NCERT Exemplar Solutions Class 11 Physics Motion in a Plane — Long Answer Questions Q29–37

NCERT Exemplar Problems, Class 11 Physics — Motion in a Plane, Long Answer Questions 29 to 37. All 9 questions solved. The book numbers them 4.29 to 4.37; it calls this Unit 4, while the rationalised textbook numbers the chapter 3.

Three ideas carry this page. Choose axes to suit the problem: for a projectile on an incline, axes along and perpendicular to the plane turn a messy intersection into two simple one-dimensional motions. Velocities add as vectors: the velocity of rain, a swimmer or a thrown ball as seen from the ground is its velocity relative to the moving body plus that body’s own velocity, and every river, rain and running-throw question is one vector triangle. Optimise honestly: when a question asks for the shortest time or the largest range, set up the quantity as a function of the free choice and find its true extremum, rather than fixing a convenient configuration first.

Key insight. In Question 29 the printed answer insists that the top of the trajectory sits over the hill. Nothing requires that; the packet can cross the hill on the way down. Dropping the extra assumption cuts the answer from 45 s to about 20 s.

Question 29

A hill is $500$ m high. Supplies are to be sent across the hill using a canon that can hurl packets at a speed of $125$ m/s over the hill. The canon is located at a distance of $800$ m from the foot of hill and can be moved on the ground at a speed of $2$ m/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 $g=10\ \text{m/s}^2$.

Solution. Treat the hill, as the printed solution does, as a peak $500$ m high standing $800$ m from the cannon, with the ground on the far side at the cannon’s level. A packet launched with components $(v_x, v_y)$, $v_x^2+v_y^2=125^2$, lands back at ground level after $T=2v_y/g$, so the flight time depends only on $v_y$.

Least vertical speed. To rise $500$ m at all, $v_y^2\ge2gh=10\,000$, so $v_y\ge100$ m/s and the flight lasts at least $20$ s. With $v_y=100$ m/s, $v_x=\sqrt{125^2-100^2}=75$ m/s, and the packet reaches exactly $500$ m at its highest point, $75\times10=750$ m from the cannon.

The printed method. The key places that highest point over the hill: the cannon is moved $800-750=50$ m closer, taking $50/2=25$ s, and the packet then takes $10$ s up and $10$ s down, giving $25+10+10=45$ s.

Why that is not the shortest. The packet need not be at its highest point when it passes the hill; it only has to be at least $500$ m up at a horizontal distance of $800$ m. From the cannon’s original position, require $$y(800)=v_y\,\frac{800}{v_x}-5\left(\frac{800}{v_x}\right)^2\ge500,\qquad v_x=\sqrt{125^2-v_y^2}.$$ Solving the equality numerically gives the smallest such $v_y$: $$v_y\approx100.24\ \text{m/s},\quad v_x\approx74.68\ \text{m/s}\quad(\text{elevation}\approx53.3^\circ).$$ The packet passes the peak at $t\approx10.71$ s, just as it starts to descend, and lands after $$T=\frac{2v_y}{g}\approx20.05\ \text{s},$$ with no time spent moving the cannon. Since no launch can take less than $20$ s, and moving the cannon only adds time, this is the minimum.

Printed key error. The key’s $45$ s is a possible time, not the shortest. Its extra condition (highest point exactly over the hill) costs $25$ s of moving the cannon, whereas raising the vertical speed from $100$ to about $100.24$ m/s lets the packet clear the hill from where the cannon already stands.

Question 30

A gun can fire shells with maximum speed $v_0$ and the maximum horizontal range that can be achieved is $R=\dfrac{v_0^2}{g}$. If a target farther away by distance $\Delta x$ (beyond $R$) has to be hit with the same gun (Fig 4.5), show that it could be achieved by raising the gun to a height at least $$h=\Delta x\left[1+\frac{\Delta x}{R}\right].$$

(Hint: This problem can be approached in two different ways: (i) Refer to the diagram: target T is at horizontal distance $x=R+\Delta x$ and below point of projection $y=-h$. (ii) From point P in the diagram: Projection at speed $v_0$ at an angle $\theta$ below horizontal with height $h$ and horizontal range $\Delta x$.)

