Key Applications of Newton’s Laws:
Understanding Everyday Motion: Understanding why objects fall, why projectiles follow curved paths, and why objects roll down inclines.
Engineering and Design: Designing structures, machines, and vehicles that can withstand forces and operate efficiently.
Space Exploration: Understanding the motion of spacecraft and celestial bodies.
Sports: Analyzing the mechanics of various sports, such as the motion of balls, the forces exerted by athletes, and the impact of equipment.
By understanding Newton’s laws, we can explain and predict the motion of objects in a wide range of situations.
(For simplicity in numerical calculations, take g = 10 m s-2)
1. Give the magnitude and direction of the net force acting on
(a) a drop of rain falling down with a constant speed,
(b) a cork of mass 10 g floating on water,
(c) a kite skillfully held stationary in the sky,
(d) a car moving with a constant velocity of 30 km/h on a rough road,
(e) a high-speed electron in space far from all material objects, and free of electric and magnetic fields.
Ans : Net Force on Various Objects:
Raindrop falling at constant speed:
Net force: 0 N
Explanation: A constant velocity implies zero acceleration, leading to zero net force.
Cork floating on water:
Net force: 0 N
Explanation: The cork is stationary, indicating no net force acting on it.
Kite held stationary in the sky:
Net force: 0 N
Explanation: The stationary kite implies zero acceleration, resulting in zero net force.
Car moving at constant velocity:
Net force: 0 N
Explanation: Constant velocity means zero acceleration, leading to zero net force.
High-speed electron in empty space:
Net force: 0 N
Explanation: The electron moves at a constant velocity without external forces, resulting in zero net force.
2. A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble,
(a) during its upward motion,
(b) during its downward motion,
(c) at the highest point where it is momentarily at rest.
Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction? (Ignore air resistance. )
Ans : Throughout the pebble’s motion, the only force acting on it is gravity. This is because we’re ignoring air resistance.
(a) During its upward motion
Direction of net force: Downward
Magnitude of net force: The magnitude of the force due to gravity is given by:
F = mg
F = 0.05 kg * 9.81 m/s²
F ≈ 0.49 N
(b) During its downward motion
Direction of net force: Downward
Magnitude of net force: The same as in (a), 0.49 N
(c)At the instant the pebble stops rising and begins to descend
Direction of net force: Downward
Magnitude of net force: 0.49 N
The direction and magnitude of the net force on the pebble remain the same throughout its motion, regardless of whether it was thrown vertically upwards or at an angle of 45° with the horizontal.
This is because the force of gravity acts vertically downward, and its magnitude depends only on the mass of the pebble. The horizontal component of the initial velocity does not affect the vertical forces acting on the pebble.
3. Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg,
(a) just after it is dropped from the window of a stationary train,
(b) just after it is dropped from the window of a train running at a constant velocity of 36 km/h,
(c ) just after it is dropped from the window of a train accelerating with 1 m s-2 ,
(d) lying on the floor of a train which is accelerating with 1 m s-2, the stone being at rest relative to the train. Neglect air resistance throughout.
Ans : The total force exerted on the 0.1 kg stone is :
(a) 0 N downwards. As the stone is not accelerating horizontally while on the train, the only force is gravity.”
(b) 0 N downwards. Because the stone isn’t accelerating horizontally relative to the train, there’s no horizontal net force.
(c) 0.1 N downwards. Due to the downward acceleration of 1 m/s², the net force on the stone is calculated as mass times acceleration.
(d) 0.1 N upwards. Since the stone is stationary relative to the train, the only horizontal force is the normal force from the floor, which is upward.
4. One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is :
[Choose the correct alternative].
Ans : T
The tension in the string, T, acts as the centripetal force, pulling the particle towards the center of the circular path. This force is necessary to keep the particle moving in a circular trajectory.
5. A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 m s-1. How long does the body take to stop ?
Ans : To determine how long it takes the body to halt, we employ the equation
v = u + at
where v is the final velocity (0 m/s), u is the initial velocity (15 m/s), a is the acceleration, and t is the time.
Newton’s second law of motion provides a means to calculate acceleration
F = ma
where F = ma
In this case, the force is a retarding force, so it is negative:
F = -50 N
Putting the values in equation,
-50 N = 20 kg * a
a = -50 N / 20 kg
a = -2.5 m/s^2
Let’s substitute the values of u, v, and a into the equation to find t.
