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NCERT Class 11 Physics Chapter 4 Laws of Motion
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Laws of Motion
Chapter: 4
| Part – I |
EXERCISE
(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.
Ans: The drop is moving with a constant speed, so according to Newton’s first law, the net force on it is zero.
(b) A cork of mass 10g floating on water.
Ans: As the cork is floating on water, its weight is balanced by the upthrust due to water. Hence net force on the cork is zero.
(c) A kite skillfully held stationary in the sky.
Ans: Kite is held stationary, according to Newton’s first law the force on the kite is zero.
(d) A car moving with a constant velocity of 30 km/h on a rough road.
Ans: The car moving with a constant velocity, the net force on the car is zero.
(e) A high-speed electron in space far from all material objects, and free of electric and magnetic fields.
Ans: The net force on electron is zero.
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.
Ans: Given:
F = net force.
m = mass of the pebble = 0.05kg.
g = 9.8 m/s.
F = 0.05 × 9.8 = 0.49N.
(b) During its downward motion.
Ans: Given:
F = net force.
m = mass of the pebble = 0.05kg.
g = 9.8 m/s.
F = 0.05 × 9.8 = 0.49N.
(c) At the highest point where it is momentarily at rest.
Ans: Given:
F= net force.
m = mass of the pebble = 0.05kg.
g= 9.8 m/s.
F = 0.05 x 9.8 = 0.49N.
Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction?
Ans: If the pebble was thrown at an angle of 45° the answers remain the same. Regardless of the angle of projection, as long as we ignore air resistance.
3. Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg. (Neglect air resistance throughout.)
(a) Just after it is dropped from the window of a stationary train.
Ans: The only force acting on the stone is gravity:
Force = mass x acceleration due to gravity (F = mg)
F = 0.1kg × 9.8 m/s-2
= 0.98 N.
(b) Just after it is dropped from the window of a train running at a constant velocity of 36 km/h.
Ans: The window of a train running at a constant velocity of 36 km/h, no force acts on the stone due to the motion of the train. Thus, Force on stone = Mg – 0.98N.
(c ) Just after it is dropped from the window of a train accelerating with 1 m s-2.
Ans: Force on the stone F = 0.1 × 1 = 0.1 N.
This force also acts vertically downwards.
(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.
Ans: The net force on the stone is given by: F = 0.1 × 1 = 0.1 N.
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:
T is the tension in the string. [Choose the correct alternative].
Ans: (i) is correct.
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: Given that:
The retarding force F = 50N
The mass m = 20 kg
Acceleration a = F/m
= 50 N / 20 kg = 2.5 m/s2
Since this is a retarding force the acceleration will be negative
a = – 2.5 m/s2
Now,
v is the final velocity
u = 15 m/s
a = – 2.5 m/s2
v = u + at
0 = 15 m/s + (- 2.5 m/s2) × t
– 15 m/s = – 2.5 m/s2 × t
t = 15m/s / 2.5 m/s2
t = 6s.
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: Given that:
u = 2 ms-1
v = 3.5 ms-1
t = 25s
v = v – u
= 3.5m/s – 2.0m/s
= 1.5 m/s
Calculate of acceleration:
a = 1.5 m/s / 25s
a = 0.06 m/s2
Newton’s second law to find the force:
F = ma
Where:
m = 3.0 kg
F = 3.0 kg x 0.06 m/s2
= 0.18 N.
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: Here F1 = 8N
F2 = 6N
F = √(F1)2 + (F2)2
= √ (8)2 + (6)2
= 10N
Therefore,
a = F/M
= 10/5 = 2ms-2
If F makes angle θ with the direction F1, then
cos θ = F1/F
= 8/10
= 0.8
= θ = cos-1 (0.8) = 36.87°.
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: Given u = 36 km/h
36km/h = 36 x 1000m/3600 = 10 m/s.
Given:
Initial speed u = 10m/s
v = 0 m/s
t = 4.0s
a = u – v /t
= 0 – 10m/s/ 4.0s
= – 2.5 m/s2
m = 400 kg + 65 kg
= 465 kg
F = ma = 465 (-2.5)
= 1162.5 N.