The figure shows the gun raised to height $h$, firing at $v_0$ and angle $\theta$; the trajectory returns to the gun’s level at P, a distance $R$ ahead, moving at $v_0$ at angle $\theta$ below the horizontal, and goes on to the target T on the ground a further $\Delta x$ ahead.

Solution. The maximum range $R=v_0^2/g$ is obtained at $\theta=45^\circ$, so keep that elevation. By symmetry the shell comes back to the level of the gun at P, a distance $R$ ahead, with the same speed $v_0$, now at $45^\circ$ below the horizontal (hint (ii)). From P it must fall $h$ while moving forward $\Delta x$. With $x$ forward and $y$ downward from P, $$y=x\tan45^\circ+\frac{g x^2}{2v_0^2\cos^2 45^\circ}=x+\frac{g x^2}{v_0^2}=x+\frac{x^2}{R}.$$ Setting $x=\Delta x$ gives the height the gun must be raised: $$h=\Delta x+\frac{(\Delta x)^2}{R}=\Delta x\left[1+\frac{\Delta x}{R}\right].$$

Hint (i) gives the same result directly: the trajectory $y=x\tan45^\circ-\dfrac{gx^2}{2v_0^2\cos^245^\circ}=x-\dfrac{x^2}{R}$ at $x=R+\Delta x$ gives $y=-\Delta x\left(1+\dfrac{\Delta x}{R}\right)$.

A remark on “at least”: this is the height needed when the gun keeps its $45^\circ$ elevation, and any greater height also works with a suitable angle. If the gun is re-aimed for the best angle from the raised position, the maximum range becomes $\dfrac{v_0}{g}\sqrt{v_0^2+2gh}$, and the height actually needed falls to $\Delta x\left(1+\dfrac{\Delta x}{2R}\right)$, so the printed $h$ is sufficient rather than the smallest possible.

Question 31

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. 4.6).

(a) 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). (b) Time of flight. (c) $\beta$ at which range will be maximum.

(Hint: (i) Point P at which particle hits the plane can be seen as intersection of its trajectory (parabola) and straight line. Remember particle is projected at an angle $(\alpha+\beta)$ w.r.t. horizontal. (ii) We can take $x$-direction along the plane and $y$-direction perpendicular to the plane. In that case resolve $g$ into $g_x$ along the plane and $g_y$ perpendicular to the plane.)

The particle is projected up the slope from a point on it, with speed $v_0$ (the figure itself does not print in the book).

Solution. Use hint (ii). Along the plane (up the slope): initial velocity $v_0\cos\beta$, acceleration $-g\sin\alpha$. Perpendicular to the plane: initial velocity $v_0\sin\beta$, acceleration $-g\cos\alpha$.

(b) The particle returns to the plane when its perpendicular displacement is zero: $$v_0\sin\beta\,T-\tfrac12 g\cos\alpha\,T^2=0\quad\Rightarrow\quad T=\frac{2v_0\sin\beta}{g\cos\alpha}.$$

(a) The distance along the plane in that time is $$L=v_0\cos\beta\,T-\tfrac12 g\sin\alpha\,T^2=\frac{2v_0^2\sin\beta}{g\cos^2\alpha}\left(\cos\beta\cos\alpha-\sin\beta\sin\alpha\right)=\frac{2v_0^2\sin\beta\cos(\alpha+\beta)}{g\cos^2\alpha}.$$

(c) Write $2\sin\beta\cos(\alpha+\beta)=\sin(\alpha+2\beta)-\sin\alpha$. Only the first term depends on $\beta$, and it is largest when $\alpha+2\beta=\dfrac{\pi}{2}$: $$\beta=\frac{\pi}{4}-\frac{\alpha}{2},$$ which gives $L_{\max}=\dfrac{v_0^2}{g(1+\sin\alpha)}$. For $\alpha=0$ this is the familiar $45^\circ$ and $v_0^2/g$.

