NCERT Exemplar Solutions Class 11 Physics Motion in a Plane — Very Short and Short Answer Questions Q16–28

NCERT Exemplar Problems, Class 11 Physics — Motion in a Plane, Very Short Answer Questions 16 to 20 and Short Answer Questions 21 to 28. All 13 questions solved. The book numbers them 4.16 to 4.28; it calls this Unit 4, while the rationalised textbook numbers the chapter 3.

Three ideas carry this page. A projectile has one acceleration: $\mathbf{g}$, downward, at every point of its flight, while its velocity is always along the tangent to the path. Motion depends on the observer: a ball tossed straight up in a moving car is a projectile to someone on the road, and a projectile can look like straight up-and-down motion to someone moving with its horizontal velocity. Circular motion needs a centripetal acceleration $v^2/r=4\pi^2r/T^2$ towards the centre, whose size explains why we do not feel the Earth’s spin.

Key insight. Split every motion into a part that is the same for all observers (the acceleration $\mathbf{g}$) and a part that depends on the observer (the horizontal velocity). Questions 18 to 24 all turn on keeping those two apart.

Question 16

A cyclist starts from centre O of a circular park of radius $1$ km and moves along the path OPRQO as shown in Fig. 4.3. If he maintains constant speed of $10\ \text{m s}^{-1}$, what is his acceleration at point R in magnitude and direction?

The figure shows the circle with centre O. P is on the circle to the right of O and Q at the top; R is on the arc between P and Q. The cyclist goes straight out from O to P, along the arc through R to Q, and straight back down from Q to O.

Solution. At R the cyclist is on the circular arc, moving at constant speed. With constant speed there is no tangential acceleration; the only acceleration is the centripetal one that keeps turning his velocity, directed from R towards the centre O: $$a=\frac{v^2}{r}=\frac{(10)^2}{1000}=0.1\ \text{m s}^{-2}.$$

Question 17

A particle is projected in air at some angle to the horizontal, moves along parabola as shown in Fig. 4.4, 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.

Solution. The velocity at any point is along the tangent to the path, in the direction of motion. At A, on the way up, it points up and forward. At B, the highest point, the vertical velocity is zero, so the velocity is purely horizontal. At C, on the way down, it points forward and downward. The acceleration is the same at all three points: $\mathbf{g}$, vertically downward, because gravity is the only force acting.

Question 18

A ball is thrown from a roof top at an angle of $45^\circ$ 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.

Solution. The horizontal velocity $u\cos45^\circ$ never changes; only the vertical velocity does, so the speed is $\sqrt{u_x^2+v_y^2}$ and is decided by $|v_y|$.

• (a) The ball ends up below its starting point, so it has fallen furthest just before it hits the ground. There $|v_y|$ is largest (energy conservation says the same: the most potential energy has been converted), so the speed is greatest just before it hits the ground.
• (b) At the highest point $v_y=0$, so the speed is just $u\cos45^\circ$, the smallest it ever is.
• (c) The acceleration is $g$, downward, throughout the flight, so it is the same at every point; there is no point of greatest acceleration.

Question 19

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

Solution. (a) Once the ball has left the foot, the only force on it (ignoring air) is its weight, so its acceleration is $g$, directed downward, at every point including the top. (b) At the highest point the ball momentarily stops rising before it starts to fall, so its velocity there is zero. A zero velocity with a non-zero acceleration is exactly what lets it turn round.

Question 20

$\mathbf{A}$, $\mathbf{B}$ and $\mathbf{C}$ are three non-collinear, non co-planar vectors. What can you say about direction of $\mathbf{A}\times(\mathbf{B}\times\mathbf{C})$?

Solution. $\mathbf{B}\times\mathbf{C}$ is perpendicular to the plane containing $\mathbf{B}$ and $\mathbf{C}$. Any vector crossed with it is perpendicular to it, so $\mathbf{A}\times(\mathbf{B}\times\mathbf{C})$ lies in the plane of $\mathbf{B}$ and $\mathbf{C}$. It is also perpendicular to $\mathbf{A}$. The expansion $\mathbf{A}\times(\mathbf{B}\times\mathbf{C})=(\mathbf{A}\cdot\mathbf{C})\mathbf{B}-(\mathbf{A}\cdot\mathbf{B})\mathbf{C}$ shows the same thing: it is a combination of $\mathbf{B}$ and $\mathbf{C}$.

Question 21

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.

