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A car of mass 1200 kg accelerates at 2.5 m s⁻². What is the net force on it?
Given: m = 1200 kg, a = 2.5 m s⁻²
Use: F = ma
F = 1200 × 2.5 = 3000 N, in the direction of the acceleration.
Answer: 3000 N
A 20 kg block on a horizontal floor is pulled by a horizontal force of 60 N. The friction is 20 N. Find the acceleration.
Given: m = 20 kg, F = 60 N, f = 20 N
Use: Fnet = F − f = ma
Net force = 60 − 20 = 40 N.
a = 40/20 = 2 m s⁻².
Answer: 2 m s⁻²
A fielder stops a 0.16 kg cricket ball moving at 25 m s⁻¹. Find the average force if the ball is stopped in (a) 0.05 s, (b) 0.5 s (hands drawn back).
Given: m = 0.16 kg, u = 25 m s⁻¹, v = 0
Use: F = Δp/Δt
Change in momentum = 0.16 × 25 = 4 kg m s⁻¹ (magnitude).
(a) F = 4/0.05 = 80 N.
(b) F = 4/0.5 = 8 N, ten times smaller.
Answer: (a) 80 N; (b) 8 N
A bullet of mass 20 g moving at 300 m s⁻¹ is stopped after penetrating 10 cm into a block. Find the average retarding force and the time of penetration.
Given: m = 0.020 kg, u = 300 m s⁻¹, s = 0.10 m
Use: v² = u² − 2as, F = ma
a = u²/(2s) = 90000/0.2 = 4.5 × 10⁵ m s⁻².
F = 0.020 × 4.5 × 10⁵ = 9000 N.
t = u/a = 300/(4.5 × 10⁵) = 6.7 × 10⁻⁴ s.
Answer: 9000 N; 6.7 × 10⁻⁴ s
A body of mass 2 kg is acted upon by a force of 6 N towards the east and a force of 8 N towards the north. Find the magnitude and direction of its acceleration.
Given: m = 2 kg, F₁ = 6 N (east), F₂ = 8 N (north)
Use: Add the forces as vectors, then use a = F/m
Net force = √(6² + 8²) = 10 N.
a = 10/2 = 5 m s⁻².
Direction: tan α = 8/6, α = 53.1° north of east (the same as the net force).
Answer: 5 m s⁻², 53° north of east
A car of mass 1000 kg starts from rest and is acted upon by a constant net force of 2000 N. Find its speed and the distance covered after 10 s.
Given: m = 1000 kg, F = 2000 N, t = 10 s
Use: a = F/m, v = at, s = ½at²
a = 2000/1000 = 2 m s⁻².
v = 2 × 10 = 20 m s⁻¹.
s = ½ × 2 × 100 = 100 m.
Answer: 20 m s⁻¹; 100 m
A 2 kg body moving at 3 m s⁻¹ is acted upon by a constant force of 10 N opposite to its motion. After how long does it stop momentarily? What is its velocity after 4 s?
Given: m = 2 kg, u = +3 m s⁻¹, F = −10 N
Use: a = F/m, v = u + at
a = −10/2 = −5 m s⁻².
Stops when 3 − 5t = 0: t = 0.6 s.
After 4 s: v = 3 − 5 × 4 = −17 m s⁻¹, so it is moving at 17 m s⁻¹ in the opposite direction.
Answer: 0.6 s; −17 m s⁻¹ (17 m s⁻¹ reversed)
A gun of mass 4 kg fires a bullet of mass 10 g at 400 m s⁻¹. Find the recoil speed of the gun.
Given: M = 4 kg, m = 0.010 kg, v = 400 m s⁻¹
Use: Conservation of momentum: total momentum before = 0
0 = m v + M V, so V = −m v/M = −0.010 × 400/4 = −1 m s⁻¹.
The gun recoils at 1 m s⁻¹ opposite to the bullet.