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: Here m = 2 × 104 kg, 𝚫v/ 𝚫t = 5 ms-2
As, 𝚫v/𝚫t = vr/m × 𝚫m/𝚫t -g
Hence vr = 𝚫m/𝚫t
= m. 𝚫v/𝚫v/𝚫t + mg
= 2 × 104 × 5 + 2 × 104 × 9.8
= 105 + 1.96 × 105
= 2.96 × 105 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: Given: Here m = 0.40 kg
u = 10 ms-1
F = – 8N (retarding force)
Time duration of force application t force = 30s
F = ma
a = F/m
= – 8.0/0.40
= – 20m/s2
Determination of t = -5s, t = 25s and t=100s.
t = – 5s
Before the velocity is applied the body’s constant velocity of 10m/s
Position at t = -5s
S-5 = 10 x (-5) + ½ x 0 x (-5)2
= -50m
(ii) Position at, t = 25s
S25 = 10 x 25 + ½ x (-20) x (25)2
= – 6000m
= – 6km
(iii) Position at t = 30s
= S30 = 10 × 30 + ½ × (-20) × (30)2
= 300 – 0.5 × 20 v 9000
= 300 – 9000
= – 8700
Velocity at t = 30s
v = u + at
v30 = 10 + (-20) × (30)
= 10 – 600
= – 590 m/s
Additional time after force stops: t additional
= 100 – 30 = 70s
S100-30 = -590 × 70 + ⅕ x 0 × (70)2
= – 41300 ms-1
Total distance = S30 + S30-100
= – 41300 – 8700
= – 50000m
= – 50km.
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) As v = u + at, when v = vx, u= 0
a = 2 ms-2, t = 10s
We get v = 0 + 2 × 10s = 20 ms-1
vy = 0 + 9.8 × 0.1 = 0.98 ms-1
= 10.1 – 10 = 0.1s
vR = √v2x + v2y
= √(20)2 + (0.98)2
= 20.02 ms-1
And tan θ = vy /vx
= 0.98 / 20
= 0.049
Or θ = tan-1 (0.049) = 2.8°
(b) Acceleration at 10.1s = g = 9.8 ms-2.
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) As the bob of a simple pendulum has no velocity at the extreme position. If the string is cut, it will fall vertically downwards.
(b) At the mean position, f the string is cut, bob is acted by vertical gravitational force = a = 9.8 𝑚𝑠−2 Hence bob will behave like a projectile and follows a parabolic path.
The bob will have a periodic path as it is having horizontal velocity.
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.
Ans: R = mg
Given m = 70kg
g = 9.8 m/s
R = mg
= 70kg × 9.8
= 686 N.
(b) Downwards with a uniform acceleration of 5 m s-2.
Ans: R’ = m (g – a)
Given:
m = 70kg
g = 9.8
a = 5ms-2
= 70 (9.8 – 5)
= 70 × 4.8
= 336 N.
(c) Upwards with a uniform acceleration of 5 m s-2. What would be the readings on the scale in each case?
Ans: R’’ = m (g + a)
Given:
m = 70kg
g = 9.8
a = 5ms-2
= 70 ( 9.8 + 5)
= 70 × 14.8
= 1036 N.
(d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?
Ans: R’’’ = m (g – g)
Given:
m = 70kg
g = 9.8
= 70 kg ( 9.8 – 9.8)
= 70 × 0
= 0 N.
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).
Ans: (a) For t < 0 and t > 4s, the particle is at rest as the position does not change w.r.t time. Clearly no force acts on the particle during these intervals.
Further, for 0 < t < 4s, the position of the particle continuously changes with respect to time. As the position – time graph is a straight line, it represents uniform motion and there is no acceleration. Hence it is also clear that no force acts on the particle during these intervals.
(b) Because the velocity is uniform from O and A, hence velocity at O = velocity at A = slope of the graph OA = ¾ ms-1
Impulse (at t= 0) = change in momentum.
= Final momentum-initial momentum
= 0 – mv
= – 4 ( ¾)
= – 3 kg ms-1.
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 the string. What is the tension in the string in each case?
