Exercises
(g = 10 m s⁻² for numerical calculations)
Q4.1. Give the magnitude and direction of the net force acting on
(a) a raindrop falling with constant speed (b) a cork of mass 10 g floating on water (c) a kite held stationary (d) a car moving at constant 30 km h⁻¹ on a rough road (e) a high-speed electron in space, free from all fields
Ans.
(a) Constant speed → zero acceleration → by Newton's 2nd law, net force = zero.
(b) Cork floating on water is in equilibrium (not accelerating) → net force = zero.
(c) Kite is stationary → zero acceleration → net force = zero.
(d) Constant velocity → zero acceleration → net force = zero.
(e) No electric, magnetic field, and far from matter (gravity negligible) → net force = zero.
2. A pebble of mass 0.05 kg is thrown vertically upwards. Net force on pebble during: (a) upward motion (b) downward motion (c) at highest point. Does answer change if thrown at 45°?
Answer
In all three cases (neglecting air resistance), only gravity acts:
(a) Force = mg = 0.05 × 10 = 0.5 N, downward
(b) Force = 0.5 N, downward
(c) At highest point (momentarily at rest): force = 0.5 N, downward
The answers do not change if thrown at 45° — gravity acts vertically downward regardless of the direction of throw.
Q4.3. Net force on a stone of mass 0.1 kg in a train
(a) dropped from stationary train (b) dropped from train at constant 36 km h⁻¹ (c) dropped from train accelerating at 1 m s⁻² (d) lying on floor of train accelerating at 1 m s⁻²
Ans.
(a) Only gravity acts: F = mg = 0.1 × 10 = 1 N, vertically downward
(b) Train at constant velocity means no horizontal force. After being dropped, only gravity acts: 1 N, vertically downward
(c) After being dropped, stone has only gravity acting (no longer in contact with train): 1 N, vertically downward. The horizontal velocity the stone had when dropped doesn't change the net force after release.
(d) Stone is at rest relative to accelerating train → stone also accelerates at 1 m s⁻² horizontally. Net force = √(F_gravity² + F_horizontal²)
F_h = ma = 0.1 × 1 = 1 N (horizontal, forward); F_g = 1 N (vertical, downward)
Net F = √(1² + 1²) = √2 ≈ 1.41 N at 45° below horizontal in the direction of acceleration.
4. A particle of mass m moves in a circle with speed v at the end of a string of length l on a smooth table. The net force (centripetal) is: (i) T (ii) T − mv²/l (iii) T + mv²/l (iv) 0
Answer
The only horizontal force on the particle is the tension T in the string, which provides the centripetal force. T = mv²/l (directed toward centre).
