Chapter Five

LAWS OF MOTION

MCQ I

5.1 A ball is travelling with uniform translatory motion. This means that

(a) it is at rest.

(b) the path can be a straight line or circular and the ball travels with uniform speed.

(c) all parts of the ball have the same velocity (magnitude and direction) and the velocity is constant.

(d) the centre of the ball moves with constant velocity and the ball spins about its centre uniformly.

5.2 A metre scale is moving with uniform velocity. This implies

(a) the force acting on the scale is zero, but a torque about the centre of mass can act on the scale.

(b) the force acting on the scale is zero and the torque acting about centre of mass of the scale is also zero.

(c) the total force acting on it need not be zero but the torque on it is zero.

(d) neither the force nor the torque need to be zero.

5.3 A cricket ball of mass $150 \mathrm{~g}$ has an initial velocity $\mathbf{u}=(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) \mathrm{m} \mathrm{s}^{-1}$ and a final velocity $\mathbf{v}=-(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) \mathrm{m} \mathrm{s}^{-1}$ after being hit. The change in momentum (final momentum-initial momentum) is (in $\mathrm{kg} \mathrm{m} \mathrm{s}^{1}$ )

(a) zero

(b) $-(0.45 \hat{\mathbf{i}}+0.6 \hat{\mathbf{j}})$

(c) $-(0.9 \hat{\mathbf{i}}+1.2 \hat{\mathbf{j}})$

(d) $-5(\hat{\mathbf{i}}+\hat{\mathbf{j}})$.

5.4 In the previous problem (5.3), the magnitude of the momentum transferred during the hit is

(a) Zero

(b) $0.75 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}$

(c) $1.5 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}$

(d) $14 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}$.

5.5 Conservation of momentum in a collision between particles can be understood from

(a) conservation of energy.

(b) Newton’s first law only.

(c) Newton’s second law only.

(d) both Newton’s second and third law.

5.6 A hockey player is moving northward and suddenly turns westward with the same speed to avoid an opponent. The force that acts on the player is

(a) frictional force along westward.

(b) muscle force along southward.

(c) frictional force along south-west.

(d) muscle force along south-west.

5.7 A body of mass $2 \mathrm{~kg}$ travels according to the law $x(t)=p t+q t^{2}+r t^{3}$ where $p=3 \mathrm{~m} \mathrm{~s}^{-1}, q=4 \mathrm{~m} \mathrm{~s}^{-2}$ and $r=5 \mathrm{~m} \mathrm{~s}^{-3}$.

The force acting on the body at $t=2$ seconds is

(a) $136 \mathrm{~N}$

(b) $134 \mathrm{~N}$

(c) $158 \mathrm{~N}$

(d) $68 \mathrm{~N}$

5.8 A body with mass $5 \mathrm{~kg}$ is acted upon by a force $\mathbf{F}=(-3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) \mathrm{N}$. If its initial velocity at $t=0$ is $\boldsymbol{v}=(6 \hat{\mathbf{i}}-12 \hat{\mathbf{j}}) \mathrm{m} \mathrm{s}^{-1}$, the time at which it will just have a velocity along the $y$-axis is

(a) never

(b) $10 \mathrm{~s}$

(c) $2 \mathrm{~s}$

(d) $15 \mathrm{~s}$

5.9 A car of mass $m$ starts from rest and acquires a velocity along east $\boldsymbol{v}=v \hat{\mathbf{i}}(v>0)$ in two seconds. Assuming the car moves with uniform acceleration, the force exerted on the car is

(a) $\frac{m v}{2}$ eastward and is exerted by the car engine.

(b) $\frac{m v}{2}$ eastward and is due to the friction on the tyres exerted by the road.

(c) more than $\frac{m v}{2}$ eastward exerted due to the engine and overcomes the friction of the road.

(d) $\frac{m v}{2}$ exerted by the engine .

MCQ II

5.10 The motion of a particle of mass $m$ is given by $x=0$ for $t<0$ $\mathrm{s}, x(t)=\mathrm{A} \sin 4 p t$ for $0<t<(1 / 4) \mathrm{s}(\mathrm{A}>\mathrm{o})$, and $x=0$ for $t>(1 / 4) \mathrm{s}$. Which of the following statements is true?

