NCERT Exemplar Solutions Class 11 Physics Laws of Motion — Long Answer Questions Q34–42

NCERT Exemplar Problems, Class 11 Physics — Laws of Motion, Long Answer Questions 34 to 42. All 9 questions solved. The book numbers them 5.34 to 5.42; here they are Questions 34 to 42.

Three ideas carry this page. Zero net force means zero acceleration, and when all forces pass through one point it also means zero torque about any point. On a slope, resolve along and across the plane: the normal force is $mg\cos\theta$, so friction is at most $\mu mg\cos\theta$, while gravity pulls down the slope with $mg\sin\theta$. On a level curve, friction supplies the centripetal force, so the top safe speed is $\sqrt{\mu g r}$.

Key insight. Questions 35 and 41 are the same incline written two ways: the rough plane only replaces $g\sin\theta$ by $g(\sin\theta – \mu\cos\theta)$. Question 37 is three separate speed limits, one for each kind of stretch of track.

Question 34

There are three forces $\mathbf{F}_1$, $\mathbf{F}_2$ and $\mathbf{F}_3$ acting on a body, all acting on a point P on the body. The body is found to move with uniform speed.

(a) Show that the forces are coplanar.

(b) Show that the torque acting on the body about any point due to these three forces is zero.

Solution. The body moves with uniform speed in a straight line (uniform velocity), so it has no acceleration and the net force is zero: $$\mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = \mathbf{0}.$$

(a) $\mathbf{F}_1$ and $\mathbf{F}_2$ both pass through P, and two intersecting lines always lie in one plane; call it plane A. Their sum $\mathbf{F}_1 + \mathbf{F}_2$ also lies in plane A. Since $\mathbf{F}_3 = -(\mathbf{F}_1 + \mathbf{F}_2)$, it too lies in plane A. So all three forces are coplanar.

(b) Take any point O. Because every force acts at P, each torque has the same arm $\overrightarrow{OP}$, and the total torque is $$\boldsymbol{\tau} = \overrightarrow{OP} \times \mathbf{F}_1 + \overrightarrow{OP} \times \mathbf{F}_2 + \overrightarrow{OP} \times \mathbf{F}_3 = \overrightarrow{OP} \times (\mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3) = \mathbf{0}.$$ (About P itself the arm is zero, so the torque is zero trivially.)

Question 35

When a body slides down from rest along a smooth inclined plane making an angle of $45^\circ$ with the horizontal, it takes time $T$. When the same body slides down from rest along a rough inclined plane making the same angle and through the same distance, it is seen to take time $pT$, where $p$ is some number greater than $1$. Calculate the co-efficient of friction between the body and the rough plane.

Solution. From rest with constant acceleration $a$, a distance $s$ takes $t = \sqrt{2s/a}$, so $t^2 \propto 1/a$.

Smooth plane. $a_1 = g\sin 45^\circ = g/\sqrt2$.

Rough plane. Friction $\mu mg\cos 45^\circ$ acts up the slope, so $$a_2 = g\sin 45^\circ – \mu g\cos 45^\circ = \frac{(1 – \mu)g}{\sqrt2}.$$

The same distance is covered in each case, so $$\frac{t_2^2}{t_1^2} = \frac{a_1}{a_2} \quad\Rightarrow\quad p^2 = \frac{1}{1 – \mu} \quad\Rightarrow\quad \mu = 1 – \frac{1}{p^2}.$$ Since $p > 1$, this gives $0 < \mu < 1$, as it must for the body to slide at all on a $45^\circ$ slope.

Question 36

Figure 5.9 shows $(v_x, t)$ and $(v_y, t)$ diagrams for a body of unit mass. Find the force as a function of time.

Fig. 5.9(a): $v_x$ rises in a straight line from $0$ at $t = 0$ to $2$ m s$^{-1}$ at $t = 1$ s, falls in a straight line back to $0$ at $t = 2$ s, and stays $0$ after that. Fig. 5.9(b): $v_y$ rises in a straight line from $0$ at $t = 0$ to $1$ m s$^{-1}$ at $t = 1$ s and stays at $1$ m s$^{-1}$ after that.

Solution. With $m = 1$ kg, the force components equal the slopes of the velocity graphs.

From (a): $v_x = 2t$ for $0 < t < 1$ s, so $F_x = 2$ N; $v_x = 2(2 – t)$ for $1 < t < 2$ s, so $F_x = -2$ N; $v_x = 0$ for $t > 2$ s, so $F_x = 0$.

From (b): $v_y = t$ for $0 < t < 1$ s, so $F_y = 1$ N; $v_y = 1$ for $t > 1$ s, so $F_y = 0$.