The net force on the particle = centripetal force = T = mv²/l
⇒ correct answer is (i) T
5. A retarding force of 50 N acts on a body of mass 20 kg moving at 15 m s⁻¹. How long to stop?
Answer
Given: F = −50 N; m = 20 kg; u = 15 m s⁻¹; v = 0
Retardation a = F/m = −50/20 = −2.5 m s⁻²
v = u + at
⇒ 0 = 15 + (−2.5)t
⇒ t = 15/2.5 = 6 s
6. A constant force on a body of mass 3.0 kg changes speed from 2.0 m s⁻¹ to 3.5 m s⁻¹ in 25 s. Find force.
Answer
Given: m = 3.0 kg; u = 2.0 m s⁻¹; v = 3.5 m s⁻¹; t = 25 s
a = (v − u)/t = (3.5 − 2.0)/25 = 1.5/25 = 0.06 m s⁻²
F = ma = 3.0 × 0.06 = 0.18 N in the direction of motion
7. A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Find acceleration magnitude and direction.
Answer
Net force F = √(8² + 6²) = √(64 + 36) = √100 = 10 N
Acceleration a = F/m = 10/5 = 2 m s⁻²
Direction: tanθ = 6/8 = 0.75
⇒ θ = arctan(0.75) ≈ 37° with the 8 N force
8. A three-wheeler at 36 km h⁻¹ stops in 4.0 s. Mass of three-wheeler = 400 kg, driver = 65 kg. Find retarding force.
Answer
Total mass = 400 + 65 = 465 kg; u = 36 km h⁻¹ = 10 m s⁻¹; v = 0; t = 4.0 s
Retardation a = (v − u)/t = (0 − 10)/4 = −2.5 m s⁻²
Retarding force = ma = 465 × 2.5 = 1162.5 N ≈ 1163 N
9. A rocket of lift-off mass 20,000 kg is blasted upward with initial acceleration 5.0 m s⁻². Find thrust.
Answer
For upward acceleration, net upward force = ma
Thrust − weight = ma
Thrust = m(a + g) = 20000 × (5.0 + 10) = 20000 × 15 = 3.0 × 10⁵ N
10. A body of mass 0.40 kg moves at 10 m s⁻¹ northward. Force of 8.0 N southward applied at t = 0. Predict position at t = −5 s, 25 s, 100 s.
Answer
Taking north as positive: u = +10 m s⁻¹; F = −8.0 N; a = −8.0/0.40 = −20 m s⁻²
At t = 0, x = 0.
t = −5 s: Before force was applied, body moved at constant 10 m s⁻¹ northward.
x = 10 × (−5) = −50 m (50 m south of x=0)
t = 25 s: First find when body stops: v = u + at = 10 − 20t = 0
⇒ t = 0.5 s
Body reverses after t = 0.5 s.
x at 25 s: x = ut + ½at² = 10(25) + ½(−20)(25)² = 250 − 6250 = −6000 m (6000 m south)
t = 100 s: x = 10(100) + ½(−20)(100)² = 1000 − 100000 = −99000 m (99 km south)
11. A truck accelerates at 2.0 m s⁻². At t = 10 s, a stone is dropped from 6 m height on the truck. Find (a) velocity and (b) acceleration of stone at t = 11 s.
Answer
At t = 10 s, truck's velocity = at = 2 × 10 = 20 m s⁻¹ (horizontal)
Stone is dropped
⇒ it has the same horizontal velocity as truck at moment of release = 20 m s⁻¹
After release, no horizontal force on stone (neglect air resistance)
⇒ horizontal velocity stays 20 m s⁻¹
Vertical: initial vertical velocity = 0; g acts downward
At t = 11 s (1 s after release):
Vertical velocity = g × 1 = 10 m s⁻¹ (downward)
Horizontal velocity = 20 m s⁻¹
(a) Speed = √(20² + 10²) = √500 ≈ 22.4 m s⁻¹ at angle arctan(10/20) = 26.6° below horizontal
(b) After release, only gravity acts
⇒ acceleration = g = 10 m s⁻², vertically downward
12. A 0.1 kg bob on a 2 m string oscillates. Speed at mean position = 1 m s⁻¹. Trajectory if string cuts at: (a) extreme position (b) mean position?
Answer
(a) At extreme position: velocity = 0. When string is cut, only gravity acts
⇒ bob falls vertically downward (straight line, under gravity from rest).
(b) At mean position: velocity is maximum (1 m s⁻¹), horizontal. When string is cut, the bob is a projectile with initial horizontal velocity 1 m s⁻¹ and acts under gravity
⇒ parabolic trajectory.
13. A man of mass 70 kg is on a weighing scale in a lift. Readings when lift moves:
Answer
(a) upward at constant 10 m s⁻¹ (b) downward with acceleration 5 m s⁻² (c) upward with acceleration 5 m s⁻² (d) in free fall
Ans.
Apparent weight W' = m(g + a) for upward acceleration; m(g − a) for downward acceleration
(a) Constant velocity
⇒ a = 0; W' = mg = 70 × 10 = 700 N
⇒ reading = 70 kg
(b) a = 5 m s⁻² downward; W' = m(g − a) = 70 × (10 − 5) = 70 × 5 = 350 N
⇒ reading = 35 kg
(c) a = 5 m s⁻² upward; W' = m(g + a) = 70 × (10 + 5) = 70 × 15 = 1050 N
⇒ reading = 105 kg
(d) Free fall: a = g downward; W' = m(g − g) = 0
⇒ reading = 0 (weightlessness)
14. Position-time graph for 4 kg particle. Find force for t < 0, t > 4 s, 0 < t < 4 s, and impulse at t = 0 and t = 4 s.
Answer
From a typical x-t graph for this problem (straight line for t<0 and t>4s, slope changes between them):
For t < 0: straight line
⇒ constant velocity
⇒ a = 0
⇒ F = 0
For t > 4 s: straight line
⇒ constant velocity
⇒ a = 0
⇒ F = 0
For 0 < t < 4 s: straight line but different slope
⇒ constant velocity
⇒ a = 0
⇒ F = 0
Note: x-t graph appears to show three straight segments with slope changes only at t = 0 and t = 4s.