(a) The force at $t=(1 / 8) \mathrm{s}$ on the particle is $-16 \pi^{2} \mathrm{~A} \mathrm{~m}$.

(b) The particle is acted upon by on impulse of magnitude $4 \pi^{2} A \mathrm{~m}$ at $t=0 \mathrm{~s}$ and $t=(1 / 4) \mathrm{s}$.

(c) The particle is not acted upon by any force.

(d) The particle is not acted upon by a constant force.

(e) There is no impulse acting on the particle.

5.11 In Fig. 5.1, the co-efficient of friction between the floor and the body $\mathrm{B}$ is 0.1 . The co-efficient of friction between the bodies B and A is 0.2. A force $\mathbf{F}$ is applied as shown on $B$. The mass of $A$ is $m / 2$ and of $B$ is $m$. Which of the following statements are true?

Fig. 5.1

(a) The bodies will move together if $F=0.25 \mathrm{mg}$.

(b) The body A will slip with respect to $\mathrm{B}$ if $F=0.5 \mathrm{mg}$.

(c) The bodies will move together if $F=0.5 \mathrm{mg}$.

(d) The bodies will be at rest if $F=0.1 \mathrm{mg}$.

(e) The maximum value of $F$ for which the two bodies will move together is $0.45 \mathrm{mg}$.

5.12 Mass $m_{1}$ moves on a slope making an angle $\theta$ with the horizontal and is attached to mass $m_{2}$ by a string passing over a frictionless pulley as shown in Fig. 5.2. The co-efficient of friction between $m_{1}$ and the sloping surface is $\mu$. Which of the following statements are true?

Fig. 5.2

(a) If $m_{2}>m_{1} \sin \theta$, the body will move up the plane.

(b) If $m_{2}>m_{1}(\sin \theta+\mu \cos \theta)$, the body will move up the plane.

(c) If $m_{2}<m_{1}(\sin \theta+\mu \cos \theta)$, the body will move up the plane.

(d) If $m_{2}<m_{1}(\sin \theta-\mu \cos \theta)$, the body will move down the plane.

5.13 In Fig. 5.3, a body A of mass $m$ slides on plane inclined at angle $\theta_{1}$ to the horizontal and $\mu_{1}$ is the coefficent of friction between $\mathrm{A}$ and the plane. A is connected by a light string passing over a frictionless pulley to another body $\mathrm{B}$, also of mass $m$, sliding on a frictionless plane inclined at angle $\theta_{2}$ to the horizontal. Which of the following statements are true?

Fig. 5.3

(a) A will never move up the plane.

(b) A will just start moving up the plane when

$$ \mu=\frac{\sin \theta_{2}-\sin \theta_{1}}{\cos \theta_{1}} $$

(c) For A to move up the plane, $\theta_{2}$ must always be greater than $\theta_{1}$.

(d) $\mathrm{B}$ will always slide down with constant speed.

5.14 Two billiard balls A and B, each of mass $50 \mathrm{~g}$ and moving in opposite directions with speed of $5 \mathrm{~m} \mathrm{~s}^{-1}$ each, collide and rebound with the same speed. If the collision lasts for $10^{-3} \mathrm{~s}$, which of the following statements are true?

(a) The impulse imparted to each ball is $0.25 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}$ and the force on each ball is $250 \mathrm{~N}$.

(b) The impulse imparted to each ball is $0.25 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}$ and the force exerted on each ball is $25 \times 10^{-5} \mathrm{~N}$.

(c) The impulse imparted to each ball is $0.5 \mathrm{Ns}$.

(d) The impulse and the force on each ball are equal in magnitude and opposite in direction.

5.15 A body of mass $10 \mathrm{~kg}$ is acted upon by two perpendicular forces, $6 \mathrm{~N}$ and $8 \mathrm{~N}$. The resultant acceleration of the body is

(a) $1 \mathrm{~m} \mathrm{~s}^{-2}$ at an angle of $\tan ^{-1}\left(\frac{4}{3}\right)$ w.r.t. $6 \mathrm{~N}$ force.

