A car travels 100 km north in 2 hours, then immediately turns around and travels 50 km south in 1 hour. What is the magnitude of the car's average velocity for the entire trip?
150 km/h
50 km/h
16.7 km/h
✓
33.3 km/h
Correct Answer
16.7 km/h
Average velocity is defined as total displacement divided by total time. The total displacement is 100 km (North) - 50 km (South) = 50 km (North). The total time is 2 h + 1 h = 3 h. So, the magnitude of average velocity is 50 km / 3 h = 16.7 km/h. Option A is the average speed for the first leg. Option B is the average speed for the entire trip. Option C is the sum of the speeds.
Question 2
A 1 kg ball and a 10 kg ball are dropped simultaneously from the same height in a vacuum. Which statement accurately describes their motion?
The 10 kg ball hits the ground first because the gravitational force on it is greater.
The 10 kg ball hits the ground first because it has greater inertia.
Both balls hit the ground at the same time because the acceleration due to gravity is independent of mass.
✓
The 1 kg ball hits the ground first because it experiences less air resistance.
Correct Answer
Both balls hit the ground at the same time because the acceleration due to gravity is independent of mass.
In a vacuum, all objects fall with the same acceleration (g), regardless of their mass, thus they hit the ground at the same time. Option A incorrectly applies the concept of inertia to falling time. Option B is incorrect because there is no air resistance in a vacuum. Option D confuses gravitational force with acceleration; while the force is greater on the 10 kg ball, its greater mass results in the same acceleration.
Question 3
A car starting from rest accelerates uniformly at 2.0 m/s² for 10 seconds. How far does it travel during this time?
100 meters
✓
20 meters
200 meters
40 meters
Correct Answer
100 meters
Using the kinematic equation d = v₀t + (1/2)at², with initial velocity v₀=0, acceleration a=2.0 m/s², and time t=10 s, the distance traveled is d = 0 + (1/2)(2.0 m/s²)(10 s)² = 100 meters. Option A incorrectly calculates final velocity. Option C omits the (1/2) factor in the equation. Option D is a miscalculation.
Question 4
An object is moving at a constant velocity. Which statement must be true about the forces acting on the object?
There are no forces acting on the object.
The net force acting on the object is zero.
✓
The forces acting on the object are balanced, but only if it is at rest.
The object is not accelerating because its mass is very small.
Correct Answer
The net force acting on the object is zero.
According to Newton's First Law, an object moving at a constant velocity has zero acceleration, which implies that the net force acting on it is zero. Option A is incorrect because forces can be present but balanced. Option C is incorrect; constant velocity implies zero acceleration regardless of mass. Option D is incorrect because balanced forces result in zero acceleration for both constant velocity and rest.
Question 5
A 5.0 kg box is pulled across a horizontal floor by a horizontal force of 20 N. If the coefficient of kinetic friction between the box and the floor is 0.20, what is the acceleration of the box? (Take g = 10 m/s²)
1.0 m/s²
0 m/s²
2.0 m/s²
✓
4.0 m/s²
Correct Answer
2.0 m/s²
The normal force is N = mg = 5.0 kg * 10 m/s² = 50 N. The kinetic friction force is f_k = μ_k * N = 0.20 * 50 N = 10 N. The net force on the box is F_net = Applied Force - f_k = 20 N - 10 N = 10 N. Using Newton's Second Law, a = F_net / m = 10 N / 5.0 kg = 2.0 m/s². Option A neglects friction. Option C implies the box is at rest or constant velocity, meaning net force is zero. Option D is a miscalculation.
Question 6
A book rests on a table. Which of the following is the Newton's Third Law action-reaction pair for the force of gravity exerted by the Earth on the book?
The force of the table pushing up on the book.
The force of the book pushing down on the table.
The force of gravity exerted by the book on the table.
The force of the book pulling up on the Earth.
✓
Correct Answer
The force of the book pulling up on the Earth.
Newton's Third Law states that for every action, there is an equal and opposite reaction acting on different bodies. The force of gravity exerted by the Earth on the book (action) has its reaction as the force of the book pulling up on the Earth. Option A is the normal force, which is a reaction to the book pushing on the table. Option B is an action force from the book on the table. Option D is incorrect; gravity acts between masses, not between a book and a table in this context.
