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Pancyprian Technical School Physics (4-hour) study materials
Pancyprian Technical School Physics (4-hour) study resources

Pancyprian Technical School Physics (4-hour) practice questions

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Question 1Mechanics: Kinematics and Dynamics

A car travels along a straight road. Its velocity-time graph is a horizontal line above the time axis. Which statement correctly describes the car's motion?

  1. The car is at rest at a fixed position.
  2. The car moves with uniformly increasing velocity.
  3. The car moves with constant nonzero acceleration.
  4. The car moves with constant velocity and zero acceleration. Correct answer.

Explanation

On a velocity-time graph the value of the plotted line is the instantaneous velocity, so a horizontal line means velocity does not change. Acceleration equals the slope of a velocity-time graph, and the slope of a horizontal line is zero. Therefore the car moves at constant velocity with zero acceleration.

Question 2Mechanics: Kinematics and Dynamics

A cyclist starts from rest and reaches 12 m/s in 6.0 s with uniform acceleration. What distance does the cyclist cover during this interval?

  1. 36 m Correct answer.
  2. 72 m
  3. 24 m
  4. 18 m

Explanation

For uniform acceleration the average velocity is the mean of initial and final velocity, here (0 + 12)/2 = 6 m/s. The distance is average velocity multiplied by time, giving 6 × 6.0 = 36 m. Equivalently, a = 2 m/s² and s = ½at² = ½ × 2 × 36 = 36 m.

Question 3Mechanics: Kinematics and Dynamics

Which pair of quantities are both vectors?

  1. Displacement and force Correct answer.
  2. Work and temperature
  3. Speed and mass
  4. Distance and time

Explanation

A vector has both magnitude and direction and must combine by vector addition. Displacement is a directed change of position and force is a directed push or pull, so both are vectors. Speed, mass, distance, time, work and temperature are scalars with magnitude only.

Question 4Mechanics: Kinematics and Dynamics

A block of mass 4.0 kg rests on a horizontal table. What is the magnitude of the normal reaction force exerted by the table on the block? Take g = 9.8 m/s².

  1. 4.0 N
  2. 39 N Correct answer.
  3. 0 N
  4. 9.8 N

Explanation

The block has no vertical acceleration, so the vertical forces balance. The normal reaction equals the weight, mg = 4.0 × 9.8 = 39.2 N, which rounds to 39 N. The reaction is not a third-law pair with the weight because both act on the same body.

Question 5Mechanics: Kinematics and Dynamics

A resultant force of 15 N acts on a body of mass 3.0 kg. What is the acceleration of the body?

  1. 0.20 m/s²
  2. 5.0 m/s² Correct answer.
  3. 45 m/s²
  4. 18 m/s²

Explanation

Newton's second law states that the resultant force equals mass times acceleration. Rearranging gives a = F/m = 15/3.0 = 5.0 m/s². The acceleration is in the same direction as the resultant force.

Question 6Mechanics: Kinematics and Dynamics

A stone is dropped from rest from a cliff and falls freely. Neglecting air resistance, what is its speed after 3.0 s? Take g = 9.8 m/s².

  1. 29 m/s Correct answer.
  2. 9.8 m/s
  3. 4.9 m/s
  4. 44 m/s

Explanation

Free fall means the only force is weight, giving a constant downward acceleration g. Using v = u + at with u = 0 gives v = 9.8 × 3.0 = 29.4 m/s, about 29 m/s. The mass of the stone does not affect the result.

Question 7Mechanics: Kinematics and Dynamics

An object moves in a straight line at constant speed. Which statement about the forces on it is correct?

  1. A constant resultant force acts in the direction of motion.
  2. The object must be in free fall.
  3. The resultant force acts opposite to the motion.
  4. The resultant force on the object is zero. Correct answer.

Explanation

Constant velocity, including constant speed in a straight line, means zero acceleration. By Newton's second law a net force would produce acceleration, so the resultant force must be zero. Individual forces may act, but they cancel.

Question 8Mechanics: Kinematics and Dynamics

A car travels 150 m due east and then 50 m due west along the same straight road. What is the magnitude of its displacement from the starting point?