Question 32

A particle falling vertically from a height hits a plane surface inclined to horizontal at an angle $\theta$ with speed $v_0$ and rebounds elastically (Fig 4.7). Find the distance along the plane where it will hit second time.

(Hint: (i) After rebound, particle still has speed $v_0$ to start. (ii) Work out angle particle speed has with horizontal after it rebounds. (iii) Rest is similar to if particle is projected up the incline.)

The figure shows the particle coming straight down onto the incline at P, bouncing off, and landing a distance $L$ further down the slope.

Solution. Take axes along the incline (down the slope) and perpendicular to it. The incoming vertical velocity $v_0$ has a component $v_0\sin\theta$ down the slope and $v_0\cos\theta$ into the surface. An elastic bounce reverses only the perpendicular component, so just after the bounce the particle has $v_0\cos\theta$ away from the surface and still $v_0\sin\theta$ down the slope. Gravity has components $g\sin\theta$ down the slope and $g\cos\theta$ into the surface.

The particle returns to the surface when $$v_0\cos\theta\,t-\tfrac12 g\cos\theta\,t^2=0\quad\Rightarrow\quad t=\frac{2v_0}{g}.$$ In that time it moves down the slope $$L=v_0\sin\theta\cdot\frac{2v_0}{g}+\frac12 g\sin\theta\left(\frac{2v_0}{g}\right)^2=\frac{2v_0^2\sin\theta}{g}+\frac{2v_0^2\sin\theta}{g}=\frac{4v_0^2\sin\theta}{g}.$$ If the particle fell from rest through a height $h$, then $v_0^2=2gh$ and $L=8h\sin\theta$.

Printed key error. The key prints $\dfrac{A v_0^2}{g}\sin\theta$, with a letter A where the numerical factor belongs. The factor is $4$.

Question 33

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 $45^\circ$ to the vertical. What is the speed of the rain? In what direction does rain fall as observed by a ground based observer?

(Hint: Assume north to be $\hat{\mathbf{i}}$ direction and vertically downward to be $-\hat{\mathbf{j}}$. Let the rain velocity $\mathbf{v}_r$ be $a\hat{\mathbf{i}}+b\hat{\mathbf{j}}$. The velocity of rain as observed by the girl is always $\mathbf{v}_r-\mathbf{v}_{girl}$.)

Solution. At $5$ m/s the rain seems vertical, so the rain’s horizontal velocity must equal hers: $$\mathbf{v}_r-5\hat{\mathbf{i}}=(a-5)\hat{\mathbf{i}}+b\hat{\mathbf{j}}\ \text{is vertical}\quad\Rightarrow\quad a=5.$$ At $10$ m/s the relative velocity is $-5\hat{\mathbf{i}}+b\hat{\mathbf{j}}$, at $45^\circ$ to the vertical, so $|b|=5$; the rain falls, so $b=-5$. Hence $$\mathbf{v}_r=5\hat{\mathbf{i}}-5\hat{\mathbf{j}},\qquad|\mathbf{v}_r|=5\sqrt2\approx7.07\ \text{m/s}.$$ To someone on the ground the rain falls at $45^\circ$ to the vertical, moving towards the north.

Question 34

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. 4.8).

(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?

The figure shows the river between parallel banks running east–west, flowing east at $3$ m/s, with A on the south bank directly opposite B on the north bank.

