speed v_o . If he throws the ball while running with speed u at an angle θ to the horizontal, find

(a) The effective angle to the horizontal at which the ball is projected in air as seen by a spectator.

(b) What will be time of flight?

(c) What is the distance (horizontal range) from the point of projection at which the ball will land ?

(d) Find q at which he should throw the ball that would maximise the horizontal range as found in (iii).

(e) How does q for maximum range change if u >v_o , u = v_o , u < v_o?

(f) How does q in (v) compare with that for u = 0 (i.e.45⁰) ?

8 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

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

u_y=velocity in y direction= v₀sin $θ$

now tan $θ = \frac{u_{y}}{u_{x}} = \frac{u_{o} s i n θ}{u + u_{o} c o s θ}$

$θ = {t a n}^{- 1} \frac{u_{o} s i n θ}{u + u_{o} c o s θ}$

b) let t be the time flight y =0 u_y=v_osin $θ, a_{y} = - g, t = T$

y= u_yt+1/2 a_yt²

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

So T = $\frac{2 u_{o} s i n θ}{g}$

c) horizontal range R, = (u+v_ocos $θ$ T= (u+v_ocos $θ$ ) $\frac{2 u_{o} s i n θ}{g}$

d) for range

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New answer posted•a year ago

Motion in a Plane Physics NCERT Exemplar Solutions Class 11th Chapter Four

A river is flowing due east with a speed 3m/s. A swimmer can swim in still water at a speed of 4 m/s .

(a) If swimmer starts swimming due north, what will be his resultant velocity (magnitude and direction)?

(b) If he wants to start from point A on south bank and reach opposite point B on north bank,

(a) Which direction should he swim?

(b) What will be his resultant speed?

(c) From two different cases as mentioned in (a) and (b) above, in which case will he reach opposite bank in shorter time?

13 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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= $\sqrt[]{v_{r}^{2} + v_{s}^{2}} = \sqrt[]{3^{2} + 4^{2}} = 5 m / s$

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 $θ = \frac{v_{r}}{v} = \frac{3}{\sqrt[]{7}}$

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New answer posted•a year ago

Motion in a Plane Physics NCERT Exemplar Solutions Class 11th Chapter Four

A girl riding a bicycle with a speed of 5 m/s towards north direction, observes rain falling vertically down. If she increases her speed to 10 m/s, rain appears to meet her at 45° to the vertical. What is the speed of the rain? In what direction does rain fall as observed by a ground based observer?

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

Explanation – V_r= a? +b?

Velocity v_g= 5m/s

Velocity of rain w.r.t girl = V_r-V_g= a? +b? -5?

= (a-5)? +b?

a-5=0, a=5

case II

v_g = 10m/s?

V_rg= V_r- V_g

= a? +b? -10? = (a-10)? +b?

Rain appear to be fall at 45 degree so = b/a-10 =

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