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Admissions tests / ESAT / Physics / Mechanics

Stretch. 15 questions, 15 marks, about 32 minutes.

ESAT Physics: Mechanics, set 4

Scalars and vectors, distance, displacement, speed and velocity, acceleration, the equations of motion, graphs of motion, forces, Newton's laws, momentum, moments, work, energy and power.

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  • Answer all questions. No calculator.
  • Each question has exactly one correct answer.
  • This is a stretch set: questions often combine two specification points, and the fastest route is rarely the longest calculation.
  1. 11 mark

    A ball of mass 0.5 kg is dropped from rest and falls freely (ignore air resistance) through a height of 20 m before it lands. Using g = 10 N/kg as the free-fall acceleration, what is the magnitude of the ball's momentum immediately before it lands?

    1. A 200 kg m/s
    2. B 100 kg m/s
    3. C 10 kg m/s
    4. D 5 kg m/s
  2. 21 mark

    A trolley of mass 5 kg, moving at 10 m/s, decelerates uniformly under a constant resultant force and comes to rest after travelling 5 m. What is the magnitude of the resultant force acting on the trolley?

    1. A 50 N
    2. B 100 N
    3. C 10 N
    4. D 20 N
  3. 31 mark

    A skydiver of mass 80 kg is falling, and at one instant the upward air resistance acting on her is 600 N. Using g = 10 N/kg, what is the magnitude and direction of her acceleration at that instant?

    1. A 10 m/s^2, downward
    2. B 0.25 m/s^2, downward
    3. C 2.5 m/s^2, upward
    4. D 2.5 m/s^2, downward
  4. 41 mark

    A spring of spring constant 200 N/m is compressed by 0.1 m and used to launch a ball of mass 0.4 kg vertically upward from ground level. Assuming all the elastic potential energy stored converts to gravitational potential energy at the ball's maximum height, and using g = 10 N/kg, what maximum height does the ball reach? (Ignore air resistance and the size of the ball.)

    1. A 0.5 m
    2. B 0.25 m
    3. C 2.5 m
    4. D 0.1 m
  5. 51 mark

    Two ice skaters, A (mass 50 kg) and B (mass 75 kg), start at rest and push off from each other. Skater A moves off at 6 m/s. Using conservation of momentum, what is skater B's speed immediately afterward, and how does the size of the force skater A exerts on skater B compare, at each instant during the push, to the size of the force skater B exerts on skater A?

    1. A 4 m/s; the forces are equal in magnitude at every instant, by Newton's third law, regardless of the skaters' different masses.
    2. B 6 m/s; the forces are equal in magnitude at every instant, by Newton's third law.
    3. C 4 m/s; skater B, being heavier, is pushed with a smaller force than the force she exerts on skater A.
    4. D 9 m/s; the forces are equal in magnitude at every instant, by Newton's third law.
  6. 61 mark

    A car accelerates uniformly from rest to 30 m/s in 2 s, travels at this constant 30 m/s for a further 6 s, then decelerates uniformly to rest in 2 s. What is the car's average speed over the whole 10 s journey?

    1. A 30 m/s
    2. B 40 m/s
    3. C 24 m/s
    4. D 21 m/s
  7. 71 mark

    A parachutist of mass 80 kg falls at a constant terminal velocity of 5 m/s for 20 s before landing. Using g = 10 N/kg, how much energy is transferred to heat and sound (by air resistance) during this 20 s fall?

    1. A 1000 J
    2. B 8000 J
    3. C 40000 J
    4. D 80000 J
  8. 81 mark

    A stationary object of total mass 12 kg splits into two fragments in an explosion. One fragment, of mass 4 kg, moves off at 15 m/s. If the push apart lasts 0.02 s, what is the magnitude of the average force exerted on the 4 kg fragment during the explosion?