Combining, $$\mathbf{F} = \begin{cases} 2\hat{\mathbf{i}} + \hat{\mathbf{j}}\ \text{N}, & 0 < t < 1\ \text{s} \\ -2\hat{\mathbf{i}}\ \text{N}, & 1 < t < 2\ \text{s} \\ \mathbf{0}, & t > 2\ \text{s}. \end{cases}$$

Question 37

A racing car travels on a track (without banking) ABCDEFA (Fig. 5.10). ABC is a circular arc of radius $2R$. CD and FA are straight paths of length $R$ and DEF is a circular arc of radius $R = 100$ m. The co-efficient of friction on the road is $\mu = 0.1$. The maximum speed of the car is $50$ m s$^{-1}$. Find the minimum time for completing one round.

Solution. On a level road, friction is the only horizontal force, so it must provide the centripetal force on each curve. The highest speed on a curve of radius $r$ is therefore $$\frac{mv^2}{r} = \mu mg \quad\Rightarrow\quad v_{\max} = \sqrt{\mu g r}.$$ Taking $g = 10$ m s$^{-2}$ as the key does:

• Arc ABC, radius $2R = 200$ m: $v = \sqrt{0.1 \times 10 \times 200} = \sqrt{200} \approx 14.14$ m s$^{-1}$. The figure shows ABC is three quarters of a full circle (the quarter between A and C is cut off by the corner), so its length is $\tfrac34 \times 2\pi(200) = 300\pi$ m and the time is $300\pi/14.14 \approx 66.64$ s.
• Arc DEF, radius $R = 100$ m: $v = \sqrt{100} = 10$ m s$^{-1}$. It is a quarter circle, of length $\tfrac14 \times 2\pi(100) = 50\pi$ m, so the time is $5\pi \approx 15.71$ s.
• Straights CD and FA, $100$ m each: the car can do its top speed of $50$ m s$^{-1}$, so the time is $200/50 = 4$ s.

Treating the speed changes between stretches as instantaneous, the minimum time for one round is $$t = 66.64 + 15.71 + 4 \approx 86.4\ \text{s}.$$ (The key gives $86.3$ s; the exact sum is $86.35$ s.)

Question 38

The displacement vector of a particle of mass $m$ is given by $\mathbf{r}(t) = \hat{\mathbf{i}}A\cos\omega t + \hat{\mathbf{j}}B\sin\omega t$.

(a) Show that the trajectory is an ellipse.

(b) Show that $\mathbf{F} = -m\omega^2\mathbf{r}$.

Solution. (a) The coordinates are $x = A\cos\omega t$ and $y = B\sin\omega t$. Eliminate $t$ using $\cos^2 + \sin^2 = 1$: $$\frac{x^2}{A^2} + \frac{y^2}{B^2} = \cos^2\omega t + \sin^2\omega t = 1,$$ which is an ellipse with semi-axes $A$ and $B$.

(b) Differentiate twice: $$\mathbf{v} = \frac{d\mathbf{r}}{dt} = -\hat{\mathbf{i}}\,\omega A\sin\omega t + \hat{\mathbf{j}}\,\omega B\cos\omega t,$$ $$\mathbf{a} = \frac{d\mathbf{v}}{dt} = -\omega^2\left(\hat{\mathbf{i}}A\cos\omega t + \hat{\mathbf{j}}B\sin\omega t\right) = -\omega^2\mathbf{r}.$$ So $\mathbf{F} = m\mathbf{a} = -m\omega^2\mathbf{r}$: the force always points towards the origin and grows with distance from it.

Question 39

A cricket bowler releases the ball in two different ways

(a) giving it only horizontal velocity, and

(b) giving it horizontal velocity and a small downward velocity.

The speed $v_s$ at the time of release is the same. Both are released at a height $H$ from the ground. Which one will have greater speed when the ball hits the ground? Neglect air resistance.

Solution. Without air resistance, the only force is gravity, and mechanical energy is conserved. In both cases the ball starts with the same kinetic energy $\tfrac12 mv_s^2$ (speed is the same, whatever the direction) and falls the same height $H$. So at the ground $$\tfrac12 mv^2 = \tfrac12 mv_s^2 + mgH \quad\Rightarrow\quad v = \sqrt{v_s^2 + 2gH}$$ in both cases. For (a) this can also be seen from components: $v_x = v_s$ and $v_z = \sqrt{2gH}$, giving the same result. Ball (b) reaches the ground sooner, but not faster.

Question 40

There are four forces acting at a point P produced by strings as shown in Fig. 5.11, which is at rest. Find the forces $\mathbf{F}_1$ and $\mathbf{F}_2$.

Solution. P is at rest, so the forces balance in each direction. Take $x$ to the right and $y$ upwards. The $2$ N string is $45^\circ$ to the left of the upward vertical and the $1$ N string is $45^\circ$ to the right of it; $\mathbf{F}_1$ points along $+x$ and $\mathbf{F}_2$ along $-y$.

Horizontal. $F_1 + 1\cos 45^\circ = 2\cos 45^\circ$, so $$F_1 = \frac{2 – 1}{\sqrt2} = \frac{1}{\sqrt2} \approx 0.71\ \text{N}.$$

Vertical. $F_2 = 2\sin 45^\circ + 1\sin 45^\circ$, so $$F_2 = \frac{2 + 1}{\sqrt2} = \frac{3}{\sqrt2} \approx 2.12\ \text{N}.$$

Question 41

A rectangular box lies on a rough inclined surface. The co-efficient of friction between the surface and the box is $\mu$. Let the mass of the box be $m$.

