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Motion in a Plane Physics - Motion in a Plane

Rajasthan Board (RBSE) - STD 11 Science - Physics (Rajasthan - English Medium). 37 questions in this chapter.

Question types

Motion in a Plane question types

37 questions across 5 question groups — pick any mix to generate a Physics paper with step-by-step answer keys.

37
Questions
5
Question groups
5
Question types
Sample Questions

Motion in a Plane questions

One sample from each question group in this chapter. Select any group above to see the full set with answer keys.

Q 1M.C.Q (1 Marks)1 Mark
It is found that $|\text{A}+\text{B}|=|\text{A}|.$ This necessarily implies,
  1. $\text{B}=0$
  2. A, B are antiparallel.
  3. A, B are perpendicular.
  4. $\text{A.B}≤0$
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Q 2M.C.Q (1 Marks)1 Mark
For two vectors A and B,$ |\text{A} + \text{B}| = |\text{A} - \text{B}|$ is always true when

  1. $|\text{A}|=|\text{B}|\neq0$

  2. $\text{A}\bot\text{B}$

  3. $|\text{A}|=|\text{B}|\neq0$ and A and B are parallel or anti parallel

  4. when either $|\text{A}|$ or $|\text{B}|$ is zero.

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Q 3M.C.Q (1 Marks)1 Mark
Figure shows the orientation of two vectors u and v in the XY plane.

If $\text{u}=\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}$ and $\text{v}=\text{p}\hat{\text{i}}+\text{q}\hat{\text{j}}$ which of the following is correct?

  1. a and p are positive while b and q are negative.
  2. a, p and b are positive while q is negative.
  3. a, q and b are positive while p is negative.
  4. a, b, p and q are all positive.
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Q 4M.C.Q (1 Marks)1 Mark
The angle between $\vec{\text{A}}=\hat{\text{i}}+\hat{\text{j}}$ and $\vec{\text{B}}=\hat{\text{i}}-\hat{\text{j}}$ is
  1. 45°
  2. 90°
  3. –45°
  4. 180°
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Q 5M.C.Q (1 Marks)1 Mark
Following are four differrent relations about displacement, velocity and acceleration for the motion of a particle in general. Choose the incorrect one (s):

  1. $\text{v}_\text{av}=\frac{1}2\big[\text{v}(\text{t}_1)+\text{v}(\text{t}_2)\big]$

  2. $\text{v}_\text{av}=\frac{r(\text{t}_2)-\text{r}(\text{t}_1)}{\text{t}_2-\text{t}_1}$

  3. $\text{r}=\frac{1}2(\text{v}(\text{t}_2)-\text{v}(\text{t}_1))(\text{t}_2-\text{t}_1)$

  4. $\text{a}_\text{av}=\frac{\text{v}(\text{t}_2)-\text{v}(\text{t}_1)}{\text{t}_2-\text{t}_1}$

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Q 61 Marks Question1 Mark
Motion in two dimensions, in a plane can be studied by expressing position, velocity and acceleration as vectors in Cartesian co-ordinates $\text{A}=\text{A}_\text{x}\hat{\text{i}}+\text{A}_\text{y}\hat{\text{j}}$ where $\hat{\text{i}}$ and $\hat{\text{j}}$ are unit vector along x and y directions, respectively and Ax and Ay are corresponding components of A Fig. Motion can also be studied by expressing vectors in circular polar co-ordinates as $\text{A}=\text{A}_\text{r}\hat{\text{r}}+\text{A}_\theta\hat{\theta}$ where $\hat{\text{r}}=\frac{\text{r}}{\text{r}}=\cos\theta\hat{\text{i}}+\sin\theta\hat{\text{j}}$ and $\hat{\theta}=-\sin\theta\hat{\text{i}}+\cos\theta\hat{\text{j}}$ are unit vectors along direction in which ‘r’ and ‘$\theta$’ are increasing.
  1. Express $\hat{\text{i}}$ and $\hat{\text{j}}$ in terms of $\hat{\text{r}}$ and $\hat{\theta}$.
  2. Show that both $\hat{\text{r}}$ and $\hat{\theta}$ are unit vectors and are perpendicular to each other.
  3. Show that $\frac{\text{d}}{\text{dt}}(\hat{\text{r}})=\omega\hat{\theta}$ where $\omega=\frac{\text{d}\theta}{\text{dt}}$ and $\frac{\text{d}}{\text{dt}}(\hat{\text{r}})=-\omega\hat{\text{r}}$
  4. For a particle moving along a spiral given by $\text{r}=\text{a}\theta\hat{\text{r}}$ , where a = 1 (unit), find dimensions of ‘a’.
  5. Find velocity and acceleration in polar vector represention for particle moving along spiral described in (d) above.