0 m/s = 15 m/s + (-2.5 m/s^2) * t
-15 m/s = -2.5 m/s^2 * t
t = 15 m/s / 2.5 m/s^2
t = 6 s
Therefore, the body takes 6 seconds to stop.
6. A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s-1 to 3.5 m s-1 in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force ?
Ans : To find the magnitude and direction of the force, we can use Newton’s second law of motion:
ma = F
where,
F = force,
m = mass,
a = acceleration.
The acceleration can be found using the equation:
a = (v – u) / t
where,
v = final velocity,
u = initial velocity,
t = time.
Substituting the values into the equations, we get:
a = (3.5 m/s – 2.0 m/s) / 25 s = 0.06 m/s²
F = 3.0 kg * 0.06 m/s²
= 0.18 N
Since the direction of the motion of the body remains unchanged, the direction of the force is also the same as the direction of the motion.
7. A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.
Ans : To find the magnitude and direction of the acceleration, we can use Newton’s second law of motion:
F_net = m * a
where F_net is the net force acting on the body, m is the mass of the body, and a is the acceleration of the body.
Since the two forces are perpendicular, we can find the net force using the Pythagorean theorem:
F_net = sqrt(F1^2 + F2^2)
where F1 and F2 represent the magnitudes of the forces acting on the body
Substituting the given values, we get:
F_net = sqrt(8 N^2 + 6 N^2)
F_net = sqrt(100 N^2)
F_net = 10 N
Now, we can find the acceleration:
a = F_net / m
a = 10 N / 5 kg
a = 2 m/s^2
The net force causes the object to accelerate in its direction. Since the two forces are perpendicular, the net force will be at an angle of θ to the 8 N force, where:
tan(θ) = F2 / F1
tan(θ) = 6 N / 8 N
θ = tan^-1(0.75)
θ ≈ 36.87°
Therefore, the magnitude of the acceleration is 2 m/s² and its direction is 36.87° to the 8 N force.
8. The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle ? The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg.
Ans : To find the average retarding force on the vehicle, we can use Newton’s second law of motion:
F = ma
where,
F = force,
m = mass,
a = acceleration.
Here, convert the speed from km/h to m/s:
36 km/h =
36 * 1000 m
————————–
3600 s
= 10 m/s
Next, we can find the acceleration using the equation:
a = (v – u) / t
where,
v = final velocity (0 m/s),
u = initial velocity (10 m/s),
t = time (4 s).
Substituting the values, we get:
a =
(0 m/s – 10 m/s)
—————————–
4 s
a = -2.5 m/s^2
The negative sign indicates that the acceleration is in the opposite direction of the initial velocity, which is what we expect for a retarding force.
Now, we can find the total mass of the three-wheeler and the driver:
m_total = m_vehicle + m_driver
m_total = 400 kg + 65 kg
m_total = 465 kg
By using Newton’s second law, we determine the Force
F = ma
F = 465 kg * (-2.5 m/s^2)
F = -1162.5 N
The magnitude of the force is 1162.5 N, and the direction is opposite to the initial direction of motion (i.e., a retarding force).
9. A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 m s-2. Calculate the initial thrust (force) of the blast.
Ans : To calculate the initial thrust of the rocket, we can use Newton’s second law of motion:
F_net = ma
where F_net is the net force acting on the rocket, m is the mass of the rocket, and a is the acceleration of the rocket.
In this case, the net force is the difference between the thrust force (F_thrust) and the weight of the rocket (W):
F_net = F_thrust – W
Here, weight of the rocket is :
W = mg
with g, the acceleration due to gravity, being 9.81 m/s².
Putting these equations into the 1st equation,
F_thrust – mg = ma
F_thrust = m(a + g)
Now, we can substitute the given values:
F_thrust = 20,000 kg * (5.0 m/s² + 9.81 m/s²)
F_thrust ≈ 296,200 N
Therefore, the initial thrust of the blast is approximately 296,200 N.
10. A body of mass 0.40 kg moving initially with a constant speed of 10 m s-1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at that time to be x = 0, and predict its position at t = –5 s, 25 s, 100 s.
Ans : To solve this problem, we can break it down into two phases:
Phase 1: Before the force is applied (t < 0)
The body is moving north at a constant speed of 10 m/s.
Since there is no net force acting on it, its velocity remains constant.
We can use the equation x = ut to find its position at any time before t = 0.