Solution. Let the car’s speed be $u$ and the ball’s upward speed, relative to the boy in the car, be $v$. To the boy in the car the ball goes straight up and comes straight down. But the ball already had the car’s horizontal velocity $u$ when it was tossed, and nothing changes that horizontal velocity during the flight. So to the boy on the footpath the ball starts with velocity components $u$ horizontally and $v$ vertically: speed $$v_0=\sqrt{u^2+v^2}\quad\text{at an angle}\quad\theta=\tan^{-1}\frac{v}{u}$$ to the horizontal. He sees an ordinary projectile parabola that starts at the car’s position and ends at the car’s later position. Because the car and the ball move forward at the same rate $u$, the ball comes down exactly into the boy’s hand.

Question 22

A boy throws a ball in air at $60^\circ$ to the horizontal along a road with a speed of $10$ m/s ($36$ km/h). Another boy sitting in a passing by car observes the ball. Sketch the motion of the ball as observed by the boy in the car, if car has a speed of ($18$ km/h). Give explanation to support your diagram.

Solution. The ball’s horizontal velocity is $10\cos60^\circ=5$ m/s $=18$ km/h, the same as the car’s speed. Take the car to be moving in the same direction as the ball (the case the question intends). Relative to the car the ball’s horizontal velocity is $5-5=0$, and its vertical velocity $10\sin60^\circ=5\sqrt3\approx8.7$ m/s is unchanged. So the boy in the car sees the ball go straight up and come straight down, rising to $\dfrac{(5\sqrt3)^2}{2g}\approx3.8$ m, while the boy on the road sees the full parabola.

Question 23

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.

Solution. Air resistance always opposes the velocity. On the way up it adds to gravity in slowing the vertical motion, so the ball does not rise as high. Throughout, it steadily reduces the horizontal velocity, so the ball covers less ground on the way down than on the way up. The result is a lower, shorter path that is no longer symmetric: the highest point is shifted towards the landing point and the descent is steeper than the ascent.

Question 24

A fighter plane is flying horizontally at an altitude of $1.5$ km with speed $720$ km/h. At what angle of sight (w.r.t. horizontal) when the target is seen, should the pilot drop the bomb in order to attack the target?

Solution. The bomb leaves the plane with the plane’s horizontal speed $v=720$ km/h $=200$ m/s and no vertical speed. It falls $H=1500$ m in $$t=\sqrt{\frac{2H}{g}}=\sqrt{\frac{2\times1500}{9.8}}\approx17.5\ \text{s},$$ and in that time moves forward $R=vt=v\sqrt{2H/g}\approx3499$ m. The pilot must release it when the target is $R$ ahead and $H$ below, that is, when the line of sight makes an angle $\phi$ below the horizontal with $$\tan\phi=\frac{H}{R}=\frac{1}{v}\sqrt{\frac{gH}{2}}=\frac{\sqrt{7350}}{200}\approx0.429,\qquad\phi\approx23.2^\circ=23^\circ12′.$$ (With $g=10\ \text{m s}^{-2}$ the angle is $23^\circ25’$.)

Question 25

(a) Earth can be thought of as a sphere of radius $6400$ km. 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 do these accelerations compare with $g=9.8\ \text{m/s}^2$?

(b) Earth also moves in circular orbit around sun once every year with an orbital radius of $1.5\times10^{11}$ m. 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 $g=9.8\ \text{m/s}^2$?

(Hint: acceleration $\dfrac{V^2}{R}=\dfrac{4\pi^2R}{T^2}$.)

Solution. (a) At the equator, $R=6.4\times10^6$ m and $T=86400$ s: $$a=\frac{4\pi^2R}{T^2}=\frac{4\pi^2\times6.4\times10^6}{(86400)^2}\approx3.4\times10^{-2}\ \text{m s}^{-2}.$$ At latitude $\theta$ the object goes round a smaller circle, of radius $R\cos\theta$, centred on the axis, in the same time, so $$a_\theta=\frac{4\pi^2R\cos\theta}{T^2}\approx3.4\times10^{-2}\cos\theta\ \text{m s}^{-2},$$ directed towards the axis. Both are tiny compared with $g$: $g/a\approx290$ at the equator, more elsewhere. That is why the Earth’s spin barely affects our weight.

(b) For the orbit, $R=1.5\times10^{11}$ m and $T=1$ year $\approx3.16\times10^7$ s: $$a=\frac{4\pi^2\times1.5\times10^{11}}{(3.16\times10^7)^2}\approx5.9\times10^{-3}\ \text{m s}^{-2},$$ about $1/1650$ of $g$.

Question 26

Given below in column I are the relations between vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ and in column II are the orientations of $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ in the $XY$ plane. Match the relation in column I to correct orientations in column II.