Answer: 1 m s⁻¹ in the opposite direction
A man of mass 60 kg jumps from a stationary boat of mass 200 kg with a horizontal speed of 3 m s⁻¹ relative to the water. Find the recoil speed of the boat (neglect water resistance).
Given: m = 60 kg, v = 3 m s⁻¹, M = 200 kg
Use: Conservation of momentum
0 = 60 × 3 + 200 V, so V = −180/200 = −0.9 m s⁻¹.
The boat moves at 0.9 m s⁻¹ in the opposite direction.
Answer: 0.9 m s⁻¹ opposite to the jump
A 2 kg body moving at 6 m s⁻¹ collides with a 4 kg body at rest and they move together. Find the common speed and the loss of kinetic energy.
Given: m₁ = 2 kg, u₁ = 6 m s⁻¹, m₂ = 4 kg, u₂ = 0
Use: Conservation of momentum
(2 + 4) V = 2 × 6, so V = 2 m s⁻¹.
Initial KE = ½ × 2 × 36 = 36 J; final KE = ½ × 6 × 4 = 12 J.
Loss = 24 J (momentum is conserved, kinetic energy is not).
Answer: 2 m s⁻¹; 24 J lost
A 3 kg body moving at 4 m s⁻¹ collides head-on with a 1 kg body moving at 2 m s⁻¹ in the opposite direction. They stick together. Find their velocity.
Given: m₁ = 3 kg, u₁ = +4 m s⁻¹, m₂ = 1 kg, u₂ = −2 m s⁻¹
Use: Conservation of momentum with signs
(3 + 1) V = 3 × 4 + 1 × (−2) = 10, so V = 2.5 m s⁻¹.
The combined body moves in the direction of the heavier body's initial velocity.
Answer: 2.5 m s⁻¹
A body of mass 6 kg at rest explodes into two parts of 4 kg and 2 kg. The 4 kg part moves at 3 m s⁻¹. Find the velocity of the other part.
Given: M = 6 kg at rest; m₁ = 4 kg, v₁ = 3 m s⁻¹; m₂ = 2 kg
Use: Total momentum remains zero
0 = 4 × 3 + 2 v₂, so v₂ = −6 m s⁻¹.
The 2 kg part moves at 6 m s⁻¹ in the opposite direction.
Answer: 6 m s⁻¹ opposite to the 4 kg part
A 0.2 kg ball strikes a wall at 10 m s⁻¹ and rebounds at 8 m s⁻¹ along the same line. The contact lasts 0.01 s. Find the impulse and the average force on the ball.
Given: m = 0.2 kg, u = +10 m s⁻¹, v = −8 m s⁻¹, Δt = 0.01 s
Use: J = Δp, F = J/Δt
Δp = 0.2 × (−8) − 0.2 × 10 = −3.6 N s.
The impulse has magnitude 3.6 N s, directed away from the wall.
F = 3.6/0.01 = 360 N.
Answer: 3.6 N s; 360 N
A 5 kg mass hangs from a string. It is pulled aside by a horizontal force of 30 N and is in equilibrium (g = 10 m s⁻²). Find the angle the string makes with the vertical and the tension.
Given: m = 5 kg, H = 30 N, W = 50 N
Use: Resolve forces: T cos θ = mg, T sin θ = H
T cos θ = 50 and T sin θ = 30.
tan θ = 30/50 = 0.6, so θ = 31.0°.
T = √(50² + 30²) = 58.3 N.
Answer: θ ≈ 31°, T ≈ 58.3 N
A man of mass 60 kg stands on a weighing machine in a lift (g = 10 m s⁻²). What does the machine read when the lift accelerates (a) upward at 2 m s⁻², (b) downward at 2 m s⁻²?
Given: m = 60 kg, a = 2 m s⁻²
Use: R − mg = ma (taking up as positive)
(a) R = m (g + a) = 60 × 12 = 720 N.
(b) R = m (g − a) = 60 × 8 = 480 N.
Answer: (a) 720 N; (b) 480 N
A lift of mass 500 kg is pulled by a cable with a tension of 6000 N (g = 10 m s⁻²). Find its acceleration.