  1. 150 m
  2. 100 m Correct answer.
  3. 50 m
  4. 200 m

Explanation

Displacement is the straight-line vector from start to finish, so opposite directions carry opposite signs. Taking east as positive gives +150 − 50 = +100 m. The magnitude is therefore 100 m, whereas the distance travelled is 200 m.

Question 9Mechanics: Kinematics and Dynamics

A ball is thrown vertically upward with an initial speed of 20 m/s. Neglecting air resistance, how long does it take to reach its highest point? Take g = 9.8 m/s².

  1. 20 s
  2. 4.1 s
  3. 2.0 s Correct answer.
  4. 0.49 s

Explanation

At the highest point the vertical velocity is momentarily zero. Using v = u − gt with v = 0 gives t = u/g = 20/9.8 ≈ 2.0 s. The acceleration remains g downward throughout the ascent.

Question 10Mechanics: Kinematics and Dynamics

A body of mass 5.0 kg is pulled along a rough horizontal floor by a horizontal force of 30 N. A constant friction force of 10 N opposes the motion. What is the acceleration of the body?

  1. 6.0 m/s²
  2. 2.0 m/s²
  3. 8.0 m/s²
  4. 4.0 m/s² Correct answer.

Explanation

The acceleration depends on the resultant force, not on the applied force alone. The resultant is 30 − 10 = 20 N, so a = F/m = 20/5.0 = 4.0 m/s². Friction must be subtracted because it acts opposite to the motion.

Question 11Mechanics: Kinematics and Dynamics

A projectile is launched horizontally from a height with an initial horizontal velocity. Neglecting air resistance, which statement about its motion is correct?

  1. The horizontal velocity decreases while the vertical velocity stays constant.
  2. Both horizontal and vertical velocity components stay constant.
  3. The horizontal velocity stays constant while the vertical velocity increases downward. Correct answer.
  4. The horizontal and vertical accelerations are both equal to g.

Explanation

With air resistance neglected the only force is weight, which acts vertically. Hence the horizontal acceleration is zero and the horizontal velocity is constant, while the vertical motion is free fall with acceleration g. The two components are independent.

Question 12Mechanics: Kinematics and Dynamics

A spring obeys Hooke's law with spring constant 200 N/m. What extension is produced by a stretching force of 12 N?

  1. 0.60 m
  2. 24 m
  3. 2.4 m
  4. 0.060 m Correct answer.

Explanation

Hooke's law states that the force is proportional to the extension, F = kx. Rearranging gives x = F/k = 12/200 = 0.060 m. The extension is proportional to the applied force only within the elastic limit.

Question 13Mechanics: Kinematics and Dynamics

Two forces of 6 N and 8 N act at a single point at right angles to each other. What is the magnitude of their resultant?

  1. 14 N
  2. 48 N
  3. 2 N
  4. 10 N Correct answer.

Explanation

Perpendicular vectors combine by the Pythagorean theorem because their components are independent. The resultant magnitude is √(6² + 8²) = √100 = 10 N. Simple addition would apply only if the forces were parallel.

Question 14Mechanics: Kinematics and Dynamics

A vehicle moving at 20 m/s brakes uniformly and stops in 5.0 s. What is the magnitude of its deceleration?

  1. 100 m/s²
  2. 4.0 m/s² Correct answer.
  3. 2.0 m/s²
  4. 0.25 m/s²

Explanation

Deceleration is the magnitude of the negative acceleration. From v = u + at, 0 = 20 + a(5.0), so a = −4.0 m/s² and the magnitude is 4.0 m/s². The negative sign simply shows the acceleration opposes the motion.

Question 15Mechanics: Kinematics and Dynamics

An object moves in a straight line with constant acceleration. Its velocity increases from 4.0 m/s to 14 m/s while covering 90 m. What is the acceleration?

  1. 0.10 m/s²
  2. 1.0 m/s² Correct answer.
  3. 1.5 m/s²
  4. 2.0 m/s²

Explanation

The time can be found from the average velocity: v_avg = (4 + 14)/2 = 9.0 m/s, so t = 90/9.0 = 10 s. Then a = (14 − 4)/10 = 1.0 m/s². The same result follows from v² = u² + 2as.