Solution. (a) His velocity relative to the ground is $4$ m/s north plus $3$ m/s east: $$v=\sqrt{4^2+3^2}=5\ \text{m/s},\quad\text{at}\ \tan^{-1}\tfrac34\approx37^\circ\ \text{east of north}.$$

(b) To land at B the resultant must point due north, so the east component of his swimming velocity must cancel the current: $4\sin\phi=3$, where $\phi$ is measured west of north. So $\sin\phi=\tfrac34$, $\phi=\tan^{-1}\dfrac{3}{\sqrt7}\approx48.6^\circ$ west of north (upstream), and his resultant speed is $$\sqrt{4^2-3^2}=\sqrt7\approx2.65\ \text{m/s}.$$

(c) The time to cross depends only on the component of velocity across the river. In (a) it is $4$ m/s; in (b) it is $\sqrt7\approx2.65$ m/s. So he reaches the other bank sooner in case (a), though he lands downstream of B.

Question 35

A cricket fielder can throw the cricket ball with a speed $v_0$. If he throws the ball while running with speed $u$ at an angle $\theta$ 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 $\theta$ at which he should throw the ball that would maximise the horizontal range as found in (iii). (e) how does $\theta$ for maximum range change if $u>v_0$, $u=v_0$, $u<v_0$? (f) how does $\theta$ in (v) compare with that for $u=0$ (i.e. $45^\circ$)?

(The question’s “(iii)” and “(v)” refer to parts (c) and (d).)

Solution. Take the throw to be forward, in the direction he is running. Relative to the ground the ball’s velocity is his throw plus his own velocity: $$v_x=v_0\cos\theta+u,\qquad v_y=v_0\sin\theta.$$

(a) The angle seen by a spectator is $$\theta’=\tan^{-1}\left(\frac{v_0\sin\theta}{v_0\cos\theta+u}\right),$$ always smaller than $\theta$ when $u>0$.

(b) The vertical motion is unaffected by running: $T=\dfrac{2v_0\sin\theta}{g}$.

(c) $R=v_xT=\dfrac{2v_0\sin\theta\,(v_0\cos\theta+u)}{g}$.

(d) Set $\dfrac{dR}{d\theta}=0$: $$v_0\cos2\theta+u\cos\theta=0\quad\Rightarrow\quad2v_0\cos^2\theta+u\cos\theta-v_0=0,$$ whose positive root is $$\cos\theta_{\max}=\frac{-u+\sqrt{u^2+8v_0^2}}{4v_0}.$$

(e) For $u=v_0$: $\cos\theta_{\max}=\dfrac{-1+3}{4}=\dfrac12$, so $\theta_{\max}=60^\circ$. For $u\ll v_0$: $\cos\theta_{\max}\approx\dfrac{1}{\sqrt2}-\dfrac{u}{4v_0}$, a little above $45^\circ$. For $u\gg v_0$: $\sqrt{u^2+8v_0^2}\approx u+\dfrac{4v_0^2}{u}$, so $\cos\theta_{\max}\approx\dfrac{v_0}{u}\to0$ and $\theta_{\max}\to90^\circ$. In every case, the faster he runs, the steeper he should throw: running already supplies horizontal speed, so the throw is better spent on time in the air.

(f) For $u=0$ the formula gives $\cos\theta_{\max}=\dfrac{1}{\sqrt2}$, the usual $45^\circ$. For any $u>0$, $\theta_{\max}>45^\circ$.

Question 36

Motion in two dimensions, in a plane can be studied by expressing position, velocity and acceleration as vectors in Cartesian co-ordinates $\mathbf{A}=A_x\hat{\mathbf{i}}+A_y\hat{\mathbf{j}}$ where $\hat{\mathbf{i}}$ and $\hat{\mathbf{j}}$ are unit vectors along $x$ and $y$ directions, respectively and $A_x$ and $A_y$ are corresponding components of $\mathbf{A}$ (Fig. 4.9). Motion can also be studied by expressing vectors in circular polar co-ordinates as $\mathbf{A}=A_r\hat{\mathbf{r}}+A_\theta\hat{\boldsymbol{\theta}}$ where $\hat{\mathbf{r}}=\dfrac{\mathbf{r}}{r}=\cos\theta\,\hat{\mathbf{i}}+\sin\theta\,\hat{\mathbf{j}}$ and $\hat{\boldsymbol{\theta}}=-\sin\theta\,\hat{\mathbf{i}}+\cos\theta\,\hat{\mathbf{j}}$ are unit vectors along direction in which $r$ and $\theta$ are increasing.