Given: m = 500 kg, T = 6000 N
Use: T − mg = ma
Weight = 500 × 10 = 5000 N.
a = (6000 − 5000)/500 = 2 m s⁻², upward.
Answer: 2 m s⁻² upward
Masses of 5 kg and 3 kg hang over a light frictionless pulley (g = 10 m s⁻²). Find the acceleration and the tension.
Given: m₁ = 5 kg, m₂ = 3 kg
Use: For each mass: m₁g − T = m₁a and T − m₂g = m₂a
a = (m₁ − m₂) g/(m₁ + m₂) = 2 × 10/8 = 2.5 m s⁻².
T = m₂ (g + a) = 3 × 12.5 = 37.5 N.
Answer: 2.5 m s⁻²; 37.5 N
Two blocks of masses 4 kg and 6 kg touch each other on a smooth horizontal floor. A horizontal force of 20 N pushes the 4 kg block. Find the acceleration and the contact force between the blocks.
Given: m₁ = 4 kg, m₂ = 6 kg, F = 20 N
Use: System: F = (m₁ + m₂) a; on the 6 kg block: N = m₂ a
a = 20/10 = 2 m s⁻².
The 6 kg block is accelerated only by the contact force: N = 6 × 2 = 12 N.
Answer: 2 m s⁻²; 12 N
A 2 kg block on a smooth horizontal table is connected by a light string over a pulley at the edge of the table to a 3 kg mass hanging freely (g = 10 m s⁻²). Find the acceleration and the tension.
Given: m₁ = 2 kg (table), m₂ = 3 kg (hanging)
Use: m₂ g − T = m₂ a, T = m₁ a
a = m₂ g/(m₁ + m₂) = 30/5 = 6 m s⁻².
T = m₁ a = 2 × 6 = 12 N.
Answer: 6 m s⁻²; 12 N
A 10 kg block on a horizontal floor just begins to move when a horizontal force of 40 N is applied (g = 10 m s⁻²). Find μₛ.
Given: m = 10 kg, F = 40 N (at the verge of motion)
Use: (fₛ)max = μₛ N
N = mg = 100 N.
μₛ = 40/100 = 0.4.
Answer: 0.4
A block on an inclined plane starts sliding when the inclination is increased to a certain angle. If μₛ = 0.5, find that angle.
Given: μₛ = 0.5
Use: tan θmax = μₛ
θmax = tan⁻¹ (0.5) = 26.6°.
The mass of the block does not affect the angle.
Answer: 26.6°
A 5 kg block is pushed along a rough horizontal floor by a horizontal force of 25 N. μₖ = 0.3 (g = 10 m s⁻²). Find the acceleration.
Given: m = 5 kg, F = 25 N, μₖ = 0.3
Use: fₖ = μₖ m g, F − fₖ = ma
fₖ = 0.3 × 5 × 10 = 15 N.
a = (25 − 15)/5 = 2 m s⁻².
Answer: 2 m s⁻²
A box lies on the floor of a train. μₛ = 0.25. What is the largest acceleration of the train for which the box does not slide (g = 10 m s⁻²)?
Given: μₛ = 0.25
Use: The static friction provides the acceleration: ma ≤ μₛ mg
a ≤ μₛ g = 0.25 × 10 = 2.5 m s⁻².
Beyond this the box slips backward relative to the floor.
Answer: 2.5 m s⁻²
A block slides down a 30° incline with μₖ = 0.2 (g = 10 m s⁻²). Find its acceleration.
Given: θ = 30°, μₖ = 0.2
Use: ma = mg sin θ − μₖ mg cos θ
a = g (sin 30° − 0.2 cos 30°) = 10 (0.5 − 0.1732) = 3.27 m s⁻².
Answer: 3.27 m s⁻² down the slope
A car moving at 20 m s⁻¹ brakes with its wheels locked on a road with μₖ = 0.5 (g = 10 m s⁻²). Find the stopping distance.