    1. A 9000 N
    2. B 3000 N
    3. C 1.2 N
    4. D 6000 N
  9. 91 mark

    A person of mass 70 kg stands on weighing scales inside a lift. Using g = 10 N/kg, if the lift accelerates upward at 2 m/s^2, what reading, in newtons, do the scales show (that is, what is the normal contact force the scales exert on the person)?

    1. A 840 N
    2. B 140 N
    3. C 560 N
    4. D 1400 N
  10. 101 mark

    A spring is stretched well beyond its limit of proportionality by a slowly increasing force, and a force-extension graph is plotted for the whole process, starting from zero extension. Which method correctly gives the total elastic energy stored in the spring at the final extension?

    1. A Use E = (1/2)Fx, taking F and x as the final force and final extension, since this formula for elastic energy applies at any extension.
    2. B Use E = (1/2)kx^2, taking k as the final force divided by the final extension, since this always gives the correct spring constant to use.
    3. C Multiply the final force by the final extension directly, E = Fx, since energy transferred always equals force times distance.
    4. D Find the area under the force-extension graph, from zero up to the final extension, whether or not that part of the graph is a straight line.
  11. 111 mark

    An electric hoist lifts a load of mass 60 kg at a constant velocity through a vertical height of 15 m in a time of 12 s. Using g = 10 N/kg, if the hoist's overall percentage efficiency is 60%, what is the total power input to the hoist, in watts?

    1. A 750 W
    2. B 450 W
    3. C 1250 W
    4. D 12.5 W
  12. 121 mark

    A ball of mass 0.4 kg moving at 10 m/s hits a wall and rebounds, moving back the way it came at 6 m/s. The collision with the wall lasts 0.02 s. What is the magnitude of the average force the wall exerts on the ball during the collision?

    1. A 80 N
    2. B 120 N
    3. C 320 N
    4. D 32 N
  13. 131 mark

    A tablecloth lies under several plates on a table. If the tablecloth is pulled out extremely quickly and horizontally, the plates barely move, sliding only a very small distance before coming to rest again. Which explanation correctly accounts for this, in terms of Newton's laws and friction?

    1. A There is no friction at all between the tablecloth and the plates, so the cloth cannot affect the plates as it is pulled away.
    2. B The plates barely move because their weight is far greater than the weight of the tablecloth, so the cloth cannot exert any force on them.
    3. C The force needed to move the plates is greater than the maximum force friction between cloth and plate can ever provide, however fast the cloth is pulled.
    4. D Because the cloth is pulled out so quickly, it is only in contact with the plates for a very short time, so the friction force between them produces only a small change in the plates' momentum before the cloth is gone; by Newton's first law the plates then continue at close to their original, near-zero velocity.
  14. 141 mark

    An astronaut of mass 90 kg on the Moon, where the gravitational field strength is 1.6 N/kg, throws a rock of mass 2 kg so that it leaves her hand at 12 m/s. Comparing this to an identical throw on Earth (same rock, same release speed of 12 m/s), where g = 10 N/kg, how does the magnitude of the rock's momentum as it leaves her hand compare on the Moon?

    1. A The momentum is exactly the same on the Moon as on Earth, because momentum depends only on the rock's mass and velocity, neither of which changes between the two locations.
    2. B The momentum on the Moon is smaller, because the rock weighs less there, so p = mv must be smaller too.
    3. C The momentum on the Moon is larger, because less force is needed to achieve the same throw where gravity is weaker, so more of the astronaut's effort goes into the rock's motion.
    4. D The momentum on the Moon is 1.6/10 times the momentum on Earth, because momentum must scale with the local value of g in the same way that weight does.
  15. 151 mark

    A trolley of mass 2 kg rests against a spring of spring constant 2 N/m on a frictionless horizontal track. The spring is compressed by 3 m and released, launching the trolley. Assuming all the elastic potential energy in the spring converts to the trolley's kinetic energy, what speed does the trolley reach once it leaves the spring?