(b) $0.2 \mathrm{~m} \mathrm{~s}^{-2}$ at an angle of $\tan ^{-1}\left(\frac{4}{3}\right)$ w.r.t. $6 \mathrm{~N}$ force.

(c) $1 \mathrm{~m} \mathrm{~s}^{-2}$ at an angle of $\tan ^{-1}\left(\frac{3}{4}\right)$ w.r.t. $8 \mathrm{~N}$ force.

(d) $0.2 \mathrm{~m} \mathrm{~s}^{-2}$ at an angle of $\tan ^{-1}\left(\frac{3}{4}\right)$ w.r.t. $8 \mathrm{~N}$ force.

VSA

5.16 A girl riding a bicycle along a straight road with a speed of $5 \mathrm{~m} \mathrm{~s}^{-1}$ throws a stone of mass $0.5 \mathrm{~kg}$ which has a speed of $15 \mathrm{~m} \mathrm{~s}^{-1}$ with respect to the ground along her direction of motion. The mass of the girl and bicycle is $50 \mathrm{~kg}$. Does the speed of the bicycle change after the stone is thrown? What is the change in speed, if so?

5.17 A person of mass $50 \mathrm{~kg}$ stands on a weighing scale on a lift. If the lift is descending with a downward acceleration of $9 \mathrm{~m} \mathrm{~s}^{-2}$, what would be the reading of the weighing scale? $\left(g=10 \mathrm{~m} \mathrm{~s}^{-2}\right)$

5.18 The position time graph of a body of mass $2 \mathrm{~kg}$ is as given in Fig. 5.4. What is the impulse on the body at $t=0 \mathrm{~s}$ and $t=4 \mathrm{~s}$.

Fig. 5.4

5.19 A person driving a car suddenly applies the brakes on seeing a child on the road ahead. If he is not wearing seat belt, he falls forward and hits his head against the steering wheel. Why?

5.20 The velocity of a body of mass $2 \mathrm{~kg}$ as a function of $t$ is given by $\mathbf{v}(t)=2 t \hat{\mathbf{i}}+t^{2} \hat{\mathbf{j}}$. Find the momentum and the force acting on it, at time $t=2 \mathrm{~s}$.

5.21 A block placed on a rough horizontal surface is pulled by a horizontal force $F$. Let $f$ be the force applied by the rough surface on the block. Plot a graph of $f$ versus $F$.

5.22 Why are porcelain objects wrapped in paper or straw before packing for transportation?

5.23 Why does a child feel more pain when she falls down on a hard cement floor, than when she falls on the soft muddy ground in the garden?

5.24 A woman throws an object of mass $500 \mathrm{~g}$ with a speed of $25 \mathrm{~m} \mathrm{~s}^{1}$.

(a) What is the impulse imparted to the object?

(b) If the object hits a wall and rebounds with half the original speed, what is the change in momentum of the object?

5.25 Why are mountain roads generally made winding upwards rather than going straight up?

SA

5.26 A mass of $2 \mathrm{~kg}$ is suspended with thread AB (Fig. 5.5). Thread CD of the same type is attached to the other end of $2 \mathrm{~kg}$ mass. Lower thread is pulled gradually, harder and harder in the downward directon so as to apply force on $A B$. Which of the threads will break and why?

Fig. 5.5

5.27 In the above given problem if the lower thread is pulled with a jerk, what happens?

5.28 Two masses of $5 \mathrm{~kg}$ and $3 \mathrm{~kg}$ are suspended with help of massless inextensible strings as shown in Fig. 5.6. Calculate $T_{1}$ and $T_{2}$ when whole system is going upwards with acceleration $=2 \mathrm{~m} \mathrm{~s}^{2}$ (use $g=9.8 \mathrm{~m} \mathrm{~s}^{-2}$ ). Fig. 5.6

5.29 Block A of weight $100 \mathrm{~N}$ rests on a frictionless inclined plane of slope angle $30^{\circ}$ (Fig. 5.7). A flexible cord attached to A passes over a frictonless pulley and is connected to block B of weight W. Find the weight W for which the system is in equilibrium.