Given: u = 20 m s⁻¹, μₖ = 0.5
Use: a = μₖ g, 0 = u² − 2as
a = 0.5 × 10 = 5 m s⁻² (deceleration).
s = u²/(2a) = 400/10 = 40 m.
Answer: 40 m
A car takes a level circular turn of radius 50 m. μₛ = 0.4 (g = 10 m s⁻²). What is the maximum safe speed?
Given: R = 50 m, μₛ = 0.4
Use: vmax = √(μₛ R g)
vmax = √(0.4 × 50 × 10) = √200 = 14.1 m s⁻¹.
In km h⁻¹ this is 14.1 × 3.6 ≈ 51 km h⁻¹.
Answer: 14.1 m s⁻¹ (≈ 51 km h⁻¹)
A road of radius 100 m is banked at 30°. Find the speed at which a car needs no friction (g = 10 m s⁻²).
Given: R = 100 m, θ = 30°
Use: vo = √(R g tan θ)
vo = √(100 × 10 × 0.577) = √577 = 24.0 m s⁻¹.
This is about 86 km h⁻¹.
Answer: 24.0 m s⁻¹
A 0.5 kg stone is whirled in a horizontal circle of radius 1 m at 4 m s⁻¹. Find the tension. The string breaks if the tension exceeds 50 N. What is the greatest speed?
Given: m = 0.5 kg, R = 1 m, v = 4 m s⁻¹, Tmax = 50 N
Use: T = mv²/R
T = 0.5 × 16/1 = 8 N.
vmax = √(Tmax R/m) = √(50 × 1/0.5) = 10 m s⁻¹.
Answer: 8 N; 10 m s⁻¹
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In each case the body is at rest or moves with constant velocity, so by the first law the acceleration is zero and the net force is zero.
With air resistance ignored the only force is the weight: F = mg = 0.05 × 10 = 0.5 N, directed vertically downward.
The answers do not change if the pebble is thrown at 45° with the horizontal: the net force is always the weight, 0.5 N downward. (At the highest point the velocity is then horizontal, not zero, but the force is the same.)
The correct alternative is (i) T. The only horizontal force on the particle is the tension, and it is directed towards the centre; so it is the net force and it provides the centripetal force: T = mv²/l. (The 'centripetal force' is not an extra force to be added or subtracted.)
Retardation a = F/m = 50/20 = 2.5 m s⁻².
From v = u + at with v = 0: t = u/a = 15/2.5 = 6 s.
a = (3.5 − 2.0)/25 = 0.06 m s⁻².
F = ma = 3.0 × 0.06 = 0.18 N, in the direction of motion (the speed increases along the same direction).
Net force F = √(8² + 6²) = 10 N, at tan⁻¹(6/8) = 36.9° to the 8 N force (towards the 6 N force).
a = F/m = 10/5 = 2 m s⁻², in the direction of the net force: at 36.9° with the 8 N force.
u = 36 km h⁻¹ = 10 m s⁻¹, v = 0, t = 4.0 s, so a = −10/4 = −2.5 m s⁻².
The mass that stops is 400 + 65 = 465 kg. F = ma = 465 × 2.5 = 1162.5 N ≈ 1.2 × 10³ N, opposite to the direction of motion.
Forces: thrust F upward and weight mg downward. F − mg = ma.
F = m (g + a) = 20,000 × (10 + 5.0) = 3.0 × 10⁵ N.
Take north as positive. The acceleration while the force acts is a = −8.0/0.40 = −20 m s⁻² (towards the south).
At t = 10 s the speed of the truck, and hence the horizontal speed of the stone when it is released, is 2.0 × 10 = 20 m s⁻¹. After release there is no horizontal force (air resistance neglected), so the horizontal velocity stays 20 m s⁻¹.
(a) In the 1 s after release, the vertical velocity is vy = −g t = −10 m s⁻¹. The velocity at t = 11 s is the vector sum: v = √(20² + 10²) = 22.4 m s⁻¹, at tan⁻¹(10/20) = 26.6° below the horizontal, in the direction of the truck's motion.