Question 16Mechanics: Kinematics and Dynamics

A crate of mass 20 kg is pushed across a horizontal floor at constant velocity by a horizontal force. The coefficient of kinetic friction is 0.25. What is the magnitude of the pushing force? Take g = 9.8 m/s².

  1. 49 N Correct answer.
  2. 5.0 N
  3. 20 N
  4. 196 N

Explanation

Constant velocity means the resultant force is zero, so the push equals the kinetic friction. Friction is μN, and on a horizontal floor N = mg = 20 × 9.8 = 196 N. Thus the push is 0.25 × 196 = 49 N.

Question 17Mechanics: Kinematics and Dynamics

Which statement about Newton's third law is correct?

  1. Action and reaction forces can act in the same direction.
  2. Action and reaction forces act on the same body and cancel.
  3. Action and reaction forces act on different bodies and are equal in magnitude and opposite in direction. Correct answer.
  4. The reaction force is always smaller than the action force.

Explanation

Newton's third law states that if body A exerts a force on body B, then B exerts an equal and opposite force on A. The two forces therefore act on different bodies, which is why they never cancel each other. They are always equal in magnitude and opposite in direction.

Question 18Mechanics: Kinematics and Dynamics

A body moves with a constant speed of 5.0 m/s around a circular path of radius 2.0 m. What is the magnitude of its centripetal acceleration?

  1. 10 m/s²
  2. 12.5 m/s² Correct answer.
  3. 0 m/s²
  4. 2.5 m/s²

Explanation

Even at constant speed, circular motion involves changing velocity direction and hence acceleration. The centripetal acceleration is v²/r = 5.0²/2.0 = 12.5 m/s², directed toward the centre. A centripetal force is required to provide it.

Question 19Mechanics: Kinematics and Dynamics

A car's velocity-time graph rises uniformly from 0 to 20 m/s between t = 0 and t = 4 s, then stays at 20 m/s until t = 10 s. What total distance does the car travel in the first 10 s?

  1. 160 m Correct answer.
  2. 80 m
  3. 200 m
  4. 120 m

Explanation

Distance travelled equals the area under a velocity-time graph. During acceleration the area is a triangle: ½ × 4 × 20 = 40 m. During constant speed the area is a rectangle: 20 × 6 = 120 m. The total is 40 + 120 = 160 m.

Question 20Mechanics: Kinematics and Dynamics

A 2.0 kg block hangs from a light string over a frictionless pulley and is connected to a 3.0 kg block resting on a frictionless horizontal table. When released, what is the acceleration of the system? Take g = 9.8 m/s².

  1. 2.0 m/s²
  2. 3.9 m/s² Correct answer.
  3. 9.8 m/s²
  4. 5.9 m/s²

Explanation

The driving force is the weight of the hanging block, 2.0 × 9.8 = 19.6 N. This accelerates the total mass of 5.0 kg because the string connects both blocks. Hence a = 19.6/5.0 ≈ 3.9 m/s².

Question 21Mechanics: Kinematics and Dynamics

A body is projected at 25 m/s at 30° above the horizontal from level ground. Neglecting air resistance, what is the maximum height reached? Take g = 9.8 m/s² and sin 30° = 0.50.

  1. 16 m
  2. 32 m
  3. 24 m
  4. 8.0 m Correct answer.

Explanation

The vertical component of velocity is 25 × 0.50 = 12.5 m/s. At the top the vertical velocity is zero, so using v² = u² − 2gh gives h = u²/(2g) = 156.25/19.6 ≈ 8.0 m. The horizontal component does not affect the height.

Question 22Mechanics: Kinematics and Dynamics

A lift of mass 500 kg accelerates upward at 2.0 m/s². What is the tension in the supporting cable? Take g = 9.8 m/s².

  1. 1000 N
  2. 3900 N
  3. 4900 N
  4. 5900 N Correct answer.

Explanation

Taking upward as positive, the resultant force is T − mg and this equals ma. So T = m(g + a) = 500 × (9.8 + 2.0) = 500 × 11.8 = 5900 N. The tension exceeds the weight because the lift is accelerating upward.