Phase 2: After the force is applied (0 ≤ t ≤ 30 s)
The body is now subject to a constant force of 8.0 N south.
This force will cause the body to decelerate.
We can use the equation v = u + at to find the velocity at any time t, and then integrate this to find the position.
Phase 3: After the force stops (t > 30 s)
The body will continue to move with its final velocity from phase 2.
We can use the equation x = ut to find its position at any time after t = 30 s.
Calculations:
Phase 1 (t < 0):
For t = -5 s, x = 10 m/s * (-5 s) = -50 m
Phase 2 (0 ≤ t ≤ 30 s):
First, find the acceleration: a = F/m = 8.0 N / 0.40 kg = -20 m/s² (negative because the force is south)
Use v = u + at to find the velocity at any time t: v = 10 m/s – 20 m/s² * t
Integrate this to find the position: x = 10t – (1/2) * 20t² = 10t – 10t²
Phase 3 (t > 30 s):
The final velocity at t = 30 s is v = 10 m/s – 20 m/s² * 30 s = -590 m/s
Use x = ut to find the position: x = -590 m/s * (t – 30 s)
Predictions:
t = -5 s: x = -50 m (the body is 50 m north of its starting position)
t = 25 s: x = 10 * 25 – 10 * 25² = -6000 m (the body is 6000 m south of its starting position)
t = 100 s: x = -590 m/s * (100 s – 30 s) = -41300 m (the body is 41300 m south of its starting position)
11. A truck starts from rest and accelerates uniformly at 2.0 m s-2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the
(a) velocity, and
(b) acceleration of the stone at t = 11s ?
(Neglect air resistance.)
Ans : (a) Horizontal Motion:
The only force acting on the stone is gravity, which acts vertically downwards. Therefore, there is no horizontal acceleration. The stone continues to move horizontally with a constant velocity of 20 m/s.
Vertical Motion:
The stone is in free fall under the influence of gravity. Using the equation of motion:
v = u + at
where u = 0 (initial vertical velocity), a = g = 10 m/s² (acceleration due to gravity), and t = 1 s (time after dropping).
v = 0 + 10 × 1 = 10 m/s
Resultant Velocity:
The horizontal and vertical components of the velocity are perpendicular to each other. Therefore, we can use the Pythagorean theorem to find the resultant velocity:
v = √(v_x² + v_y²) = √(20² + 10²) = √500 ≈ 22.36 m/s
The angle θ between the resultant velocity and the horizontal can be found using:
tan θ = v_y / v_x = 10 / 20 = 0.5
θ = tan⁻¹(0.5) ≈ 26.56°
(b) After being dropped from the car, the only force acting on the stone is gravity. This results in a constant downward acceleration of 10 m/s², which is also the net acceleration of the stone.
12 A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 m s-1 . What is the trajectory of the bob if the string is cut when the bob is
(a) at one of its extreme positions,
(b) at its mean position.
Ans : (a) When the string is cut at one of its extreme positions, the bob will fall vertically downwards. This is because at the extreme positions, the bob has no horizontal velocity. The only force acting on it after the string is cut is gravity, which pulls it straight downwards.
(b) When the string is cut at its mean position, the bob will follow a parabolic path. This is because at the mean position, the bob has a horizontal velocity. When the string is cut, the bob is no longer constrained to move in a circular path, so it will continue to move horizontally with its initial velocity while also being subject to the force of gravity, which will cause it to accelerate downwards. This combination of horizontal motion and downward acceleration results in a parabolic trajectory.
13. A man of mass 70 kg stands on a weighing scale in a lift which is moving
(a) upwards with a uniform speed of 10 m s-1 ,
(b) downwards with a uniform acceleration of 5 m s-2 ,
(c) upwards with a uniform acceleration of 5 m s-2 . What would be the readings on the scale in each case?
(d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity ?
Ans : (a) Uniform Upward Motion
* When the lift moves upward with a constant speed, there’s no acceleration.
* The reading on the scale, which measures the normal force (R), is equal to the weight of the person:
R = mg
= 70 kg × 10 m/s²
= 700 N
(b)Downward Acceleration:
* When the lift accelerates downward at 5 m/s², the net downward acceleration experienced by the person is (g – a).
R = m(g – a) = 70 kg × (10 m/s² – 5 m/s²)
= 350 N
(c) Upward Acceleration:
* When the lift accelerates upward at 5 m/s², the net upward acceleration experienced by the person is (g + a).