(b) The only force is gravity, so the acceleration is 10 m s⁻² vertically downward. The horizontal acceleration is zero (the stone does not carry the truck's acceleration).
(a) At an extreme position the bob is momentarily at rest. If the string is cut, only gravity acts on the bob starting from rest: it falls vertically downward in a straight line.
(b) At the mean position the velocity is 1 m s⁻¹ horizontal. After the string is cut the bob has a horizontal velocity and falls under gravity: it follows a parabolic path (like a horizontally projected body).
The scale reads the normal reaction R. Take upward as positive: R − mg = ma. Here mg = 700 N.
From Fig. 4.16: x = 0 for t < 0 (at rest); a straight line from x = 0 at t = 0 to x = 3 m at t = 4 s; and x = 3 m (at rest) for t > 4 s.
Let A = 10 kg and B = 20 kg. The whole system accelerates with a = 600/(10 + 20) = 20 m s⁻² in both cases.
Let the 12 kg mass move down and the 8 kg mass up with a common acceleration a and tension T.
12 g − T = 12 a and T − 8 g = 8 a. Adding: 4 g = 20 a, so a = 4 × 10/20 = 2 m s⁻².
T = 8 (g + a) = 8 × 12 = 96 N (check: 12 (g − a) = 12 × 8 = 96 N).
The nucleus is at rest, so its initial momentum is zero. No external force acts on it during the disintegration, so by the law of conservation of momentum the total momentum of the products is also zero: p₁ + p₂ = 0, i.e. p₁ = −p₂.
The two momenta are equal and opposite, so the two nuclei move in opposite directions (along the same line), with speeds in the inverse ratio of their masses. If the products were not on a line, their vector sum could not be zero.
Take the initial direction of ball A as positive. Initial momentum of A = +0.05 × 6 = +0.3 kg m s⁻¹; after the collision it moves back: −0.3 kg m s⁻¹.
Impulse on A = Δp = −0.3 − 0.3 = −0.6 kg m s⁻¹ (magnitude 0.6 N s, opposite to its initial motion). For ball B the impulse is +0.6 kg m s⁻¹, equal and opposite, as the third law requires.
Momentum before firing = 0. After firing: 0.020 × 80 + 100 V = 0.
V = −1.6/100 = −0.016 m s⁻¹. The recoil speed of the gun is 0.016 m s⁻¹, opposite to the shell. (The mass of the gun is large, so the recoil is small.)
u = 54 km h⁻¹ = 15 m s⁻¹. The ball is deflected through 45° (its direction of motion turns by 45°), with the same speed.
Initial velocity along x: u î. Final velocity: u cos 45° î + u sin 45° ĵ. Change in velocity: Δv = u (cos 45° − 1) î + u sin 45° ĵ = u (−0.293 î + 0.707 ĵ).
|Δv| = u √(0.293² + 0.707²) = 0.765 u = 11.5 m s⁻¹, which equals 2u sin 22.5°.
Impulse = m |Δv| = 0.15 × 11.5 ≈ 1.7 N s. Its direction makes an angle of 112.5° with the initial direction of the ball, i.e. it is perpendicular to the line bisecting the incoming and outgoing directions.
ω = 2π × 40/60 = 4.19 rad s⁻¹. The tension provides the centripetal force: T = mω²R = 0.25 × (4.19)² × 1.5 = 6.6 N.
For a maximum tension of 200 N: Tmax = mv²/R, so vmax = √(Tmax R/m) = √(200 × 1.5/0.25) = √1200 = 34.6 m s⁻¹ (about 35 m s⁻¹).
The correct statement is (b). Once the string breaks, the tension (the only horizontal force) vanishes. The velocity at that instant is along the tangent to the circle, and with no force (in the horizontal plane) the stone continues in that direction with that velocity: it flies off tangentially. (Gravity makes it fall, so it follows a parabola, but the horizontal motion is along the tangent.)