Question 23Mechanics: Kinematics and Dynamics

An object is thrown vertically upward and returns to the launch point. Neglecting air resistance, which statement about its motion is correct?

  1. The acceleration is zero at the highest point.
  2. The acceleration is upward on the way up and downward on the way down.
  3. The acceleration is larger on the way down than on the way up.
  4. The acceleration is g downward throughout the flight. Correct answer.

Explanation

Once released, the only force on the object is its weight, which acts downward at every instant. Therefore the acceleration has magnitude g and points downward for the whole flight. At the highest point the velocity is zero but the acceleration is still g.

Question 24Mechanics: Kinematics and Dynamics

A 4.0 kg block on a rough horizontal surface is connected by a light string over a frictionless pulley to a hanging 6.0 kg block. The coefficient of kinetic friction between the 4.0 kg block and the surface is 0.20. What is the acceleration of the system? Take g = 9.8 m/s².

  1. 5.9 m/s²
  2. 7.8 m/s²
  3. 5.1 m/s² Correct answer.
  4. 3.9 m/s²

Explanation

The driving force is the hanging weight, 6.0 × 9.8 = 58.8 N. Friction on the sliding block is μmg = 0.20 × 4.0 × 9.8 = 7.84 N. The resultant is 58.8 − 7.84 = 50.96 N acting on a total mass of 10 kg, so a ≈ 5.1 m/s².

Question 25Mechanics: Kinematics and Dynamics

A skydiver falls from rest and eventually reaches terminal velocity. Which statement correctly explains why the acceleration becomes zero at terminal velocity?

  1. The air resistance acts downward and cancels the weight from the start.
  2. The weight of the skydiver becomes zero as speed increases.
  3. Gravity stops acting on the skydiver at high speed.
  4. The air resistance increases with speed until it equals the weight, so the resultant force is zero. Correct answer.

Explanation

Air resistance grows as the skydiver's speed increases, while the weight stays constant. Terminal velocity is reached when the upward drag equals the downward weight, making the resultant force zero. With zero resultant force the acceleration is zero and the speed stays constant.

Question 26Mechanics: Kinematics and Dynamics

A cyclist moves along a straight road with velocity increasing uniformly from 2.0 m/s to 10 m/s in 4.0 s. What distance does the cyclist cover in this interval?

  1. 32 m
  2. 48 m
  3. 24 m Correct answer.
  4. 12 m

Explanation

For uniform acceleration the average velocity is the mean of the initial and final velocities, (2.0 + 10)/2 = 6.0 m/s. Multiplying by the time interval gives s = 6.0 × 4.0 = 24 m. Equivalently, s = ut + ½at² with a = 2.0 m/s² gives the same result.

Question 27Mechanics: Kinematics and Dynamics

A resultant force acts on a body of fixed mass. According to Newton's second law, which quantity does the resultant force equal?

  1. The product of the body's mass and its speed
  2. The product of the body's weight and its velocity
  3. The rate of change of the body's kinetic energy
  4. The rate of change of the body's momentum Correct answer.

Explanation

Newton's second law states that the resultant force equals the rate of change of momentum, F = Δp/Δt. For constant mass this reduces to F = ma, since momentum is mass times velocity. The other expressions mix up force with energy or with speed rather than acceleration.

Question 28Mechanics: Kinematics and Dynamics

A ball is kicked from level ground with a velocity of 20 m/s at 60° above the horizontal. Neglecting air resistance, what is the speed of the ball at the highest point of its path?

  1. 10 m/s Correct answer.
  2. 0 m/s
  3. 17 m/s
  4. 20 m/s

Explanation

At the highest point the vertical component of velocity is zero, so the speed equals the horizontal component. That component is u·cos60° = 20 × 0.50 = 10 m/s and stays constant because no horizontal force acts. The vertical component u·sin60° ≈ 17 m/s has been reduced to zero at that instant.

Question 29Mechanics: Kinematics and Dynamics

A box of mass 8.0 kg rests on a horizontal floor. A horizontal force of 20 N is applied but the box does not move. Which statement correctly describes the friction acting on the box?