    1. A 3 sqrt(2) m/s
    2. B 3 m/s
    3. C sqrt(3) m/s
    4. D 4.5 m/s

Worked solutions

Every question below carries the reasoning, not just the answer. The official material for this test publishes a correct option letter and nothing else.

  1. Question 1Answer: C

    1. There is no time given, so the route to velocity is the equation of motion that skips time: v^2 = u^2 + 2as.
    2. Here u = 0 (dropped from rest), a = g = 10 m/s^2 and s = 20 m, so v^2 = 0 + (2 x 10 x 20) = 400.
    3. Taking the square root, v = sqrt(400) = 20 m/s.
    4. Momentum is p = mv = 0.5 kg x 20 m/s = 10 kg m/s, so the answer is C.
    • Why not A: This correctly finds v^2 = u^2 + 2as = 0 + (2 x 10 x 20) = 400 but then reports 400 as v itself and multiplies straight into p = mv, giving 0.5 x 400 = 200. 400 is v^2 in (m/s)^2, not v in m/s; the square root was never taken.
    • Why not B: This substitutes the 20 m distance directly into v = u + at as if it were a time in seconds, giving v = 0 + (10 x 20) = 200 m/s, then p = 0.5 x 200 = 100 kg m/s. Distance and time are different quantities with different units, and mixing them here skips the equation that actually links distance to velocity, v^2 = u^2 + 2as.
    • Why not D: This correctly finds v = 20 m/s from v^2 = u^2 + 2as, but then uses p = (1/2)mv, borrowing the one-half from the kinetic energy formula. Momentum has no factor of one-half: p = mv, not (1/2)mv.

  2. Question 2Answer: A

    1. With no time given, use v^2 = u^2 + 2as to find the deceleration: 0 = 10^2 + (2 x a x 5), so 0 = 100 + 10a.
    2. Solving, a = -100/10 = -10 m/s^2 (a deceleration of magnitude 10 m/s^2).
    3. Newton's second law, F = ma, then gives the resultant force: F = 5 kg x 10 m/s^2 = 50 N.
    4. The magnitude of the resultant force is 50 N, so the answer is A.
    • Why not B: This drops the factor of 2 from the equation of motion, treating it as v^2 = u^2 + as rather than v^2 = u^2 + 2as: 0 = 100 + (a x 5), giving a = -20 m/s^2 and F = 5 x 20 = 100 N, exactly double the correct force.
    • Why not C: This correctly finds the deceleration as 10 m/s^2 using v^2 = u^2 + 2as, but then reports that number of m/s^2 directly as the force in newtons, forgetting to multiply by the mass at all. Newton's second law is F = ma; a number of m/s^2 is not a number of newtons until it is multiplied by the mass.
    • Why not D: This divides by s^2 instead of 2s when rearranging the equation of motion for acceleration: a = (u^2 - v^2)/s^2 = 100/25 = 4 m/s^2, giving F = 5 x 4 = 20 N. The equation of motion has a factor of 2s, a distance, not s^2, a distance squared.

  3. Question 3Answer: D

    1. Her weight is w = mg = 80 kg x 10 N/kg = 800 N, acting downward.
    2. The resultant force is the difference between the two opposing vertical forces: 800 N (down) - 600 N (up) = 200 N, acting downward, since weight is still the larger force.
    3. By Newton's second law, a = F/m = 200 N / 80 kg = 2.5 m/s^2.
    4. The skydiver's acceleration is 2.5 m/s^2, directed downward (she is still speeding up, just more slowly than in free fall), so the answer is D.
    • Why not A: This assumes the skydiver is in free fall, so acceleration must simply equal g, ignoring the air resistance force entirely. Air resistance is a real upward force here, so the resultant force, and hence the acceleration, must be smaller than in free fall.
    • Why not B: This correctly finds the resultant force as 800 - 600 = 200 N, but divides by the weight (800 N) instead of the mass (80 kg) to find acceleration, giving 200/800 = 0.25 m/s^2. Newton's second law needs mass in kg, not weight in newtons, in a = F/m.
    • Why not C: This gets the correct magnitude (2.5 m/s^2) but assumes that because air resistance opposes the fall, the net acceleration must point upward. Weight (800 N down) is still larger than air resistance (600 N up), so the resultant force, and the acceleration, still point downward, just reduced in size from free fall.