Question 7
A person holds a heavy briefcase stationary for 30 minutes. Which statement correctly describes the work done on the briefcase by the person?
Negative work is done because the person is expending energy.
Work is done, but it is zero because the person's force balances gravity.
Positive work is done because the briefcase has potential energy.
No work is done because there is no displacement in the direction of the force.
✓
Correct Answer
No work is done because there is no displacement in the direction of the force.
Work done on an object by a force is defined as W = F * d * cos(θ). Since the briefcase is held stationary, its displacement (d) is zero. Therefore, no work is done on the briefcase. Option A is incorrect; potential energy relates to position, not work done by holding it. Option B is incorrect; human energy expenditure is not the same as physical work done on an object. Option D correctly states zero work but gives an incomplete reason; the primary reason is zero displacement.
Question 8
A 2.0 kg object is released from rest at a height of 5.0 meters above the ground. Neglecting air resistance, what is its speed just before it hits the ground? (Take g = 10 m/s²)
7.1 m/s
5 m/s
20 m/s
10 m/s
✓
Correct Answer
10 m/s
Using the principle of conservation of energy, the initial gravitational potential energy (PE) is converted into kinetic energy (KE) just before impact. So, mgh = (1/2)mv². The mass 'm' cancels out, leaving gh = (1/2)v². Solving for v: v = sqrt(2gh) = sqrt(2 * 10 m/s² * 5.0 m) = sqrt(100) = 10 m/s. Option B would be sqrt(gh). Option C would be 2gh (missing the square root). Option D would be sqrt(gh) if h was 50m.
Question 9
A motor lifts a 50 kg mass vertically at a constant speed of 0.50 m/s. What is the power output of the motor? (Take g = 9.8 m/s²)
49 Watts
245 Watts
✓
100 Watts
25 Watts
Correct Answer
245 Watts
Power (P) is the rate at which work is done, or P = Force * velocity. To lift the mass at a constant speed, the motor's upward force must equal the object's weight (F = mg). F = 50 kg * 9.8 m/s² = 490 N. Therefore, P = 490 N * 0.50 m/s = 245 Watts. Option B is a miscalculation (e.g., 50 * 9.8 / 10). Option C is mass times velocity, ignoring gravity. Option D is a miscalculation.
Question 10
In which of the following scenarios is the total linear momentum of the system conserved?
A ball is thrown vertically upwards and falls back down.
A car skids to a stop on a rough road.
A rocket accelerates by expelling exhaust gases.
A bullet embeds itself in a wooden block resting on a frictionless surface.
✓
Correct Answer
A bullet embeds itself in a wooden block resting on a frictionless surface.
Total linear momentum of a system is conserved if there are no net external forces acting on it. Option A involves external friction. Option B involves external forces like gravity acting on the rocket+exhaust system. Option C involves external gravity acting on the ball. Option D describes an isolated system (bullet + block) where external forces (gravity, normal force) are balanced and perpendicular to the motion, and friction is absent, ensuring horizontal momentum conservation.
Question 11
A 2.0 kg object moving at 5.0 m/s collides head-on with a stationary 3.0 kg object. If they stick together after the collision, what is their combined speed immediately after impact?
2.0 m/s
✓
1.0 m/s
5.0 m/s
2.5 m/s
Correct Answer
2.0 m/s
In a collision, total momentum is conserved. For an inelastic collision where objects stick together: m₁v₁ + m₂v₂ = (m₁ + m₂)v_f. (2.0 kg)(5.0 m/s) + (3.0 kg)(0 m/s) = (2.0 kg + 3.0 kg)v_f. 10 kg·m/s = (5.0 kg)v_f. Solving for v_f gives v_f = 2.0 m/s. Option A is a result of incorrect momentum calculation. Option C is a common miscalculation. Option D implies no momentum transfer.
Question 12
A 0.15 kg baseball is thrown with a speed of 40 m/s. After being hit by a bat, it leaves the bat with a speed of 60 m/s in the opposite direction. What is the magnitude of the impulse imparted to the ball by the bat?
