Chapter 08 Force and Laws of Motion Exercise
Exercises
1. An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on
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Answer
According to the Newton’s 1st law of motion, no net force is required to move an object which is moving with a constant velocity. So, when an object experiences a net zero external unbalanced force, then it can move with a non-zero velocity. Also, if an object is initially at rest not net force acts upon it, thus the object may not move at all.
2. When a carpet is beaten with a stick, dust comes out of it, Explain.
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Answer
When a carpet is beaten with a stick, then the fibers of the carpet attain the state of motion while the dust particles remain in rest due to inertia of rest. Therefore, the dust particles fall down.
3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?
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Answer
It is advised to tie any luggage kept on the roof of a bus because when the bus stops suddenly, it readily comes in the state of rest but the luggage remains in the state of motion. So, due to inertia of motion, the luggage will move forward and would fall down from the roof of the bus.
If the bus starts suddenly, then bus comes in the state of motion but luggage remains in the state of rest. Due to inertia of rest, the luggage moves in the backward direction and will fall down.
4. A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because
(a) the batsman did not hit the ball hard enough.
(b) velocity is proportional to the force exerted on the ball.
(c) there is a force on the ball opposing the motion.
(d) there is no unbalanced force on the ball, so the ball would want to come to rest.
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Answer
(c) because the ball slows down to rest as the force of friction acting between the ground and the ball opposes the motion of the ball.
5. A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of $400 m$ in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 tonnes (Hint: 1 tonne $=1000 kg$.)
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Answer
The truck starts from rest, so initial velocity $u=0$, Distances $=400 m, t=20 s, m=7$ tons $=7 \times 1000=7000 kg$,
From 2nd equation of motion,
$ \begin{aligned} s & =u t+\frac{1}{2} a t^{2} \\ 400 & =0 \times 20+\frac{1}{2} \times a \times(20)^{2} \\ 400 & =200 a \\ a & =2 ms^{-2} \end{aligned} $
From Newton’s 2nd law of motion, force acting on the truck
$ \begin{aligned} F & =m a \\ & =7000 \times 2=14000 N \\ & =1.4 \times 10^{4} N \end{aligned} $
6. A stone of $1 kg$ is thrown with a velocity of $20 m s^{-1}$ across the frozen surface of a lake and comes to rest after travelling a distance of $50 m$. What is the force of friction between the stone and the ice?
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Answer
Mass of stone $m=1 kg$, initial velocity $u=20 ms-1$,
Final velocity $v=O$ (Therefore, the stone comes to rest), distance covered $s=50 m$.
From third equation of motion,
$v 2=u 2+2 as$
(0) $2=(20) 2+2 a(50)$
$100 a=-400$
$a=-4 m s-2$
Here negative sign shows that there is retardation in the motion of stone.
Force of friction between stone and ice = Force required to stop the stone
$ \begin{aligned} & =ma \\ & =1 \times-4=-4 N \text{ OR } 4 N \end{aligned} $
7. A $8000 kg$ engine pulls a train of 5 wagons, each of $2000 kg$, along a horizontal track. If the engine exerts a force of 40000 $N$ and the track offers a friction force of $5000 N$, then calculate:
(a) the net accelerating force and
(b) the acceleration of the train.
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Answer
(a) Net accelerating force $=$ Force exerted by engine-Friction force (Here frictional force is subtracted because it opposes the motion) $=40000-5000=35000 N$
$=3.5 \times 10^{4} N$
(b) From Newton’s second law of motion,
Accelerating force $=$ Mass of the train $\times$ Acceleration of train
$a = \frac{F}{m}$
Mass of train $=$ Mass of engine + Mass of all wagons
$=8000+5 \times 2000$
$=8000+10000=18000 kg$
Acceleration $=35000=35=1.95 ms^{-2}$
18000 18
8. An automobile vehicle has a mass of $1500 kg$. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of $1.7 m s^{-2}$ ?