  4. Question 4Answer: B

    1. Elastic potential energy stored in the spring: E = (1/2)kx^2 = (1/2) x 200 N/m x (0.1 m)^2 = (1/2) x 200 x 0.01 = 1 J.
    2. This energy converts entirely to gravitational potential energy at maximum height: E = mgh.
    3. Substituting, 1 = 0.4 kg x 10 N/kg x h = 4h, so h = 1/4 = 0.25 m.
    4. The ball reaches a maximum height of 0.25 m, so the answer is B.
    • Why not A: This drops the one-half from the elastic energy formula, computing E = kx^2 = 200 x 0.1^2 = 2 J instead of E = (1/2)kx^2 = 1 J. Using this doubled energy in E = mgh gives h = 2/4 = 0.5 m, exactly twice the correct height.
    • Why not C: This uses x instead of x^2 in the elastic energy formula, computing E = (1/2)k x = (1/2) x 200 x 0.1 = 10 J instead of squaring the extension first. The formula needs the extension squared, since the force needed to compress the spring grows as the extension grows.
    • Why not D: This correctly finds the elastic energy as 1 J, but then drops the mass from the gravitational potential energy formula, using E = gh instead of E = mgh: 1 = 10 x h, giving h = 0.1 m. Both mass and g are needed to convert an energy in joules into a height in metres.

  5. Question 5Answer: A

    1. Before the push, total momentum is zero (both skaters at rest).
    2. After the push, the two momenta must be equal and opposite so the total remains zero: m_A v_A = m_B v_B, so 50 x 6 = 75 x v_B.
    3. Solving, v_B = 300/75 = 4 m/s.
    4. By Newton's third law, the force skater A exerts on skater B and the force skater B exerts on skater A are equal in magnitude at every instant of the push, even though the resulting speeds differ because the skaters have different masses, so the answer is A.
    • Why not B: This assumes 'equal and opposite' means the skaters recoil at equal speeds, so it simply repeats A's speed for B. Newton's third law equalises the forces on the two skaters at every instant, not their resulting speeds; with different masses, equal forces produce different accelerations and so different final speeds.
    • Why not C: This gets B's speed right (4 m/s, from momentum conservation) but wrongly reasons that a larger mass means a smaller force is exerted on it, or by it. Newton's third law forces are equal in magnitude regardless of the two objects' masses; it is the resulting acceleration, F/m, that differs between the two skaters, not the force itself.
    • Why not D: This inverts the momentum equation, computing v_B = (m_B/m_A) x v_A = (75/50) x 6 = 9 m/s instead of v_B = (m_A/m_B) x v_A. Conservation of momentum requires m_A v_A = m_B v_B, so the heavier skater must move off with the SMALLER speed, not the larger one.

  6. Question 6Answer: C

    1. Average speed = total distance / total time, and the total distance is the area under the velocity-time graph.
    2. The acceleration phase is a triangle of area (1/2) x 2 x 30 = 30 m; the cruising phase is a rectangle of area 6 x 30 = 180 m; the deceleration phase is a second triangle of area (1/2) x 2 x 30 = 30 m.
    3. Total distance = 30 + 180 + 30 = 240 m, over a total time of 2 + 6 + 2 = 10 s.
    4. Average speed = 240/10 = 24 m/s, so the answer is C.
    • Why not A: This treats 'average speed' as simply the constant cruising speed, since the car spends the most time at 30 m/s. But the two ramps, where the car is slower than 30 m/s throughout, pull the true average below the cruising speed.
    • Why not B: This correctly finds the total distance (240 m, the area under the velocity-time graph) but divides by only the 6 s cruising time instead of the full 10 s journey time, giving 240/6 = 40 m/s. Average speed needs the total distance over the TOTAL time, not just one phase of it.
    • Why not D: This correctly keeps the full 10 s journey time, but forgets to include the final deceleration phase's distance (a triangle of area 30 m) in the total, using 240 - 30 = 210 m instead of 240 m, giving 210/10 = 21 m/s.