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Q 71 Marks Question1 Mark
A cricket fielder can throw the cricket ball with a speed V0. If he throws the ball while running with speed u at an angle $\theta$ to the horizontal, find
  1. The effective angle to the horizontal at which the ball is projected in air as seen by a spectator.
  2. What will be time of flight?
  3. What is the distance (horizontal range) from the point of projection at which the ball will land?
  4. Find $\theta$ at which he should throw the ball that would maximise the horizontal range as found in (iii).
  5. how does $\theta$ for maximum range change if $\text{u}>\upsilon_\text{o},\text{u}=\upsilon_\text{o},\text{ u}<\upsilon_\text{o}$?
  6. How does $\theta$ in (v) compare with that for u = 0 (i.e.45° )?
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Q 81 Marks Question1 Mark
A particle is projected in air at an angle $\beta$ to a surface which itself is inclined at an angle $\alpha$ to the horizontal (Fig.).

  1. Find an expression of range on the plane surface (distance on the plane from the point of projection at which particle will hit the surface).
(Hint: This problem can be solved in two different ways:

  1. Point P at which particle hits the plane can be seen as intersection of its trajectory (parobola) and straight line. Remember particle is projected at an angle $(\alpha+\beta)$ w.r.t. horizontal.
  2. We can take x-direction along the plane and y-direction perpendicular to the plane. In that case resolve g (acceleration due to gravity) in two differrent components, gx along the plane and gy perpendicular to the plane. Now the problem can be solved as two independent motions in x and y directions respectively with time as a common parameter.)
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Q 91 Marks Question1 Mark
A gun can fire shells with maximum speed v0 and the maximum horizontal range that can be achieved is $\text{R}=\frac{\text{v}^2_0}{\text{g}}.$

If a target farther away by distance $\Delta\text{x}$ (beyond R) has to be hit with the same gun (Fig), show that it could be achieved by raising the gun to a height at least $\text{h}=\Delta\text{x}\Big[1+\frac{\Delta\text{x}}{\text{R}}\Big]$

(Hint: This problem can be approached in two different ways:

  1. Refer to the diagram: target T is at horizontal distance $\text{x}=\text{R}+\Delta\text{x}$ and below point of projection y = – h.
  2. From point P in the diagram: Projection at speed v0 at an angle $\theta$ below horizontal with height h and horizontal range $\Delta\text{x}$.)
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Q 101 Marks Question1 Mark
A river is flowing due east with a speed 3 m/ s. A swimmer can swim in still water at a speed of 4 m/ s Fig.

  1. If swimmer starts swimming due north, what will be his resultant velocity (magnitude and direction)?
  2. If he wants to start from point A on south bank and reach opposite point B on north bank,
  • which direction should he swim?
  • what will be his resultant speed?
  1. From two different cases as mentioned in (a) and (b) above, in which case will he reach opposite bank in shorter time?

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Q 122 Marks Questions2 Marks
A cyclist starts from centre O of a circular park of radius 1km and moves along the path OPRQO as shown Fig. If he maintains constant speed of 10ms–1, what is his acceleration at point R in magnitude and direction?