R = m(g + a) = 70 kg × (10 m/s² + 5 m/s²)
= 1050 N
(d)Free Fall:
* If the lift falls freely, both the person and the lift experience the same acceleration due to gravity (g).
* The net acceleration relative to the lift is zero.
* Therefore, the reading on the scale will be zero.
14. Figure 4.16 shows the position-time graph of a particle of mass 4 kg. What is the
(a) force on the particle for t < 0, t > 4 s, 0 < t < 4 s?
(b) impulse at t = 0 and t = 4 s ?
(Consider one-dimensional motion only). Fig. 4.16
Ans : (a) When t < 0: The particle is at rest as the position-time graph is horizontal.
Hence, the force acting on the particle is 0.
When 0 < t < 4 s: The particle moves with a constant velocity of 0.75 m/s as the graph OA has a constant slope. Since there’s no change in velocity, the acceleration is zero, and hence the force acting on the particle is zero.
When t > 4 s: The particle remains at a constant position, indicating zero velocity and zero acceleration.
Hence, the force acting on the particle is 0.
(b) Impulse at t = 0 s:
Initial velocity, u = 0 m/s
Final velocity, v = 0.75 m/s
Mass, M = 4 kg
Impulse = Change in momentum = M(v – u) = 4(0.75 – 0) = 3 kg m/s
Impulse at t = 4 s:
Initial velocity, u = 0.75 m/s
Final velocity, v = 0 m/s
Mass, M = 4 kg
Impulse = Change in momentum = M(v – u) = 4(0 – 0.75)
= -3 kg m/s
15. Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force F = 600 N is applied to
(i) A,
(ii) B along the direction of string.
What is the tension in the string in each case?
Ans : (i) Force applied on A:
Let T be the tension in the string.
The net force acting on A is F – T.
Using Newton’s second law:
F – T = m₁a
600 – T = 10a —(1)
i.e net force acting on B = T.
Using Newton’s second law:
T = m₂a
T = 20a —(2)
Solving equations (1) and (2):
600 – 20a = 10a
30a = 600
a = 20 m/s²
Substituting a in equation (2):
T = 20 * 20 = 400 N
Therefore, the tension in the string is 400 N.
(ii) Force applied on B:
The net force acting on A is T.
Using Newton’s second law:
T = m₁a
T = 10a —(1)
The net force acting on B is F – T.
Using Newton’s second law:
F – T = m₂a
600 – T = 20a —(2)
Solving equations (1) and (2):
600 – 10a = 20a
30a = 600
a = 20 m/s²
Substituting a in equation (1):
T = 10 * 20 = 200 N
Hence, the tension in the string = 200 N.
16. Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.
Ans : Let’s consider the following:
A= Mass of block (m1) = 8 kg
* Mass of block B (m2) = 12 kg
* Acceleration of the system = a
* Tension in the string = T
When the masses are released, they will move with the same acceleration ‘a’ in opposite directions.
For block A:
Net force acting on A = T – m1g
Using Newton’s second law,
T – m1g = m1a
T – 8g = 8a —-(1)
For block B:
Net force acting on B = m2g – T
Using Newton’s second law,
m2g – T = m2a
12g – T = 12a —-(2)
Adding equations (1) and (2):
4g = 20a
a = g/5
= 9.8/5
= 1.96 m/s²
Substituting the value of ‘a’ in equation (1):
T – 8g = 8a
T – 8*9.8 = 8*1.96
T = 78.4 + 15.68
= 94.08 N
hence, the acceleration of the masses is 1.96 m/s² and the tension in the string is 94.08 N.
17. A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.
Ans : Suppose we have a nucleus at rest with a mass ( M). It disintegrates into two smaller nuclei with masses m1 and m2, and velocities v1 and v2 respectively.
Following the principle of conservation of momentum:
Initial momentum = Final momentum
As the nucleus is initially at rest, its initial momentum is zero. Therefore, to conserve momentum, the total final momentum of the two smaller nuclei must also be zero.
Therefore,
m1v1 + m2v2 = 0
This equation implies that either m1 or m2 (or both) must be zero, or v1 and v2 must have opposite signs.
Since masses cannot be zero, it follows that v1 and v2 must have opposite signs.
This means that the two smaller nuclei must move in opposite directions to conserve momentum.
18. Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 m s -1 collide and rebound with the same speed. What is the impulse imparted to each ball due to the other ?
Ans : Impulse = Change in momentum
= Final momentum – Initial momentum
For each ball:
Initial momentum = 0.05 kg * 6 m/s = 0.3 kg m/s (let’s consider it positive for one ball and negative for the other)
Final momentum = 0.05 kg * (-6 m/s) = -0.3 kg m/s (the direction of motion has reversed)
Impulse = (-0.3 kg m/s) – (0.3 kg m/s)
= -0.6 kg m/s
Therefore, the impulse imparted to each ball due to the other is 0.6 kg m/s in the opposite direction of their initial motion.
19. A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m s-1, what is the recoil speed of the gun ?
Ans : follow the principle of conservation of linear momentum.
Initial momentum of the system (gun + shell) = 0 (both are at rest)
Final momentum of the system = Momentum of shell + Momentum of gun
Following the principle of conservation of momentum, …
Initial momentum = Final momentum
0 = m_shell * v_shell + m_gun * v_gun
Here:
* m_shell = 0.020 kg (mass of the shell)
* v_shell = 80 m/s (velocity of the shell)
* m_gun = 100 kg (mass of the gun)
* v_gun = recoil velocity of the gun (to be found)
Substituting the values:
0 = (0.020 kg)(80 m/s) + (100 kg)(v_gun)
v_gun = -0.016 m/s
The negative sign indicates that the gun moves in the opposite direction to the shell when it recoils.
Therefore, the recoil speed of the gun is 0.016 m/s.
20. A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball ?
(Mass of the ball is 0.15 kg.)
Ans :
21. A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string ? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N ?
Ans : Given:
* (m)Mass of the stone is 0.25 kg
* Radius of the circle, r = 1.5 m
* Angular speed, ω = 40 rev/min = (40 * 2π) / 60 rad/s = 4π/3 rad/s
Tension in the string:
The tension in the string provides the necessary centripetal force to keep the stone moving in a circle.
Centripetal force, Fc = mv² / r
Also, Fc = T (tension in the string)
Therefore,
T = mv² / r = mω²r
Substituting the values:
T = 0.25 * (4π/3)² * 1.5 ≈ 16.58 N
Maximum speed:
Given maximum tension, T_max = 200 N
Using the formula for centripetal force:
T_max = mv_max² / r
Solving for v_max:
v_max = √(T_max * r / m) = √(200 * 1.5 / 0.25)
≈ 34.64 m/s
Therefore, the maximum speed with which the stone can be whirled is approximately 34.64 m/s.
22 .If, in Exercise 4.21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks :
(a) the stone moves radially outwards,
(b) the stone flies off tangentially from the instant the string breaks,
(c) the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle ?
Ans : (b)
Explanation:
When the string breaks, the centripetal force that was keeping the stone moving in a circular path is suddenly removed. In the absence of any external force acting on the stone, it will continue to move in the direction of its instantaneous velocity, which is tangential to the circular path. This is in accordance with Newton’s first law of motion, which states that an object in motion continues to move in a straight line at a constant speed unless acted upon by an external force.
23. Explain why?
(a) a horse cannot pull a cart and run in empty space,
(b) passengers are thrown forward from their seats when a speeding bus stops suddenly,
(c) it is easier to pull a lawn mower than to push it,
(d) a cricketer moves his hands backwards while holding a catch.
Ans : (a) Newton’s third law states that forces always come in pairs. When a horse pulls a cart, it exerts a force on the cart, and the cart simultaneously exerts an equal and opposite force on the horse. In empty space, there is no ground or any other object to push against. So, the horse cannot exert a force on the cart, and hence, cannot pull it.
(b) This is due to inertia. When the bus stops suddenly, the lower part of the passenger’s body, in contact with the seat, comes to rest. However, the upper part of the body, due to inertia, tends to continue moving forward, causing the passenger to be thrown forward.
(c) When pulling a lawn mower, the force is applied at an angle that tends to lift the front wheels off the ground, reducing friction. This makes it easier to pull. On the other hand, when pushing a lawn mower, the force tends to press the front wheels down, increasing friction and making it harder to move.
(d) By moving his hands backward, the cricketer increases the time taken for the ball to come to rest. This reduces the rate of change of momentum, and hence, the force exerted on the cricketer’s hands. This helps to avoid injury.