  7. Question 7Answer: D

    1. At terminal velocity the parachutist's kinetic energy is not changing, since her speed is constant, so all of the gravitational potential energy she loses must be transferred to heat and sound by air resistance.
    2. The distance fallen at constant velocity is s = vt = 5 m/s x 20 s = 100 m.
    3. The gravitational potential energy lost is E = mgh = 80 kg x 10 N/kg x 100 m = 80000 J.
    4. So 80000 J (80 kJ) is transferred to heat and sound during the fall, so the answer is D.
    • Why not A: This computes (1/2)mv^2 = (1/2) x 80 x 5^2 = 1000 J, as if all the parachutist's lost gravitational potential energy converted into kinetic energy from a standing start. But her velocity is constant throughout (terminal velocity), so her kinetic energy does not change at all; every joule of lost gravitational potential energy must instead go to heat and sound.
    • Why not B: This correctly finds the distance fallen (100 m) but drops g from the gravitational potential energy formula, computing E = mh = 80 x 100 = 8000 J instead of E = mgh.
    • Why not C: This finds the distance fallen using s = (1/2)vt = (1/2) x 5 x 20 = 50 m, the formula for distance under constant acceleration from rest. But the parachutist moves at a CONSTANT velocity throughout, so the correct distance is simply s = vt = 5 x 20 = 100 m, twice as far.

  8. Question 8Answer: B

    1. The 4 kg fragment starts at rest (the whole object is stationary) and ends up moving at 15 m/s, so its momentum changes by (4 kg x 15 m/s) - 0 = 60 kg m/s.
    2. Force is the rate of change of momentum: F = (change in momentum) / (time taken).
    3. Substituting, F = 60 kg m/s / 0.02 s = 3000 N.
    4. The magnitude of the average force on the 4 kg fragment is 3000 N, so the answer is B.
    • Why not A: This uses the total mass of the original object (12 kg) rather than the 4 kg fragment the question actually asks about, giving a momentum change of 12 x 15 = 180 kg m/s and a force of 180/0.02 = 9000 N. Only the 4 kg fragment's own momentum change determines the force on it.
    • Why not C: This correctly finds the 4 kg fragment's momentum change as 60 kg m/s, but multiplies by the time instead of dividing by it: 60 x 0.02 = 1.2 N. Force is the RATE of change of momentum, so the momentum change must be divided by the time taken, not multiplied by it.
    • Why not D: This uses the mass of the OTHER fragment (12 - 4 = 8 kg) together with the 4 kg fragment's velocity, giving a momentum change of 8 x 15 = 120 kg m/s and a force of 120/0.02 = 6000 N. The 15 m/s given belongs to the 4 kg fragment, and must be paired with its own mass, not the other fragment's.

  9. Question 9Answer: A

    1. The person's weight is w = mg = 70 kg x 10 N/kg = 700 N, acting downward.
    2. Taking upward as positive, Newton's second law gives: N - w = ma, where N is the normal contact force from the scales.
    3. Rearranging, N = w + ma = 700 + (70 x 2) = 700 + 140 = 840 N.
    4. The scales read 840 N, so the answer is A.
    • Why not B: This reports only ma = 70 x 2 = 140 N, dropping the person's weight entirely. The scale reading is the TOTAL normal contact force, which must supply both the person's weight and the extra force needed to accelerate her upward.
    • Why not C: This subtracts the extra force from the weight instead of adding it, giving N = 700 - 140 = 560 N, as though accelerating upward reduces the scale reading. An upward acceleration means the resultant force is upward, so the normal contact force must exceed the weight, not fall below it.
    • Why not D: This multiplies the weight by the acceleration instead of adding ma to it, giving N = 700 x 2 = 1400 N. The two contributions to the normal force (supporting the weight, and providing the extra resultant force) must be added, not multiplied.

