Class 11 Physics
Chapter - 4 Dev Library
Laws of Motion
Part - I
EXERCISE
(For simplicity in numerical calculations, take g = 10 m s-2)
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1. Give the magnitude and direction of the net force acting on:
(a) A drop of rain falling down with a constant speed.
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ns: The drop is moving with a constant speed, so according to Newton's first law,
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the net force on it is zero.
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(b) A cork of mass 10g floating on water.
ns: As the cork is floating on water, its weight is balanced by the upthrust due to
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water. Hence net force on the cork is zero.
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(c) A kite skillfully held stationary in the sky.
ns: Kite is held stationary, according to Newton's first law the force on the kite is
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zero.
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(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.
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( 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.
. A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction
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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:
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F = net force.
m = mass of the pebble = 0.05kg.
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g = 9.8 m/s.
F = 0.05 × 9.8 = 0.49N.
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(c) At the highest point where it is momentarily at rest.
Ans:Given:
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F= net force.
m = mass of the pebble = 0.05kg.
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g= 9.8 m/s.
F = 0.05 x 9.8 = 0.49N.
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o your answers change if the pebble was thrown at an angle of 45° with the
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horizontal direction?
ns: If the pebble was thrown at an angle of 45° the answers remain the same.
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Regardless of the angle of projection, as long as we ignore air resistance.
. Give the magnitude and direction of the net force acting on a stone of mass
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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.
ns: The window of a train running at a constant velocity of 36 km/h, no force acts
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on the stone due to the motion of the train. Thus, Force on stone = Mg - 0.98N.
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( c ) Just after it is dropped from the window of a train accelerating with 1 m
s-2.
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Ans: Force on the stone F = 0.1 × 1 = 0.1 N.
This force also acts vertically downwards.
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( 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.
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. One end of a string of length l is connected to a particle of mass m and the
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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:
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T is the tension in the string. [Choose the correct alternative].
Ans: (i) is correct.
. A constant retarding force of 50 N is applied to a body of mass 20 kg moving
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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,
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v is the final velocity
u = 15 m/s
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a = - 2.5 m/s2
v = u + at
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0 = 15 m/s + (- 2.5 m/s2) × t
- 15 m/s = - 2.5 m/s2 × t
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t = 15m/s / 2.5 m/s2
t = 6s.
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. A constant force acting on a body of mass 3.0 kg changes its speed from 2.0
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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?
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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:
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m = 3.0 kg
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F = 3.0 kg x 0.06 m/s2
= 0.18 N.
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. 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
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F2 = 6N
F = √(F1) 2+ (F2) 2
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= √ (8)2 + (6)2
= 10N
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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°.
. The driver of a three-wheeler moving with a speed of 36 km/h sees a child
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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.
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Given:
Initial speed u = 10m/s
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v = 0 m/s
t = 4.0s
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a = u - v /t
= 0 - 10m/s/ 4.0s
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= - 2.5 m/s2
m = 400 kg + 65 kg
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= 465 kg
F = ma = 465 (-2.5)
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= 1162.5 N.
. A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial
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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.
0. A body of mass 0.40 kg moving initially with a constant speed of 10 m s-1
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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
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u = 10 ms-1
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F = - 8N (retarding force)
Time duration of force application tforce = 30s
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F = ma
a = F/m
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= - 8.0/0.40
= - 20m/s2
Determination of t = -5s, t = 25s and t=100s.
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t = - 5s
Before the velocity is applied the body’s constant velocity of 10m/s
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Position at t = -5s
S-5 = 10 x (-5) + ½ x 0 x (-5)2
= -50m
(ii) Position at, t = 25s
S2 5 = 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
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Velocity at t = 30s
v = u + at
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v30 = 10 + (-20) × (30)
= 10 - 600
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= - 590 m/s
Additional time after force stops: tadditional
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= 100 - 30 = 70s
S1 00-30 = -590 × 70 + ⅕ x 0 × (70)2
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= - 41300 ms-1
Total distance = S30 + S30-100
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= - 41300 - 8700
= - 50000m
= - 50km.
1. A truck starts from rest and accelerates uniformly at 2.0 m s-2. At t = 10 s, a
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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
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= 20.02 ms-1
And tan θ = vy /vx
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= 0.98 / 20
= 0.049
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Or θ = tan-1 (0.049) = 2.8°
(b) Acceleration at 10.1s = g = 9.8 ms-2.
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2. A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long
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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.