  1. Kinetic friction of 20 N acts opposite to the applied force
  2. Static friction of 20 N acts opposite to the applied force Correct answer.
  3. Static friction greater than 20 N acts opposite to the applied force
  4. No friction acts because the box is at rest

Explanation

While the box remains at rest the net horizontal force must be zero, so static friction adjusts to equal the applied force, giving 20 N in the opposite direction. Static friction can take any value up to its limiting maximum, which here is not exceeded. Kinetic friction only arises once sliding begins.

Question 30Mechanics: Kinematics and Dynamics

A block slides down a smooth incline of angle 30° to the horizontal. A second identical block slides down a smooth incline of angle 60°. Neglecting friction, how do the magnitudes of the blocks' accelerations along the inclines compare?

  1. Both blocks have the same acceleration, equal to g
  2. The accelerations are equal because both inclines are smooth
  3. The 60° incline gives the larger acceleration, in the ratio √3 : 1 Correct answer.
  4. The 30° incline gives the larger acceleration, in the ratio √3 : 1

Explanation

On a smooth incline the only force along the slope is the component of weight, mg·sinθ, so the acceleration is g·sinθ, independent of mass. Since sin60° = √3/2 and sin30° = 1/2, the accelerations are in the ratio √3 : 1 with the steeper incline giving the larger value. Friction is neglected, so no other force along the slope needs to be considered.

Question 31Work, Energy, Power and Momentum

A crate is pushed 8 m along a horizontal floor by a constant horizontal force of 250 N. What is the work done by this force on the crate?

  1. 258 J
  2. 2000 J Correct answer.
  3. 31.25 J
  4. 0 J

Explanation

When force and displacement are parallel, work equals the product of force magnitude and displacement: W = F·s = 250 N × 8 m = 2000 J. Because the angle between them is zero, the full force contributes to the work.

Question 32Work, Energy, Power and Momentum

A ball of mass 0.4 kg is thrown vertically upward with an initial speed of 10 m/s. Taking g = 10 m/s² and neglecting air resistance, what is its kinetic energy at the moment of release?

  1. 2 J
  2. 4 J
  3. 20 J Correct answer.
  4. 40 J

Explanation

Kinetic energy is KE = ½mv². Substituting m = 0.4 kg and v = 10 m/s gives KE = ½ × 0.4 × 100 = 20 J. The gravitational acceleration is not needed because the question asks only about the energy at release.

Question 33Work, Energy, Power and Momentum

A machine lifts a 60 kg load vertically through 5 m in 3 s at constant speed. Taking g = 10 m/s², what is the average useful power output of the machine?

  1. 300 W
  2. 1000 W Correct answer.
  3. 360 W
  4. 1800 W

Explanation

The useful work equals the gain in gravitational potential energy, mgh = 60 × 10 × 5 = 3000 J. Average power is work divided by time: P = 3000 J / 3 s = 1000 W. Constant speed means no extra kinetic energy is gained.

Question 34Work, Energy, Power and Momentum

Which quantity is a vector that is conserved in an isolated system during a collision?

  1. Kinetic energy
  2. Linear momentum Correct answer.
  3. Mechanical power
  4. Work done

Explanation

Linear momentum is a vector quantity equal to mass times velocity, and in an isolated system with no external resultant force it is conserved in every collision. Kinetic energy is a scalar and is conserved only in elastic collisions.

Question 35Work, Energy, Power and Momentum

A spring stores 0.9 J of elastic potential energy when compressed by 0.06 m. What is the spring constant of the spring?

  1. 15 N/m
  2. 250 N/m
  3. 500 N/m Correct answer.
  4. 30 N/m

Explanation

Elastic potential energy is E = ½kx². Rearranging gives k = 2E/x² = (2 × 0.9)/(0.06²) = 1.8/0.0036 = 500 N/m. The squared extension is essential in this calculation.

Question 36Work, Energy, Power and Momentum

A force of 50 N acts on a body but the body does not move at all. What is the work done by this force?

  1. 50 J
  2. Infinite
  3. Cannot be determined without the mass
  4. 0 J Correct answer.