 
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Q 133 Marks Question3 Marks
Earth can be thought of as a sphere of radius 6400km. Any object (or a person) is performing circular motion around the axis of earth due to earth’s rotation (period 1 day). What is acceleration of object on the surface of the earth (at equator) towards its centre? what is it at latitude $\theta?$ How does these accelerations compare with g = 9.8m/ s2?
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Q 143 Marks Question3 Marks
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.

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Q 153 Marks Question3 Marks
In dealing with motion of projectile in air, we ignore effect of air resistance on motion. This gives trajectory as a parabola as you have studied. What would the trajectory look like if air resistance is included? Sketch such a trajectory and explain why you have drawn it that way.
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Q 163 Marks Question3 Marks
A boy travelling in an open car moving on a levelled road with constant speed tosses a ball vertically up in the air and catches it back. Sketch the motion of the ball as observed by a boy standing on the footpath. Give explanation to support your diagram.
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Q 173 Marks Question3 Marks
Earth also moves in circular orbit around sun once every year with on orbital radius of 1.5 × 1011m. What is the acceleration of earth (or any object on the surface of the earth) towards the centre of the sun? How does this acceleration compare with g = 9.8m/ s2?

$\Big(\text{Hint: acceleration}\frac{\text{V}^2}{\text{R}}=\frac{4\pi^2\text{R}}{\text{T}^2}\Big)$

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Q 185 Marks Questions5 Marks
A hill is 500m high. Supplies are to be sent across the hill using a canon that can hurl packets at a speed of 125m/ s over the hill. The canon is located at a distance of 800m from the foot of hill and can be moved on the ground at a speed of 2m/ s; so that its distance from the hill can be adjusted. What is the shortest time in which a packet can reach on the ground across the hill? Take g =10m/ s2.
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Q 195 Marks Questions5 Marks
A particle is projected in air at an angle $\beta$ to a surface which itself is inclined at an angle $\alpha$ to the horizontal (Fig.).

  1. Time of flight.
(Hint: This problem can be solved in two different ways:

  1. Point P at which particle hits the plane can be seen as intersection of its trajectory (parobola) and straight line. Remember particle is projected at an angle $(\alpha+\beta)$ w.r.t. horizontal.
  2. We can take x-direction along the plane and y-direction perpendicular to the plane. In that case resolve g (acceleration due to gravity) in two differrent components, gx along the plane and gy perpendicular to the plane. Now the problem can be solved as two independent motions in x and y directions respectively with time as a common parameter.)
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Q 205 Marks Questions5 Marks
A particle is projected in air at an angle $\beta$ to a surface which itself is inclined at an angle $\alpha$ to the horizontal (Fig.).

  1. β at which range will be maximum.
(Hint: This problem can be solved in two different ways:

  1. Point P at which particle hits the plane can be seen as intersection of its trajectory (parobola) and straight line. Remember particle is projected at an angle $(\alpha+\beta)$ w.r.t. horizontal.
  2. We can take x-direction along the plane and y-direction perpendicular to the plane. In that case resolve g (acceleration due to gravity) in two differrent components, gx along the plane and gy perpendicular to the plane. Now the problem can be solved as two independent motions in x and y directions respectively with time as a common parameter.)
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Q 215 Marks Questions5 Marks
A particle falling vertically from a height hits a plane surface inclined to horizontal at an angle $\theta$ with speed $\text{v}_0$ and rebounds elastically. Find the distance along the plane where if will hit second time.

(Hint:

  1. After rebound, particle still has speed Vo to start.
  2. Work out angle particle speed has with horizontal after it rebounds.
  3. Rest is similar to if particle is projected up the incline.
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Q 225 Marks Questions5 Marks
A man wants to reach from A to the opposite corner of the square C Fig. The sides of the square are 100m. A central square of 50m × 50m is filled with sand. Outside this square, he can walk at a speed 1m/ s. In the central square, he can walk only at a speed of $\upsilon\text{m/ s}(\upsilon<1)$. 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?

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