(a) Express $\hat{\mathbf{i}}$ and $\hat{\mathbf{j}}$ in terms of $\hat{\mathbf{r}}$ and $\hat{\boldsymbol{\theta}}$. (b) Show that both $\hat{\mathbf{r}}$ and $\hat{\boldsymbol{\theta}}$ are unit vectors and are perpendicular to each other. (c) Show that $\dfrac{d}{dt}(\hat{\mathbf{r}})=\omega\hat{\boldsymbol{\theta}}$ where $\omega=\dfrac{d\theta}{dt}$ and $\dfrac{d}{dt}(\hat{\boldsymbol{\theta}})=-\omega\hat{\mathbf{r}}$. (d) For a particle moving along a spiral given by $\mathbf{r}=a\theta\hat{\mathbf{r}}$, 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.

The figure shows a point P at $(x, y)=(r,\theta)$, with $\hat{\mathbf{r}}$ pointing outward along OP and $\hat{\boldsymbol{\theta}}$ perpendicular to it in the direction of increasing $\theta$.

Solution. (a) Solve the two defining equations. Multiply the $\hat{\mathbf{r}}$ equation by $\cos\theta$ and the $\hat{\boldsymbol{\theta}}$ equation by $\sin\theta$ and subtract, then repeat with $\sin\theta$ and $\cos\theta$ and add: $$\hat{\mathbf{i}}=\cos\theta\,\hat{\mathbf{r}}-\sin\theta\,\hat{\boldsymbol{\theta}},\qquad\hat{\mathbf{j}}=\sin\theta\,\hat{\mathbf{r}}+\cos\theta\,\hat{\boldsymbol{\theta}}.$$

(b) $\hat{\mathbf{r}}\cdot\hat{\mathbf{r}}=\cos^2\theta+\sin^2\theta=1$ and $\hat{\boldsymbol{\theta}}\cdot\hat{\boldsymbol{\theta}}=\sin^2\theta+\cos^2\theta=1$, so both are unit vectors. $\hat{\mathbf{r}}\cdot\hat{\boldsymbol{\theta}}=-\cos\theta\sin\theta+\sin\theta\cos\theta=0$, so they are perpendicular.

(c) Since $\hat{\mathbf{i}}$ and $\hat{\mathbf{j}}$ are fixed, $$\frac{d\hat{\mathbf{r}}}{dt}=\left(-\sin\theta\,\hat{\mathbf{i}}+\cos\theta\,\hat{\mathbf{j}}\right)\frac{d\theta}{dt}=\omega\hat{\boldsymbol{\theta}},\qquad\frac{d\hat{\boldsymbol{\theta}}}{dt}=\left(-\cos\theta\,\hat{\mathbf{i}}-\sin\theta\,\hat{\mathbf{j}}\right)\frac{d\theta}{dt}=-\omega\hat{\mathbf{r}}.$$

(d) $\theta$ is an angle, a dimensionless ratio, and $\mathbf{r}$ is a length, so $a$ has the dimensions of length, $[L]$ ($a=1$ m, say).