5.30 A block of mass $M$ is held against a rough vertical wall by pressing it with a finger. If the coefficient of friction between the block and the wall is $\mu$ and the acceleration due to gravity is $g$, calculate the minimum force required to be applied by the finger to hold the block against the wall?

Fig. 5.7

5.31 A $100 \mathrm{~kg}$ gun fires a ball of $1 \mathrm{~kg}$ horizontally from a cliff of height $500 \mathrm{~m}$. It falls on the ground at a distance of $400 \mathrm{~m}$ from the bottom of the cliff. Find the recoil velocity of the gun. (acceleration due to gravity $=10 \mathrm{~m} \mathrm{~s}^{-2}$ )

5.32 Figure 5.8 shows $(x, t),(y, t)$ diagram of a particle moving in 2-dimensions.

(a)

(b)

Fig. 5.8

If the particle has a mass of $500 \mathrm{~g}$, find the force (direction and magnitude) acting on the particle.

5.33 A person in an elevator accelerating upwards with an acceleration of $2 \mathrm{~m} \mathrm{~s}^{-2}$, tosses a coin vertically upwards with a speed of $20 \mathrm{~m}$ $\mathrm{s}^{1}$. After how much time will the coin fall back into his hand? $\left(g=10 \mathrm{~m} \mathrm{~s}^{-2}\right).$

LA

5.34 There are three forces $\mathbf{F_1}$, $\mathbf{F_2}$ and $\mathbf{F}_{3}$ acting on a body, all acting on a point $\mathrm{P}$ 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.

5.35 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 $p T$, where $p$ is some number greater than 1 . Calculate the co-efficient of friction between the body and the rough plane.

5.36 Figure 5.9 shows $\left(v_{x}, t\right)$, and $\left(v_{y}, t\right)$ diagrams for a body of unit mass. Find the force as a function of time.

(a)

(b)

Fig. 5.9

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

Fig. 5.10

5.38 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 $\mathbf{F}=-m \omega^{2} \mathbf{r}$.

5.39 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 $v_{\mathrm{s}}$ 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.

5.40 There are four forces acting at a point $\mathrm{P}$ produced by strings as shown in Fig. 5.11, which is at rest. Find the forces $\mathbf{F_1}$ and $\mathbf{F_2}$.

Fig. 5.11

5.41 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 $a$ ?

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

(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.

Solutions 5

5.1 (c)

5.2 (b)

5.3 (c)

5.4 (c)

5.5 (d)

5.6 (c)

5.7 (a)

5.8 (b)

5.9 (b)

5.10 (a), (b) and (d)

5.11 (a), (b), (d) and (e)

5.12 (b) and (d)

5.13 (b), (c)

5.14 (c), (d)

5.15 (a), (c)

5.16 Yes, due to the principle of conservation of momentum.

Initial momentum $=50.5 \times 5 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}$

Final momentum $=(50 v+0.5 \times 15) \mathrm{kg} \mathrm{m} \mathrm{s}{ }^{-1}$

$$ v=4.9 \mathrm{~m} \mathrm{~s}^{-1} \text {, change in speed }=0.1 \mathrm{~m} \mathrm{~s}^{-1} $$

5.17 Let $R$ be the reading of the scale, in newtons.

Effective downward acceleration $=\frac{50 g-R}{50}=g$

$R=5 g=50 \mathrm{~N}$. (The weighing scale will show $5 \mathrm{~kg}$ ).

5.18 Zero; $-\frac{3}{2} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}$

5.19 The only retarding force that acts on him, if he is not using a seat belt comes from the friction exerted by the seat. This is not enough to prevent him from moving forward when the vehicle is brought to a sudden halt.

5.20 $\mathbf{p}=8 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}, \quad \mathbf{F}=(4 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}) \mathrm{N}$

5.21 $f=F$ until the block is stationary.

$f$ remains constant if $F$ increases beyond this point and the block starts moving.

5.22 In transportation, the vehicle say a truck, may need to halt suddenly. To bring a fragile material, like porcelain object to a sudden halt means applying a large force and this is likely to damage the object. If it is wrapped up in say, straw, the object can travel some distance as the straw is soft before coming to a halt. The force needed to achieve this is less, thus reducing the possibility of damage.