Ans: Given:
Horizontal force F = 600 N
Mass of body, A, M1 = 10 kg
Mass of body B, m2 = 20 kg
Total mass of the system m = m1 + m2 = 30 kg
Here, a = 600 / 10 + 20
= 20 ms-2
Force is applied on 10 kg mass,
= 600 – T = 10 × 20
T = 400 N.
Force applied on 20 kg mass,
= 600 – T = 20 × 20
T = 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 m1, = 8 kg and m2 = 12
For mass m1, m1a = T – M1g
8a = T – 8 × 9.8 (i)
For mass m2 = mg – T
= 12a = 12 × 9.8 – T (ii)
Adding equation of (i) and (ii)
(8 + 12) a = T – 8 × 9.8 + 12 × 9.8 – T
20a = 4 × 9.8
a = 4 × 9.8 / 20
= 1.96 ms-2
Equation (i)
T = 8 × 1.96 MS-2 + 8 × 9.8
= 94.04 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. 4.18 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: Let m = initial mass of the nucleus.
m1 and m2 are masses after disintegration and v1 and v2 are their respective velocities.
Now, initial monuments of the nucleus
= m × 0 = 0
Final momentum of the nucleus
= m1v1 + m2v2
Using laws of conservation of momentum i.e., initial momentum of the system.
= Final momentum of the systems,
= 0 = m1v1 + m2v2
= m2v2 = m1v1
Or v2 = (-) m1v1 / m2.
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: Initial momentum of ball
= mu = 0.05 × 6
= 0.3 kg ms-1.
Final momentum of ball
= mv = 0.05 × (-6)
= – 0.3 kg ms-1
Change in momentum
Impulse J imparted to each ball is the change in momentum
J1 = Pfinal1 – Pinitial
J1 = 0. 3 kg m/s – 0. 3 kg m/s
J1 = – 0.6 kg m/s
J2 = Pfinal2 – Pinitial
J2 = 0.3 kg m/s – 0.3 kg m/s
J2 = 0 kg m/s
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: Given:
m shell = 0.020 kg
Mgun = 100 kg
v shell = 80 m/s
v gun to determine: ?
According to the conservation of momentum:
= m shell x v shell = – m gun × v gun
= 0.020 × 80 = – 100 × v gun
= 1.6 = – 100 × v gun
= vgun = 1.6 / – 100
= v gun = – 0.0016 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: Suppose the point O as the position of the bat. AO line shows the path along which the ball strikes the bat with velocity v and OB is the path showing defection such that <AOB = 45°
Here, Initial momentum of the ball = mu cos θ
= 0.15 × 54 × 1000 × 22.5 / 3600°
= 0.15 × 15 × 0. 9239 along NO
Final momentum of the ball = mu cos θ – (- mu cos θ)
2 mu cos θ
= 2 × 0.25 × 15 × 0.9239
= 4.16 kg ms-1.
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: M = 0.25 kg , r = 1.5
v = 40 rev.min-1 = 40/60 rev s-1
ω = 2rv = 2𝜋 × 40 / 60
= 1.33𝜋 rad s-1
Here, tension = centripetal force = Mv2 / r
= Mrω = 0.25 x 1× (1.33𝜋)2
= 5.26 N
The string can withstand a maximum tension of 200 N. if v max be the maximum speed of the stone, then,
200 = Mv2max / r
= v max =
= 25.82ms-1.
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: Option b is correct. When the string breaks the stone will move in the direction of the velocity at that instant. It is because the speed of the stone at any instant is directed along a tangent to the circular path at the point.
23. Explain why:
(a) A horse cannot pull a cart and run in empty space.
Ans: An empty space has no such response force. As a result, a horse cannot draw a cart and run in open space.
(b) Passengers are thrown forward from their seats when a speeding bus stops suddenly.
Ans: It is because of the Inertia of motion.
(c) It is easier to pull a lawn mower than to push it.
Ans: Pulling creates a downward force on the handles, increasing friction between the mower and the ground, which improves traction.
(d) A cricketer moves his hands backwards while holding a catch.
Ans: By increasing time, force is reduced.