(a) At what angle of inclination $\theta$ of the plane to the horizontal will the box just start to slide down the plane?

(b) What is the force acting on the box down the plane, if the angle of inclination of the plane is increased to $\alpha > \theta$?

(c) What is the force needed to be applied upwards along the plane to make the box either remain stationary or just move up with uniform speed?

(d) What is the force needed to be applied upwards along the plane to make the box move up the plane with acceleration $a$?

Solution. On an incline at angle $\phi$, the normal force is $mg\cos\phi$, so friction is at most $\mu mg\cos\phi$, and gravity pulls down the slope with $mg\sin\phi$.

(a) The box just starts to slide when the pull down the slope equals the largest friction: $$mg\sin\theta = \mu mg\cos\theta \quad\Rightarrow\quad \tan\theta = \mu \quad\Rightarrow\quad \theta = \tan^{-1}\mu.$$ This is the angle of repose.

(b) At $\alpha > \theta$ the box slides, friction is $\mu mg\cos\alpha$ up the slope, and the net force down the plane is $$F = mg\sin\alpha – \mu mg\cos\alpha = mg(\sin\alpha – \mu\cos\alpha).$$

(c) With the plane at $\alpha$, for the box to move up at uniform speed (or be on the point of moving up), the applied force must balance the pull down the slope and the full friction, which now acts down the slope: $$F = mg(\sin\alpha + \mu\cos\alpha).$$

(d) To give it acceleration $a$ up the plane, add $ma$: $$F = mg(\sin\alpha + \mu\cos\alpha) + ma.$$ (The key writes this part with $\theta$ in place of $\alpha$; the formula holds for whatever angle the plane is at, so the two agree.)

Question 42

A helicopter of mass $2000$ kg rises with a vertical acceleration of $15$ m s$^{-2}$. The total mass of the crew and passengers is $500$ kg. Give the magnitude and direction of the ($g = 10$ m s$^{-2}$)

(a) force on the floor of the helicopter by the crew and passengers.

(b) action of the rotor of the helicopter on the surrounding air.

(c) force on the helicopter due to the surrounding air.

Solution. (a) Apply the second law to the crew and passengers ($500$ kg). The floor pushes up with $N$; the weight acts down: $$N – 500 \times 10 = 500 \times 15 \quad\Rightarrow\quad N = 12\,500\ \text{N}.$$ By the third law, they push down on the floor with $1.25 \times 10^4$ N.

(c) Now take the helicopter with everyone inside, $2500$ kg. The only upward force is the push $R$ of the air: $$R – 2500 \times 10 = 2500 \times 15 \quad\Rightarrow\quad R = 62\,500\ \text{N},$$ so the air pushes the helicopter up with $6.25 \times 10^4$ N.

(b) By the third law, the rotor pushes the air downwards with a force of the same size, $6.25 \times 10^4$ N.

Common mistakes

• Saying the three forces in Question 34 are coplanar “because they meet at a point”. Three concurrent forces need not be coplanar; it is the zero sum that forces $\mathbf{F}_3$ into the plane of the other two.
• Comparing times instead of their squares in Question 35. $t \propto 1/\sqrt{a}$, so $p^2 = a_1/a_2$, not $p$.
• Taking ABC as a semicircle in Question 37. The figure cuts out only the quarter between A and C, so ABC is three quarters of the circle, $300\pi$ m long.
• Using $50$ m s$^{-1}$ on the curves in Question 37. On a level curve the limit is $\sqrt{\mu g r}$, which is $14.1$ and $10$ m s$^{-1}$ here.
• Expecting the ball thrown downwards to land faster in Question 39. Both start with the same speed and fall the same height, so energy gives the same final speed.
• Leaving the friction direction unchanged in Question 41(c). When the box is pushed up the plane, friction acts down it, so the terms add.
• Using $2000$ kg instead of $2500$ kg in Question 42(b) and (c). The air holds up the helicopter and everyone inside it.

Practise next

• A block slides from rest down a smooth plane inclined at $30^\circ$ and takes time $T$. On a rough plane at the same angle, it covers the same distance in $2T$. Find the coefficient of friction.

• A car takes an unbanked curve of radius $50$ m. The coefficient of friction between the tyres and the road is $0.2$. What is the highest speed at which it can take the curve without skidding? ($g = 10$ m s$^{-2}$)

• Exemplar Laws of Motion: Very Short and Short Answer Questions — the preceding questions on lifts, recoil, impulse and strings.
• Exemplar Laws of Motion: Multiple Choice Questions — the same ideas in objective form.
• NCERT textbook Chapter 4 — Laws of Motion for the theory of the three laws, momentum, friction and circular motion.