Explanation

Work requires a displacement in the direction of the force, W = F·s·cosθ. If the displacement s is zero, the work done is zero regardless of how large the force is. The mass of the body is irrelevant to this conclusion.

Question 37Work, Energy, Power and Momentum

A 2 kg trolley moving at 3 m/s collides head-on with a stationary 4 kg trolley and they stick together. What is their common velocity immediately after the collision?

  1. 1 m/s Correct answer.
  2. 0.5 m/s
  3. 1.5 m/s
  4. 2 m/s

Explanation

Momentum is conserved in the perfectly inelastic collision: m₁u₁ = (m₁ + m₂)v. Thus v = (2 × 3)/(2 + 4) = 6/6 = 1 m/s in the original direction of motion. Kinetic energy is not conserved here, so it must not be used.

Question 38Work, Energy, Power and Momentum

A block slides down a rough incline, losing 400 J of gravitational potential energy while 150 J is dissipated by friction. What is the kinetic energy gained by the block?

  1. 550 J
  2. 400 J
  3. 150 J
  4. 250 J Correct answer.

Explanation

By energy conservation, the potential energy lost is split between kinetic energy gained and work done against friction. Hence the kinetic energy gained is 400 J − 150 J = 250 J. Friction removes energy from the mechanical store rather than adding to it.

Question 39Work, Energy, Power and Momentum

A pump raises water through a height of 10 m at a steady rate of 5 kg per second. Taking g = 10 m/s², what minimum power must the pump deliver?

  1. 100 W
  2. 500 W Correct answer.
  3. 5000 W
  4. 50 W

Explanation

Each second the pump gives 5 kg of water a potential energy increase of mgh = 5 × 10 × 10 = 500 J. Power is energy per unit time, so the minimum power is 500 J / 1 s = 500 W. The mass flow rate already incorporates the time factor.

Question 40Work, Energy, Power and Momentum

A force of 80 N is applied to a box at 60° above the horizontal, and the box moves 5 m horizontally. What is the work done by this force? (cos60° = 0.5)

  1. 346 J
  2. 0 J
  3. 200 J Correct answer.
  4. 400 J

Explanation

Only the component of force along the displacement does work: W = F·s·cosθ = 80 × 5 × 0.5 = 200 J. The vertical component of the force is perpendicular to the horizontal displacement and contributes no work.

Question 41Work, Energy, Power and Momentum

A 0.15 kg ball hits a wall horizontally at 12 m/s and rebounds horizontally at 8 m/s. What is the magnitude of the impulse exerted by the wall on the ball?

  1. 1.8 N·s
  2. 0.6 N·s
  3. 3.0 N·s Correct answer.
  4. 1.2 N·s

Explanation

Impulse equals the change in momentum, taking direction into account. Choosing the initial direction as positive, Δp = m(v_final − v_initial) = 0.15 × (−8 − 12) = −3.0 kg·m/s, so the magnitude is 3.0 N·s. The rebound reverses the velocity sign, which must be included.

Question 42Work, Energy, Power and Momentum

Two skaters, initially at rest on frictionless ice, push apart. Skater A has mass 40 kg and moves at 1.5 m/s. What is the speed of skater B, whose mass is 60 kg?

  1. 2.25 m/s
  2. 0.67 m/s
  3. 1.5 m/s
  4. 1.0 m/s Correct answer.

Explanation

The total momentum before the push is zero, so the momenta after must be equal and opposite: m_A·v_A = m_B·v_B. Thus v_B = (40 × 1.5)/60 = 1.0 m/s. The heavier skater moves more slowly to keep the total momentum zero.

Question 43Work, Energy, Power and Momentum

A car of mass 1000 kg accelerates uniformly from rest to 20 m/s in 8 s. What is the average power delivered to the car?

  1. 20000 W
  2. 2500 W
  3. 40000 W
  4. 25000 W Correct answer.

Explanation

The gain in kinetic energy is ½mv² = ½ × 1000 × 400 = 200000 J. Average power is this energy divided by the time: 200000/8 = 25000 W. Using the change in kinetic energy correctly accounts for the acceleration.

Question 44Work, Energy, Power and Momentum

A ball is dropped from rest from a height h and falls freely, with air resistance neglected. Which statement about its energy during the fall is correct?