5.23 The body of the child is brought to a sudden halt when she/he falls on a cement floor. The mud floor yields and the body travels some distance before it comes to rest, which takes some time. This means the force which brings the child to rest is less for the fall on a mud floor, as the change in momentum is brought about over a longer period.

5.24 $\begin{array}{ll}\text { (a) } 12.5 \mathrm{~N} \mathrm{~s} & \text { (b) } 18.75 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}\end{array}$

5.25 $f=\mu R=\mu \mathrm{mg} \cos \theta$ is the force of friction, if $\theta$ is angle made by the slope. If $\theta$ is small, force of friction is high and there is less chance of skidding. The road straight up would have a larger slope.

5.26 $\mathrm{AB}$, because force on the upper thread will be equal to sum of the weight of the body and the applied force.

5.27 If the force is large and sudden, thread CD breaks because as CD is jerked, the pull is not transmitted to AB instantaneously (transmission depends on the elastic properties of the body). Therefore, before the mass moves, CD breaks.

5.28 $T_{1}=94.4 \mathrm{~N}, T_{2}=35.4 \mathrm{~N}$

5.29 $\mathrm{~W}=50 \mathrm{~N}$

5.30 If $F$ is the force of the finger on the book, $F=N$, the normal reaction of the wall on the book. The minimum upward frictional force needed to ensure that the book does not fall is $M g$. The frictional force $=\mu \mathrm{N}$. Thus, minimum value of $F=\frac{M g}{\mu}$.

5.31 $0.4 \mathrm{~m} \mathrm{~s}^{-1}$

5.32 $x=t, y=t^{2}$

$a_{x}=0, \quad a_{y}=2 \mathrm{~m} \mathrm{~s}^{-1}$

$\mathrm{F}=0.5 \times 2=1 \mathrm{~N}$. along $y$-axis.

5.33 $t=\frac{2 \mathrm{~V}}{g+a}=\frac{2 \times 20}{10+2}=\frac{40}{12}=\frac{10}{3}=3.33 \mathrm{~s}$.

5.34 (a) Since the body is moving with no acceleration, the sum of the forces is zero $\mathbf{F_1}+\mathbf{F_2}+\mathbf{F_3}=0$. Let $\mathbf{F_1}, \mathbf{F_2}, \mathbf{F_3}$ be the three forces passing through a point. Let $\mathbf{F_1}$ and $\mathbf{F_2}$ be in the plane A (one can always draw a plane having two intersecting lines such that the two lines lie on the plane). Then $\mathbf{F_1}+\mathbf{F_2}$ must be in the plane A.

Since $\mathbf{F_3}=-\left(\mathbf{F_1}+\mathbf{F_2}\right), \mathbf{F_3}$ is also in the plane A.

(b) Consider the torque of the forces about P. Since all the forces pass through $\mathrm{P}$, the torque is zero. Now consider torque about another point 0 . Then torque about 0 is

Torque $=\mathbf{O P} \times\left(\mathbf{F_1}+\mathbf{F_2}+\mathbf{F_3}\right)$

Since $\mathbf{F_1}+\mathbf{F_2}+\mathbf{F_3}=0$, torque $=0$

5.35 General case

$s=\frac{1}{2} a t^{2} \Rightarrow t=\sqrt{2 s / a}$

Smooth case

Acceleration $a=g \sin \theta=g / \sqrt{2}$

$\therefore t_{1}=\sqrt{2 \sqrt{2} \mathrm{~s} / g}$

Exemplar Problems-Physics

Rough case

Acceleration $a=g \sin \theta-\mu g \cos \theta$

$$ =(1-\mu) g / \sqrt{2} $$

$\therefore t_{2}=\sqrt{\frac{2 \sqrt{2} s}{(1-\mu) g}}=p t_{1}=p \sqrt{\frac{2 \sqrt{2} s}{g}}$