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ns: (a) As the bob of a simple pendulum has no velocity at the extreme position. If
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the string is cut, it will fall vertically downwards.
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( 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:
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m = 70kg
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g = 9.8
a = 5ms-2
br
= 70 (9.8 - 5)
= 70 × 4.8
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= 336 N.
( c) Upwards with a uniform acceleration of 5 m s-2. What would be the
readings on the scale in each case?
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Ans: R’’ = m (g + a)
Given:
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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)
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= 70 × 0
= 0 N.
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4. Figure 4.16 shows the position-time graph of a particle of mass 4 kg. What
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is the:
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(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).
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ns:(a) For t < 0 and t > 4s, the particle is atrest as the position does not change
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w.r.t time. Clearly no force acts on the particle during these intervals.
urther, for 0 < t < 4s, the position of the particle continuously changes with respect
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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.
5. Two bodies of masses 10 kg and 20 kg respectively kept on a smooth,
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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?
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Ans:Given:
Horizontal force F = 600 N
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Mass of body, A, M1 = 10 kg
Mass of body B, m2= 20 kg
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Total mass of the system m = m1 + m2 = 30 kg
Here, a = 600 / 10 + 20
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= 20 ms-2
Force is applied on 10 kg mass,
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= 600 - T = 10×20
T = 400 N.
Force applied on 20 kg mass,
= 600 - T = 20×20
T = 200 N.
6. Two masses 8 kg and 12 kg are connected at the two ends of a light
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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)
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(8 + 12) a = T - 8×9.8 + 12×9.8 - T
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20a = 4×9.8
a = 4×9.8 / 20
br
= 1.96 ms-2
Equation (i)
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T = 8×1.96 MS-2 + 8×9.8
= 94.04 N.
7. A nucleus is at rest in the laboratory frame of reference. Show that if it
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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?
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Ans: Let m = initial mass of the nucleus.
1 and m2 are masses after disintegration and v1 andv2 are their respective
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velocities.
Now, initial monuments of the nucleus
= m×0 = 0
Final momentum of the nucleus
= m1v 1 + m2v 2
Using laws of conservation of momentum i.e., initial momentum of the system.
= Final momentum of the systems,
= 0 = m1v 1 + m2v 2
= m2v 2 = m1v 1
Or v2 = (-) m1v 1 / m2.
8. Two billiard balls each of mass 0.05 kg moving in opposite directions with
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speed 6 m s -1 collide and rebound with the same speed. What is the impulse
imparted to each ball due to the other?
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Ans: Initial momentum of ball
= mu = 0.05×6
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= 0.3 kg ms-1.
Final momentum of ball
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= mv = 0.05×(-6)
= - 0.3 kg ms-1
Change in momentum
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Impulse J imparted to each ball is the change in momentum
J1 = Pfinal1 - Pinitial
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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
9. A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle
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speed of the shell is 80 m s-1, what is the recoil speed of the gun?
Ans:Given:
mshell = 0.020 kg
Mgun = 100 kg
vshell = 80 m/s
vgun to determine: ?
According to the conservation of momentum:
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= mshellx vshell = - mgun × vgun
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= 0.020×80 = - 100×vgun
= 1.6 = - 100×vgun
br
= vgun = 1.6 / - 100
= vgun = - 0.0016 m/s
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0. A batsman deflects a ball by an angle of 45° without changing its initial
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speed which is equal to 54 km/h. What is the impulse imparted to the ball ?
(Mass of the ball is 0.15 kg.)
ns: Suppose the point O as the position of the bat. AO line shows the path along
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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
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= 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.
1. A stone of mass 0.25 kg tied to the end of a string is whirled round in a
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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
y
= 1.33𝜋 rad s-1
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Here, tension = centripetal force = Mv2 / r
= Mrω = 0.25 x 1×(1.33𝜋)2
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= 5.26 N
he string can withstand a maximum tension of 200 N. if vmaxbe the maximum
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speed of the stone, then,
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200 = Mv2max / r
= vmax =
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= 25.82ms-1.
2. If, in Exercise 4.21, the speed of the stone is increased beyond the
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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?
ns: Option b is correct. When the string breaks the stone will move in the direction
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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.
ns: An empty space has no such response force. As a result, a horse cannot draw
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a cart and run in open space.
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( b) Passengers are thrown forward from their seats when a speeding bus
stops suddenly.
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Ans: It is because of the Inertia of motion.
(c) It is easier to pull a lawn mower than to push it.
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ns: 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.
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Ans: By increasing time, force is reduced.
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