Impulse at t = 0: J = Δp = m(v₂ − v₁) = 4 × (slope at 0<t<4 − slope at t<0)
If slopes are 3 m/s (t<0) and 0 m/s (0<t<4): J = 4(0−3) = −12 N·s
Impulse at t = 4 s: J = 4(3 − 0) = +12 N·s
15. Two bodies 10 kg and 20 kg on smooth horizontal surface tied by string. Force F = 600 N applied to: (i) A (ii) B. Find tension in each case.
Answer
Total mass = 30 kg; Acceleration = F/total mass = 600/30 = 20 m s⁻²
(i) Force applied to A (10 kg):
T acts on 20 kg body: T = 20 × 20 = 400 N
(ii) Force applied to B (20 kg):
T acts on 10 kg body: T = 10 × 20 = 200 N
16. Two masses 8 kg and 12 kg connected by string over frictionless pulley. Find acceleration and tension when released.
Answer
Let 12 kg go down (heavier). Net force = (m₂ − m₁)g = (12 − 8) × 10 = 40 N
Total mass = 12 + 8 = 20 kg
Acceleration a = 40/20 = 2 m s⁻²
For 12 kg: 12g − T = 12a
⇒ T = 12(g − a) = 12(10 − 2) = 12 × 8 = 96 N
Check with 8 kg: T − 8g = 8a
⇒ T = 8(g + a) = 8(10 + 2) = 96 N ✓
17. Show that if a nucleus at rest disintegrates into two smaller nuclei, the products must move in opposite directions.
Answer
Initial momentum of nucleus at rest = 0
By conservation of momentum: total final momentum = total initial momentum = 0
If nucleus
⇒ product 1 (mass m₁, velocity v₁) + product 2 (mass m₂, velocity v₂)
m₁v₁ + m₂v₂ = 0 → m₁v₁ = −m₂v₂
Since m₁ and m₂ are positive, v₁ and v₂ must have opposite signs (opposite directions). Therefore, the two products must move in opposite directions.
18. Two billiard balls (0.05 kg each) moving oppositely at 6 m s⁻¹ collide and rebound with same speed. Impulse imparted to each?
Answer
Take initial direction of ball 1 as positive: u₁ = +6 m s⁻¹
⇒ after collision: v₁ = −6 m s⁻¹
Impulse on ball 1 = m(v₁ − u₁) = 0.05 × (−6 − 6) = 0.05 × (−12) = −0.6 N·s (magnitude = 0.6 N·s)
Similarly for ball 2: impulse = +0.6 N·s
Impulse imparted to each ball = 0.6 N·s in the direction opposite to their initial motion.
19. A shell of mass 0.020 kg is fired at 80 m s⁻¹ from a gun of mass 100 kg. Find recoil speed of gun.
Answer
Initial momentum = 0 (system at rest)
By conservation of momentum: m_shell × v_shell + m_gun × v_gun = 0
0.020 × 80 + 100 × v_gun = 0
v_gun = −(0.020 × 80)/100 = −1.6/100 = −0.016 m s⁻¹
Recoil speed = 0.016 m s⁻¹ (opposite to direction of shell)
20. A batsman deflects a ball by 45° without changing its speed of 54 km h⁻¹. Mass = 0.15 kg. Find impulse.
Answer
v = 54 km h⁻¹ = 15 m s⁻¹; angle between initial and final velocity = 45°
Change in momentum (impulse): |Δp| = m × |Δv|
|Δv| = √(v₁² + v₂² − 2v₁v₂ cos45°) = √(15² + 15² − 2 × 15 × 15 × cos45°)
= √(225 + 225 − 450 × (1/√2)) = √(450 − 318.2) = √131.8 ≈ 11.48 m s⁻¹
Impulse = 0.15 × 11.48 ≈ 4.16 N·s
21. A stone of mass 0.25 kg on a string of radius 1.5 m makes 40 rev/min. Find tension. Also find maximum speed if string can withstand 200 N.