$\Rightarrow \frac{1}{1-\mu}=p^{2} \Rightarrow \mu=1-\frac{1}{p^{2}}$

5.36

$$ \begin{aligned} & v_{x}=2 t \quad 0<t \leq 1 \quad v_{y}=t \quad 0<t<1 \mathrm{~s} \\ & =2(2-t) \quad 1<t<2 \quad=11<t \\ & =0 \quad 2<t \\ & F_{x}=2 ; \quad 0<t<1 \quad F_{y}=1 \quad 0<t<1 \mathrm{~s} \\ & =-2 ; \quad 1 \mathrm{~s}<t<2 \mathrm{~s} \quad=0 \quad 1 \mathrm{~s}<t \\ & =0 ; \quad 2 \mathrm{~s}<t \\ & \mathbf{F}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}} \quad 0<t<1 \mathrm{~s} \\ & =-2 \hat{\mathbf{i}} 1 \mathrm{~s}<t<2 \mathrm{~s} \\ & =0 \quad 2 \mathrm{~s}<t \end{aligned} $$

5.37 For DEF

$$ \begin{aligned} & \not h \frac{v^{2}}{R}=\not h g \mu \\ & v_{\max }=\sqrt{g \mu R}=\sqrt{100}=10 \mathrm{~m} \mathrm{~s}^{-1} \end{aligned} $$

For $\mathrm{ABC}$

$$ \frac{v^{2}}{2 R}=g \mu, v=\sqrt{200}=14.14 \mathrm{~m} \mathrm{~s}^{-1} $$

Time for DEF $=\frac{\pi}{2} \times \frac{100}{10}=5 \pi \mathrm{s}$

Time for $A B C=\frac{3 \pi}{2} \frac{200}{14.14}=\frac{300 \pi}{14.14} \mathrm{~s}$

For FA and DC $=2 \times \frac{100}{50}=4 \mathrm{~s}$

Total time $=5 \pi+\frac{300 \pi}{14.14}+4=86.3 \mathrm{~s}$

5.38 $\frac{d \mathbf{r}}{d t}=\mathbf{v}=-\hat{\mathbf{i}} \omega A \sin \omega t+\hat{\mathbf{j}} \omega B \cos \omega t$

$$ \frac{d \mathbf{v}}{d t}=\mathbf{a}=-\omega^{2} \mathbf{r} ; \mathbf{F}=-m \omega^{2} \mathbf{r} $$

$x=A \cos \omega t, y=B \sin \omega t \Rightarrow \frac{x^{2}}{A^{2}}+\frac{y^{2}}{B^{2}}=1$

5.39 For (a) $\frac{1}{2} v_{z}{ }^{2}=g H \quad v_{z}=\sqrt{2 g H}$

Speed at ground $=\sqrt{v_{s}^{2}+v_{z}^{2}}=\sqrt{v_{s}^{2}+2 g H}$

For (b) also $\left[\frac{1}{2} m v_{s}{ }^{2}+m g H\right]$ is the total energy of the ball when it hits the ground.

So the speed would be the same for both (a) and (b).

5.40 $F_{2}=\frac{F_{3}+F_{4}}{\sqrt{2}}=\frac{2+1}{\sqrt{2}}=\frac{3}{\sqrt{2}} \mathrm{~N}$

$F_{1}+\frac{F_{3}}{\sqrt{2}}=\frac{F_{4}}{\sqrt{2}}$

$F_{1}=\frac{F_{4}-F_{3}}{\sqrt{2}}=\frac{1}{\sqrt{2}} \mathrm{~N}$

5.41 (a) $\theta=\tan ^{-1} \mu$

(b) $\quad m g \sin \alpha-\mu m g \cos \alpha$

(c) $\quad m g(\sin \alpha+\mu \cos \alpha)$

(d) $m g(\sin \theta+\mu \cos \theta)+m a$.

5.42 (a) $F-(500 \times 10)=(500 \times 15)$ or $F=12.5 \times 10^{3} \mathrm{~N}$, where $F$ is the upward reaction of the floor and is equal to the force downwards on the floor, by Newton’s 3rd law of motion

(b) $\mathrm{R}-(2500 \times 10)=(2500 \times 15)$ or $R=6.25 \times 10^{4} \mathrm{~N}$, action of the air on the system, upwards. The action of the rotor on the surrounding air is $6.25 \times 10^{4} \mathrm{~N}$ downwards.

(c) Force on the helicopter due to the air $=6.25 \times 10^{4} \mathrm{~N}$ upwards.