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Answer
Mass, $m=1500 kg$, Acceleration, $a=-1.7 ms-2$
From Newton’s second law of motion,
F ma
$=1500 \times(-1.7)$
$=-2550 N$
9. What is the momentum of an object of mass $m$, moving with a velocity $v$ ?
(a) $(m v)^{2}$
(b) $m v^{2}$
(c) $1 / 2 m v^{2}$
(d) $m v$
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Answer: -(d) because the momentum of an object of mass $m$ moving with a velocity $v$.
$p=m v$
10. Using a horizontal force of $200 N$, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?
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Answer
The cabinet will move across the floor with constant velocity, if there is no net external force is applied on it.
Here a horizontal force of $200 N$ is applied on the cabinet, so for the net force to be zero, an external force of $200 N$ should be applied on the cabinet in opposite direction.
Thus, the frictional force $=200 N$
(Frictional force always acts in the direction opposite to direction of motion.)
11. According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.
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Answer
The logic given by the student is not correct because two equal and opposite forces cancel each other only in the case if they act on the same body. According to Newton’s third law, action and reaction force always act on two different bodies, so they cannot cancel each other. When a massive truck is pushed, then the truck may not move because the force applied is not sufficient to move the truck.
12. A hockey ball of mass $200 g$ travelling at $10 m s^{-1}$ is struck by a hockey stick so as to return it along its original path with a velocity at $5 m s^{-1}$. Calculate the magnitude of change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
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Answer
Mass of hockey ball
$ m=200 g=0.2 kg $
Initial velocity, $u=10 ms-1$
Final velocity, $v=-5 m s-1$
(Since, final velocity of the ball is in the direction opposite to the direction of initial velocity.)
Change in momentum of the ball = Final momentum - Initial momentum
$=m v-m u$
$=m(v-u)$
$=0.2(-5-10)$
$=0.2 \times-15=-3 kg ms-1$
13. A bullet of mass $10 g$ travelling horizontally with a velocity of $150 m s^{-1}$ strikes a stationary wooden block and comes to rest in $0.03 s$. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.
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Answer
Mass of bullet, $m=10 g=\frac{10}{1000} kg=0.001 kg$
Initial velocity, $u=150 ms^{-1}$
Final velocity, $v=0$
Time, $\quad t=0.03 s$
From equation of motion,
$ \begin{aligned} v & =u+a t \\ 0 & =150+a \times 0.03 \\ a & =\frac{-150}{0.03} \\ & =-5000 ms^{-2} \end{aligned} $
Magnitude of the force applied by the bullet on the block, $F=m a$
$ =0.01 \times -5000 ms^{-2} $
$F = -50 N$
14. An object of mass $1 kg$ travelling in a straight line with a velocity of $10 m s^{-1}$ collides with, and sticks to, a stationary wooden block of mass $5 kg$. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.
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Answer
Given, mass of the object $(m_1) = 1 kg$
Mass of the block $(m_2) = 5 kg$
Initial velocity of the object $(u_1) = 10 m/s$
Initial velocity of the block $(u_2) = 0$
Mass of the resulting object $= m_1 + m_2 = 6 kg$
Velocity of the resulting object $(v) =?$
Total momentum before the collision $= m_1u_1 + m_2u_2 = (1kg) × (10m/s) + 0 = 10 kg.m.s^{-1}$
As per the law of conservation of momentum, the total momentum before the collision is equal to the total momentum post the collision. Therefore, the total momentum post the collision is also $10 kg.m.s^{-1}$
Now, $(m_1 + m_2) × v = 10kg.m.s-1$
Therefor, $v = \frac{10 kg.m.s^{-1}}{6 kg} =1.66 ms^{-1}$
The resulting object moves with a velocity of $1.66$ meters per second.
15. An object of mass $100 kg$ is accelerated uniformly from a velocity of $5 m s^{-1}$ to $8 m s^{-1}$ in $6 s$. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.
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Answer
Given, mass of the object $(m) = 100kg$
Initial velocity $(u) = 5 m/s$
Terminal velocity $(v) = 8 m/s$
Time period $(t) = 6s$
Now, initial momentum $(m \times u) = 100kg \times 5m/s = 500 kg.m.s^{-1}$
Final momentum $(m \times v) = 100kg × 8m/s = 800 kg.m.s^{-1}$
Acceleration of the object (a) $=\frac{v-u}{t}= \frac{8-5}{6}ms^{-2}$
Therefore, the object accelerates at $0.5 ms^{-2}$. This implies that the force acting on the object $(F = ma)$ is equal to:
$F = (100kg) × (0.5 ms^{-2}) = 50 N$
Therefore, a force of 50 N is applied on the 100kg object, which accelerates it by $0.5 ms^{-2}$.
16. Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result the insect died. Rahul while putting an entirely new explanation
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Answer
The explanation given by Rahul is correct, as there is no external force on the system, so both the insect and motorcar experienced the same force and a change in momentum.
17. How much momentum will a dumb-bell of mass $10 \mathrm{~kg}$ transfer to the floor if it falls from a height of $80 \mathrm{~cm}$ ? Take its downward acceleration to be $10 \mathrm{~m} \mathrm{~s}^{-2}$.
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missingAdditional Exercises
A1. The following is the distance-time table of an object in motion:
| Time in seconds | Distance in metres |
|---|---|
| 1 | 1 |
| 2 | 8 |
| 3 | 27 |
| 4 | 64 |
| 5 | 125 |
| 6 | 216 |
| 7 | 343 |
(a) What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, or zero?
(b) What do you infer about the forces acting on the object?
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Answer
(a) The distance covered by the object at any time interval is greater than any of the distances covered in previous time intervals. Therefore, the acceleration of the object is increasing.
(b) As per the second law of motion, force = mass $\times$ acceleration. Since the mass of the object remains constant, the increasing acceleration implies that the force acting on the object is increasing as well.
A2. Two persons manage to push a motorcar of mass $1200 kg$ at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of $0.2 m s^{-2}$. With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort.)
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Answer
Given, mass of the car $(m) = 1200kg$
When the third person starts pushing the car, the acceleration (a) is $0.2 ms^{-2}$ . Therefore, the force applied by the third person $(F = ma)$ is given by:
$F = 1200kg × 0.2 ms^{-2} = 240N$
The force applied by the third person on the car is 240 N. Since all 3 people push with the same muscular effort, the force applied by each person on the car is $240 N$.
A3. A hammer of mass $500 g$, moving at $50 m s^{-1}$, strikes a nail. The nail stops the hammer in a very short time of $0.01 s$. What is the force of the nail on the hammer?
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Answer
Given, mass of the hammer $(m) = 500g = 0.5kg$
Initial velocity of the hammer $(u) = 50 m/s$
Terminal velocity of the hammer $(v) = 0$ (the hammer is stopped and reaches a position of rest).
Time period $(t) = 0.01s$
Therefore, the acceleration of the hammer is given by: $a = \frac{v-u}{t} = \frac{0-50ms^{-1}}{0.01 s}$
$a = -5000 ms^{-2}$
Therefore, the force exerted by the hammer on the nail $(F = ma)$ can be calculated as:
$F = (0.5kg) \times (-5000 ms^{-2} ) = -2500 N$
As per the third law of motion, the nail exerts an equal and opposite force on the hammer. Since the force exerted on the nail by the hammer is $-2500 N$, the force exerted on the hammer by the nail will be $+2500 N$.
A4. A motorcar of mass $1200 kg$ is moving along a straight line with a uniform velocity of $90 km / h$. Its velocity is slowed down to $18 km / h$ in $4 s$ by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.
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Answer
Given, mass of the car $(m) = 1200kg$
Initial velocity $(u) = 90 \text{ km/hour} = 25 \text{ meters/sec}$
Terminal velocity $(v) = 18 \text{ km/hour} = 5 \text{ meters/sec}$
Time period $(t) = 4 \text{ seconds}$
The acceleration of the car can be calculated with the help of the formula: $a = \frac{v-u}{t}$
$a= \frac{5-25}{4} ms^{-2} = -5ms^{-2}$
Therefore, the acceleration of the car is $-5 ms^{-2}.$
Initial momentum of the car $= m × u = (1200kg) × (25m/s) = 30,000 kg.m.s^{-1}$
Final momentum of the car $= m × v = (1200kg) × (5m/s) = 6,000 kg.m.s^{-1}$