  10. Question 10Answer: D

    1. Elastic potential energy stored equals the work done stretching the spring, which is always the area under its force-extension graph.
    2. Within the limit of proportionality, that area happens to be a triangle, which is why E = (1/2)Fx = (1/2)kx^2 works there.
    3. Beyond the limit of proportionality, the graph curves (force is no longer proportional to extension), so the area is no longer a simple triangle and those formulas no longer apply.
    4. The one method that always works, straight-line region or not, is to read the actual area under the graph up to the final extension, so the answer is D.
    • Why not A: The formula E = (1/2)Fx only gives the correct area when the graph is a straight line through the origin (Hooke's law region). Beyond the limit of proportionality the graph curves, so treating the region as a single triangle using the final force and extension no longer matches the true area underneath it.
    • Why not B: This assumes a single spring constant k describes the whole stretch, calculated from the final point alone. Beyond the limit of proportionality, force is no longer proportional to extension, so k is not constant across the graph, and using one 'k' calculated at the end misrepresents everything that happened before it.
    • Why not C: This confuses the energy stored in a spring, where the force needed grows as the extension grows, with the work done by a CONSTANT force moving through a distance (W = Fx). Using the final force for the whole extension overestimates the energy, by a factor of two even in the straight-line region alone, let alone beyond it.

  11. Question 11Answer: C

    1. The useful energy output is the gravitational potential energy gained: E = mgh = 60 kg x 10 N/kg x 15 m = 9000 J.
    2. The useful power output is this energy divided by the time taken: 9000 J / 12 s = 750 W.
    3. Percentage efficiency = (useful power output / total power input) x 100, so 60 = (750 / P) x 100, giving P = 750/0.6.
    4. P = 1250 W, so the total power input to the hoist is 1250 W, and the answer is C.
    • Why not A: This correctly finds the useful power output (750 W) but reports it directly as the total power input, forgetting that the hoist is only 60% efficient and so must be supplied with MORE power than it usefully outputs.
    • Why not B: This multiplies the useful power by the efficiency instead of dividing by it: 750 x 0.6 = 450 W. A useful output that is a FRACTION of the input means the input must be found by dividing the output by that fraction, not by multiplying.
    • Why not D: This divides the useful power by 60 directly, rather than by the decimal fraction 0.6, giving 750/60 = 12.5 W. The percentage must be converted to a fraction (60% = 0.6) before it is used in the efficiency formula.

  12. Question 12Answer: C

    1. Taking the ball's initial direction as positive, its momentum changes from (0.4 x 10) = 4.0 kg m/s to (0.4 x -6) = -2.4 kg m/s, since it reverses direction.
    2. The change in momentum is -2.4 - 4.0 = -6.4 kg m/s, so its magnitude is 6.4 kg m/s.
    3. Force is the rate of change of momentum: F = 6.4 kg m/s / 0.02 s.
    4. F = 320 N, so the magnitude of the average force the wall exerts on the ball is 320 N, and the answer is C.
    • Why not A: This treats the rebound as if the ball kept moving in the same direction, just slower, using only the SPEED difference: 0.4 x (10 - 6) = 1.6 kg m/s, giving 1.6/0.02 = 80 N. But the ball reverses direction, so its velocity changes sign, not just size, and the momentum change must account for that reversal.
    • Why not B: This uses only the final momentum (0.4 x 6 = 2.4 kg m/s) to find the change, as though the ball started from rest, giving 2.4/0.02 = 120 N. The change in momentum must be the difference between the final AND initial momentum, and the ball had momentum before the collision too.
    • Why not D: This correctly finds the momentum change as 6.4 kg m/s, but misreads the collision time as 0.2 s instead of 0.02 s, giving 6.4/0.2 = 32 N, ten times too small.