9.0 N·s
15 N·s
✓
3.0 N·s
6.0 N·s
Correct Answer
15 N·s
Impulse is equal to the change in momentum (Δp = mΔv). Let the initial direction be positive. Then v_i = +40 m/s and v_f = -60 m/s. The change in velocity Δv = v_f - v_i = -60 m/s - 40 m/s = -100 m/s. Impulse = 0.15 kg * (-100 m/s) = -15 N·s. The magnitude of the impulse is 15 N·s. Option A is the initial momentum. Option B is the final momentum. Option C is a miscalculation, perhaps if the ball stopped.
Question 13
A car is traveling at a constant speed around a circular track. Which statement best describes the net force acting on the car?
The net force is directed radially inward towards the center of the track.
✓
The net force is directed tangent to the track in the direction of motion.
The net force is directed radially outward, away from the center of the track.
The net force is zero because the speed is constant.
Correct Answer
The net force is directed radially inward towards the center of the track.
For an object in uniform circular motion, even with constant speed, its velocity is continuously changing direction, meaning it is accelerating. This centripetal acceleration is directed towards the center of the circle. By Newton's Second Law (F=ma), the net force must also be directed towards the center of the track (radially inward). Option A is incorrect because constant speed does not mean constant velocity in circular motion. Option B describes a force that would change the speed. Option D describes a fictitious centrifugal force.
Question 14
A child rides a carousel that completes one revolution every 10 seconds. If the child is 4.0 meters from the center of the carousel, what is their centripetal acceleration?
0.4π m/s²
1.6π² m/s²
0.8π m/s²
0.16π² m/s²
✓
Correct Answer
0.16π² m/s²
First, calculate the tangential speed (v): v = 2πr/T = 2π(4.0 m) / 10 s = 0.8π m/s. Then, calculate the centripetal acceleration (a_c): a_c = v²/r = (0.8π m/s)² / 4.0 m = (0.64π² m²/s²) / 4.0 m = 0.16π² m/s². Option A is the value of 0.5v. Option B is the tangential speed, not acceleration. Option C results from a calculation error, such as using (2v)^2/r.
Question 15
Which action will result in the largest magnitude of torque about a pivot point?
Applying a large force close to the pivot, parallel to the lever arm.
Applying a large force far from the pivot, perpendicular to the lever arm.
✓
Applying a small force far from the pivot, perpendicular to the lever arm.
Applying a large force far from the pivot, at an angle of 45 degrees to the lever arm.
Correct Answer
Applying a large force far from the pivot, perpendicular to the lever arm.
Torque (τ) is calculated as τ = rFsinθ, where r is the lever arm, F is the force, and θ is the angle between the force and the lever arm. Torque is maximized when the force is large, the lever arm is large, and the angle is 90 degrees (sin(90°) = 1). Option A uses a small force. Option B uses a force parallel to the lever arm (sin(0°) = 0), resulting in zero torque. Option C uses an angle of 45 degrees (sin(45°) ≈ 0.707), which is less than sin(90°).
Question 16
A uniform plank of length 4.0 m and mass 20 kg is supported at its ends by two vertical ropes. A 60 kg person stands 1.0 m from the left end of the plank. What is the tension in the right rope? (Take g = 9.8 m/s²)
245 N
✓
196 N
392 N
539 N
Correct Answer
245 N
To find the tension in the right rope (T_R), take torques about the left end of the plank. The plank's weight (20 kg * 9.8 m/s² = 196 N) acts at its center (2.0 m from the left end). The person's weight (60 kg * 9.8 m/s² = 588 N) acts at 1.0 m from the left end. The right rope's tension acts at 4.0 m from the left end. For equilibrium, sum of torques is zero: (196 N * 2.0 m) + (588 N * 1.0 m) = T_R * 4.0 m. 392 N·m + 588 N·m = T_R * 4.0 m. 980 N·m = T_R * 4.0 m. T_R = 245 N. Option B is the weight of the plank. Option C is twice the plank's weight. Option D is the tension in the left rope.
Question 17
Which of the following forces is a conservative force?
Kinetic friction
Gravitational force
✓
Tension in a string
Air resistance
Correct Answer
Gravitational force
A conservative force is one for which the work done in moving an object between two points is independent of the path taken, or equivalently, the work done in a closed loop is zero. Gravitational force is a conservative force, allowing for the definition of gravitational potential energy. Kinetic friction and air resistance are non-conservative forces because the work they do depends on the path. Tension in a string, while often doing no work, is generally not considered a conservative force in the same way gravity or springs are.