Column I: (a) $\mathbf{a}+\mathbf{b}=\mathbf{c}$ (b) $\mathbf{a}-\mathbf{c}=\mathbf{b}$ (c) $\mathbf{b}-\mathbf{a}=\mathbf{c}$ (d) $\mathbf{a}+\mathbf{b}+\mathbf{c}=0$

Column II:

Solution. In each figure, follow the arrows head to tail. A vector drawn from the start of a chain to its end is the sum of the chain.

• (i) $\mathbf{a}$ runs up from the origin and $\mathbf{c}$ continues from its head; $\mathbf{b}$ goes from the origin straight to the end. So $\mathbf{a}+\mathbf{c}=\mathbf{b}$, that is, $\mathbf{b}-\mathbf{a}=\mathbf{c}$: relation (c).
• (ii) $\mathbf{a}$, then $\mathbf{b}$, then $\mathbf{c}$ bring us back to the origin. The loop closes, so $\mathbf{a}+\mathbf{b}+\mathbf{c}=0$: relation (d).
• (iii) $\mathbf{c}$ runs left from the origin and $\mathbf{b}$ continues from its head to the top of $\mathbf{a}$; $\mathbf{a}$ goes from the origin straight there. So $\mathbf{c}+\mathbf{b}=\mathbf{a}$, that is, $\mathbf{a}-\mathbf{c}=\mathbf{b}$: relation (b).
• (iv) $\mathbf{a}$ runs left and $\mathbf{b}$ continues downward; $\mathbf{c}$ goes from the origin straight to the end. So $\mathbf{a}+\mathbf{b}=\mathbf{c}$: relation (a).

Question 27

If $|\mathbf{A}|=2$ and $|\mathbf{B}|=4$, then match the relations in column I with the angle $\theta$ between $\mathbf{A}$ and $\mathbf{B}$ in column II.

Solution. $\mathbf{A}\cdot\mathbf{B}=AB\cos\theta=8\cos\theta$. So $\cos\theta=0$ gives $90^\circ$; $\cos\theta=1$ gives $0$; $\cos\theta=\tfrac12$ gives $60^\circ$; $\cos\theta=-1$ gives $180^\circ$.

Question 28

If $|\mathbf{A}|=2$ and $|\mathbf{B}|=4$, then match the relations in column I with the angle $\theta$ between $\mathbf{A}$ and $\mathbf{B}$ in column II.

Solution. $|\mathbf{A}\times\mathbf{B}|=AB\sin\theta=8\sin\theta$. So $\sin\theta=0$ gives $0^\circ$; $\sin\theta=1$ gives $90^\circ$; $\sin\theta=\tfrac12$ gives $30^\circ$; $\sin\theta=\tfrac{1}{\sqrt2}$ gives $45^\circ$. (Among the angles offered, each value of $\sin\theta$ picks out just one.)

Common mistakes

• Adding a tangential acceleration in Question 16. The speed is constant, so at R only the centripetal $v^2/r$ acts.
• Drawing the acceleration along the path in Question 17. It is $\mathbf{g}$, straight down, at every point; only the velocity follows the curve.
• Saying the acceleration is zero at the top in Questions 18 and 19. The velocity (or its vertical part) is zero there; the acceleration is still $g$.
• Forgetting the ball keeps the car’s velocity in Question 21. It was moving forward with the car when tossed, so it is a projectile to the roadside observer.
• Drawing a symmetric curve in Question 23. Drag removes horizontal velocity all the time, so the descent is steeper and shorter than the ascent.
• Measuring the angle of sight from the vertical in Question 24. The question asks for the angle with the horizontal, $\tan\phi=H/R$.
• Using $R$ instead of $R\cos\theta$ at latitude $\theta$ in Question 25. The circle is about the Earth’s axis, not its centre.
• Reading a figure tail-to-tail in Question 26. A sum is read head to tail; the closing vector of a chain runs from its first tail to its last head.

Practise next

• A plane flies horizontally at $360$ km/h at a height of $2000$ m. At what angle of sight below the horizontal should the pilot release a bomb to hit a target on the ground? Take $g=10\ \text{m s}^{-2}$.

• Rain falls vertically at $6$ m/s. A man walks at $8$ m/s. At what angle to the vertical should he hold his umbrella, and how fast does the rain hit him?

• Exemplar Motion in a Plane: Multiple Choice Questions — the same ideas in objective form, with two printed key errors corrected.
• Exemplar Motion in a Plane: Long Answer Questions — projectiles over a hill and on inclines, river crossing, rain and polar coordinates.
• NCERT textbook Chapter 3 — Motion in a Plane for the theory of vectors, projectiles and uniform circular motion.