  13. Question 13Answer: D

    1. Friction between the cloth and the plates does act on the plates as the cloth is pulled away, and this force is what changes the plates' momentum.
    2. Force is the rate of change of momentum, so for a given frictional force, a very short contact time produces only a very small total change in momentum.
    3. A small change in momentum, for the plates' mass, means only a small change in velocity, so the plates barely start moving before the cloth is already gone from underneath them.
    4. By Newton's first law, with the cloth (and its friction) gone, the plates then continue at close to their small, near-zero velocity, sliding only a little before friction with the table itself brings them to rest, so the answer is D.
    • Why not A: There is friction between cloth and plates, and it is exactly what makes the plates move at all (they do slide a small distance). The trick depends on that friction force acting for only a very short TIME, not on friction being absent.
    • Why not B: The relevant comparison is not the plates' weight against the cloth's weight; weight has no direct bearing on how much friction acts between their surfaces or for how long. The trick works because of the very short contact TIME, whatever the plates weigh.
    • Why not C: Friction from the moving cloth clearly does act on the plates, since they do slide a small distance rather than staying perfectly still. It is not that friction is incapable of moving the plates at all, but that it acts for such a short time that it changes their momentum, and hence their velocity, by only a small amount.

  14. Question 14Answer: A

    1. Momentum is defined as p = mv, depending only on an object's mass and its velocity.
    2. The rock's mass (2 kg) is a fixed property of the rock and does not change between the Moon and the Earth; only its weight, w = mg, changes, because g differs.
    3. The release speed is also given as identical in both cases, at 12 m/s.
    4. Since both mass and velocity are unchanged, the momentum is identical in both cases: p = 2 kg x 12 m/s = 24 kg m/s on the Moon and on Earth, so the answer is A.
    • Why not B: Momentum is defined as p = mv, which contains mass, not weight. The rock's mass (2 kg) does not change between the Moon and the Earth, only its weight does, and weight simply does not appear in the definition of momentum.
    • Why not C: Whatever force or effort the astronaut needs to use to achieve the throw, the question fixes both the rock's mass (2 kg) and its release speed (12 m/s) as identical in both cases. Since momentum is calculated directly from these two values alone, it cannot depend on how much force was needed to produce them.
    • Why not D: This mistakenly applies the formula for WEIGHT (w = mg, which genuinely does scale with g) to momentum (p = mv), a completely different quantity that does not involve g at all.

  15. Question 15Answer: B

    1. Elastic potential energy stored in the spring: E = (1/2)kx^2 = (1/2) x 2 N/m x (3 m)^2 = (1/2) x 2 x 9 = 9 J.
    2. This converts entirely into the trolley's kinetic energy: E = (1/2)mv^2, so 9 = (1/2)(2)v^2 = v^2.
    3. Solving, v^2 = 9, so v = sqrt(9) = 3 m/s.
    4. The trolley's speed as it leaves the spring is 3 m/s, so the answer is B.
    • Why not A: This drops the one-half from the elastic energy formula, computing E = kx^2 = 2 x 3^2 = 18 J instead of E = (1/2)kx^2 = 9 J. Setting (1/2)(2)v^2 = 18 gives v^2 = 18, so v = sqrt(18) = 3 sqrt(2) m/s, larger than the true speed because the starting energy was doubled.
    • Why not C: This forgets to square the extension in the elastic energy formula, computing E = (1/2)kx = (1/2) x 2 x 3 = 3 J instead of (1/2)kx^2 = 9 J. Setting (1/2)(2)v^2 = 3 gives v^2 = 3, so v = sqrt(3) m/s.
    • Why not D: This correctly finds the elastic energy as 9 J, but then sets it equal to E = mv (as if energy and momentum used the same formula) instead of E = (1/2)mv^2, giving 9 = 2v, so v = 4.5 m/s. Kinetic energy needs velocity squared, and a factor of one-half; momentum does not.

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