Question 18
A 70 kg person stands on a scale in an elevator. If the elevator accelerates upwards at 2.0 m/s², what does the scale read? (Take g = 9.8 m/s²)
686 N
140 N
826 N
✓
546 N
Correct Answer
826 N
The scale reads the normal force (N) exerted by the scale on the person. Using Newton's Second Law, with upward acceleration: N - mg = ma. So, N = m(g + a) = 70 kg * (9.8 m/s² + 2.0 m/s²) = 70 kg * 11.8 m/s² = 826 N. Option A is the person's actual weight (mg). Option B is the force required to accelerate the person (ma). Option D is the apparent weight if the elevator were accelerating downwards.
Question 19
Which statement is true for a perfectly elastic collision?
Only momentum is conserved, kinetic energy is lost.
Both momentum and kinetic energy are conserved.
✓
Only kinetic energy is conserved, momentum is lost.
Both momentum and kinetic energy are lost.
Correct Answer
Both momentum and kinetic energy are conserved.
In a perfectly elastic collision, both the total linear momentum and the total kinetic energy of the system are conserved. Option B describes an inelastic collision. Option C is incorrect; momentum is always conserved in an isolated system, while kinetic energy can be lost. Option D is incorrect as momentum is always conserved in an isolated system.
Question 20
A ball is thrown horizontally from the top of a 45 m tall building with an initial speed of 15 m/s. How long does it take for the ball to hit the ground? (Neglect air resistance, take g = 9.8 m/s²)
3.0 seconds
✓
4.5 seconds
9.8 seconds
1.5 seconds
Correct Answer
3.0 seconds
The time it takes for the ball to hit the ground depends only on its vertical motion. Using the kinematic equation d = v₀t + (1/2)at², where vertical displacement d = 45 m, initial vertical velocity v₀ = 0, and vertical acceleration a = g = 9.8 m/s². So, 45 m = 0 + (1/2)(9.8 m/s²)t². This gives t² = 45 / 4.9 ≈ 9.18, so t ≈ 3.03 seconds. The closest option is 3.0 seconds. Option A is a miscalculation. Option C is a miscalculation. Option D is the value of g, not time.
Question 21
A car is designed to go around a frictionless circular turn with a radius of 120 m at a speed of 20 m/s. At what angle should the road be banked? (Take g = 10 m/s²)
18.4 degrees
✓
26.6 degrees
9.5 degrees
45.0 degrees
Correct Answer
18.4 degrees
For a frictionless banked curve, the banking angle (θ) is given by the formula tan(θ) = v²/rg. Substituting the given values: tan(θ) = (20 m/s)² / (120 m * 10 m/s²) = 400 / 1200 = 1/3. Therefore, θ = arctan(1/3) ≈ 18.4 degrees. Option B is a common miscalculation. Option C would require a higher speed or smaller radius. Option D would imply v² = rg, which is not the case here.
Question 22
A 10 kg uniform rod, 2.0 m long, is pivoted at its center. A 4.0 kg mass is hung 0.50 m from the left end. Where must a 2.0 kg mass be hung to balance the rod?
0.25 m from the right end of the rod.
1.0 m from the right end of the rod.
At the right end of the rod.
✓
0.50 m from the right end of the rod.
Correct Answer
At the right end of the rod.
The rod is uniform and pivoted at its center, so its own weight creates no net torque. The 4.0 kg mass is hung 0.50 m from the left end. Since the pivot is at the center (1.0 m from the left end), this mass is 0.50 m to the left of the pivot. The counter-clockwise torque it creates is (4.0 kg * g) * 0.50 m = 2.0g N·m. To balance this, the 2.0 kg mass must create a clockwise torque of 2.0g N·m. If this mass is hung at a distance 'x' to the right of the pivot, its torque is (2.0 kg * g) * x. Setting these equal: 2.0g = 2.0gx, which means x = 1.0 m. A distance of 1.0 m to the right of the center is exactly at the right end of the 2.0 m rod. Option B would create insufficient torque. Option C is at the pivot, creating no torque. Option D would create insufficient torque.