(e) With $\mathbf{r}=a\theta\hat{\mathbf{r}}$ and the results of (c), $$\mathbf{v}=\frac{d\mathbf{r}}{dt}=a\omega\hat{\mathbf{r}}+a\theta\omega\hat{\boldsymbol{\theta}}.$$ Differentiating again, with $\dfrac{d\omega}{dt}=\dfrac{d^2\theta}{dt^2}$, $$\mathbf{a}=a\left(\frac{d^2\theta}{dt^2}-\omega^2\theta\right)\hat{\mathbf{r}}+a\left(\theta\frac{d^2\theta}{dt^2}+2\omega^2\right)\hat{\boldsymbol{\theta}}.$$ With $a=1$ these are $\mathbf{v}=\omega\hat{\mathbf{r}}+\omega\theta\hat{\boldsymbol{\theta}}$ and $\mathbf{a}=\left(\ddot\theta-\omega^2\theta\right)\hat{\mathbf{r}}+\left(\theta\ddot\theta+2\omega^2\right)\hat{\boldsymbol{\theta}}$.

Question 37

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 $50\ \text{m}\times50\ \text{m}$ 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?

Solution. Put A at the origin, so the sand occupies $25\le x,y\le75$, with P at $(25,25)$, R at $(75,25)$ and Q at $(75,75)$.

Straight path through the sand. The diagonal AC passes through P and Q. It covers $AP+QC=25\sqrt2+25\sqrt2=50\sqrt2$ m on firm ground and $PQ=50\sqrt2$ m in sand, taking $$T_{\text{sand}}=50\sqrt2\left(1+\frac1v\right)\ \text{s}.$$

Best path avoiding the sand. The shortest route round the sand bends at a corner of it, R (or, symmetrically, the opposite corner): A to R to C. Each leg is $\sqrt{75^2+25^2}=25\sqrt{10}$ m, so $$T_{\text{outside}}=50\sqrt{10}\approx158.1\ \text{s}.$$ (Going round the edge, A to B to C, takes $200$ s, which is worse.)

Condition. The sand route is faster when $$50\sqrt2\left(1+\frac1v\right)<50\sqrt{10}\quad\Rightarrow\quad1+\frac1v<\sqrt5\quad\Rightarrow\quad v>\frac{1}{\sqrt5-1}=\frac{\sqrt5+1}{4}\approx0.81\ \text{m/s}.$$

Common mistakes

• Putting the top of the trajectory over the hill in Question 29. The packet only has to be above $500$ m at the hill; it can cross on the way down, and that saves the $25$ s spent moving the cannon.
• Changing the elevation from $45^\circ$ in Question 30 and still using the printed formula. The stated $h$ assumes the $45^\circ$ shot; with re-aiming, less height is needed.
• Using $\beta$ as the elevation in the parabola in Question 31. The angle above the horizontal is $\alpha+\beta$; along-the-plane axes avoid the mix-up.
• Reversing the whole velocity in the bounce in Question 32. An elastic bounce reverses only the component normal to the surface; the $v_0\sin\theta$ along the slope survives.
• Taking the rain’s speed as $5$ m/s in Question 33. The first observation fixes only the horizontal part; the second gives the vertical part, and the speed is $5\sqrt2$ m/s.
• Thinking the shortest path gives the shortest time in Question 34. Crossing time depends only on the velocity across the river, which is largest when he swims straight across.
• Forgetting the running speed adds only horizontally in Question 35. The time of flight is the same as for a standing throw.
• Comparing with the edge route A to B to C in Question 37. The shortest dry route skirts a corner of the sand and is only $158$ m.

Practise next

• A river $200$ m wide flows at $3$ m/s. A swimmer who can swim at $5$ m/s in still water wants to land directly opposite his starting point. In which direction should he swim, and how long does the crossing take?

• A particle is projected up a plane inclined at $30^\circ$ with speed $20$ m/s at $30^\circ$ to the plane. Find its range along the plane and its time of flight. Take $g=10\ \text{m s}^{-2}$.

• Exemplar Motion in a Plane: Very Short and Short Answer Questions — the shorter questions on projectiles, circular motion and vector matching.
• Exemplar Motion in a Plane: Multiple Choice Questions — the same ideas in objective form.
• NCERT textbook Chapter 3 — Motion in a Plane for the theory of vectors, projectiles and uniform circular motion.