Answer
ω = 40 rev/min = 40 × 2π/60 = 4π/3 rad s⁻¹
Tension = centripetal force = mω²r = 0.25 × (4π/3)² × 1.5
= 0.25 × (16π²/9) × 1.5 = 0.25 × 1.5 × 16π²/9 = 0.25 × 1.5 × 17.546 ≈ 6.57 N
For maximum speed: T_max = mv²_max/r
⇒ v_max = √(T_max × r/m) = √(200 × 1.5/0.25) = √1200 ≈ 34.6 m s⁻¹
22. If the string in 4.21 breaks, which correctly describes the stone's trajectory?
Answer
(a) moves radially outward (b) flies off tangentially (c) flies off at angle with tangent
Ans.
(b) The stone flies off tangentially.
At the moment the string breaks, the stone has a velocity directed tangentially to the circle. No force now acts horizontally (string is gone). By Newton's 1st law, the stone continues moving in the direction of its instantaneous velocity — which is the tangent to the circle at that point.
Q4.23. Explain why
(a) a horse cannot pull a cart in empty space
(b) passengers thrown forward when bus stops suddenly
(c) easier to pull a lawn mower than push
(d) cricketer moves hands backward while catching
Ans.
(a) A horse pulls the cart by pushing its feet against the ground (reaction force propels it forward). In empty space, there is no ground to push against, so no reaction force and no motion is possible.
(b) By Newton's 1st law (law of inertia): the passengers' bodies tend to maintain their forward motion when the bus brakes suddenly. The lower body is decelerated by friction with the seat, but the upper body continues forward — hence passengers are thrown forward.
(c) When pulling, the force has an upward component, reducing the normal force (friction). When pushing, the force has a downward component, increasing the normal force (friction). Less friction when pulling → easier.
(d) By extending the time of impact (Δt), the impulsive force F = Δp/Δt is reduced. Moving hands backward increases Δt, thus decreasing F, reducing the chance of injury to the hands.
Important Extra Questions
Q. State Newton's three laws of motion and give one example of each.
Ans.
1st Law (Law of Inertia): A body at rest remains at rest and a body in uniform motion continues in uniform motion in a straight line unless acted upon by an external force.
Example: A ball rolling on a frictionless surface keeps moving indefinitely.
2nd Law: The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction of the force: F = ma.
Example: A car accelerates when engine provides force; heavier the car, more force needed for same acceleration.
3rd Law: To every action there is an equal and opposite reaction. Forces always occur in pairs.
Example: A rocket expels gas backward (action); gas pushes rocket forward (reaction).
Q. Define impulse. A force of 100 N acts on a body for 0.01 s. Find the impulse and change in momentum.
Ans.
Impulse = Force × Time = F × Δt. It equals the change in momentum of the body.
Impulse = 100 × 0.01 = 1 N·s
By impulse-momentum theorem: Δp = impulse = 1 kg m s⁻¹
Q. What is the principle of conservation of momentum? State its conditions.
Answer
If no external force acts on a system of particles, the total linear momentum of the system remains constant (conserved).
Mathematically: if F_ext = 0, then dp/dt = 0
⇒ p_total = constant
Condition: The system must be isolated (no net external force). Internal forces (action-reaction pairs between particles within the system) cancel out in pairs and don't change total momentum.
Examples: rocket propulsion, recoil of gun, collision between billiard balls.
Q. A 2 kg block on a rough horizontal surface (μ = 0.4) is pulled by a horizontal force of 12 N. Find acceleration. (g = 10 m s⁻²)
Ans.
Normal force N = mg = 2 × 10 = 20 N
Friction f = μN = 0.4 × 20 = 8 N
Net force = 12 − 8 = 4 N
Acceleration = F_net/m = 4/2 = 2 m s⁻²
Q. Explain the working of a frictionless Atwood machine with masses m₁ and m₂ (m₂ > m₁). Derive expressions for acceleration and tension.
Ans.
In an Atwood machine, two masses hang over a frictionless pulley connected by a string.
For mass m₂ (going down): m₂g − T = m₂a ... (1)
For mass m₁ (going up): T − m₁g = m₁a ... (2)
Adding (1) and (2): (m₂ − m₁)g = (m₁ + m₂)a
a = (m₂ − m₁)g/(m₁ + m₂)
From (2): T = m₁(g + a) = m₁ × [g + (m₂−m₁)g/(m₁+m₂)]