  1. Total mechanical energy increases as the ball speeds up
  2. Kinetic energy is constant while potential energy decreases
  3. Kinetic energy increases while gravitational potential energy decreases by the same amount Correct answer.
  4. Both kinetic and potential energy decrease

Explanation

In free fall the only force doing work is gravity, which is a conservative force. The loss in gravitational potential energy is converted entirely into kinetic energy, so their sum, the total mechanical energy, stays constant.

Question 45Work, Energy, Power and Momentum

A force acts on a body and its magnitude is plotted against the displacement along the force direction. How is the work done by the force obtained from this graph?

  1. From the reciprocal of the slope
  2. From the maximum value of the force
  3. From the gradient of the line
  4. From the area under the line Correct answer.

Explanation

Work is the integral of force with respect to displacement, W = ∫F·ds. On a force–displacement graph this integral corresponds to the area under the curve between the initial and final positions. The gradient would instead give the rate of change of force with displacement.

Question 46Work, Energy, Power and Momentum

A 3 kg object moving at 4 m/s collides elastically head-on with a stationary 1 kg object. After the collision the 3 kg object moves at 2 m/s in its original direction. What is the speed of the 1 kg object?

  1. 6 m/s Correct answer.
  2. 4 m/s
  3. 2 m/s
  4. 10 m/s

Explanation

Conserve momentum: 3 × 4 = 3 × 2 + 1 × v, so 12 = 6 + v and v = 6 m/s. Checking kinetic energy: initial ½ × 3 × 16 = 24 J, final ½ × 3 × 4 + ½ × 1 × 36 = 6 + 18 = 24 J, confirming the elastic condition.

Question 47Work, Energy, Power and Momentum

A pump must raise water through 20 m and deliver it with a speed of 4 m/s at a rate of 2 kg per second. Taking g = 10 m/s², what minimum power is required?

  1. 400 W
  2. 800 W
  3. 16 W
  4. 416 W Correct answer.

Explanation

Each second, 2 kg gains potential energy mgh = 2 × 10 × 20 = 400 J and kinetic energy ½mv² = ½ × 2 × 16 = 16 J. The total energy per second, and hence the minimum power, is 400 + 16 = 416 W. Both the height gain and the exit speed must be included.

Question 48Work, Energy, Power and Momentum

A bullet of mass 20 g travelling at 400 m/s embeds itself in a 1.98 kg block resting on a frictionless surface. The combined mass then compresses a spring of constant 200 N/m. What is the maximum compression of the spring?

  1. 0.40 m
  2. 0.10 m
  3. 0.80 m
  4. 0.20 m Correct answer.

Explanation

First conserve momentum in the inelastic impact: 0.02 × 400 = 2.0 × v, so v = 4 m/s. Then conserve energy in the spring compression: ½ × 2.0 × 16 = ½ × 200 × x², giving x² = 0.04 and x = 0.20 m. Kinetic energy is lost in the impact and must not be carried through that stage.

Question 49Work, Energy, Power and Momentum

A 5 kg block is pulled 4 m up a 30° incline at constant speed by a force parallel to the incline. The friction force is 10 N. Taking g = 10 m/s², what is the work done by the pulling force?

  1. 240 J
  2. 40 J
  3. 100 J
  4. 140 J Correct answer.

Explanation

At constant speed the pulling force balances both the component of weight along the incline and friction: F = mg·sin30° + f = 5 × 10 × 0.5 + 10 = 35 N. The work done is F·s = 35 × 4 = 140 J. The constant-speed condition is what justifies equating the forces.

Question 50Work, Energy, Power and Momentum

A ball of mass m is released from rest at height h above the ground and bounces. It rebounds to a height of h/4. What fraction of its kinetic energy just before impact is retained as kinetic energy just after impact?

  1. 1/2
  2. 1/4 Correct answer.
  3. 1/16
  4. 3/4

Explanation

Just before impact the kinetic energy equals mgh, and just after impact it equals mg(h/4). The ratio is (h/4)/h = 1/4, so one quarter of the impact kinetic energy is retained. The remaining three quarters are dissipated in deformation and sound during the bounce.