4.5 Forces — coverage pack

Tier 1 · Easy

1. 1. A trolley has a mass of $6.0\,\text{kg}$ and moves with a velocity of $2.5\,\text{m/s}$ east. State which of these two quantities is a vector.[1 mark]

   Answer

   • Velocity is the vector quantity.

   Method: Mass needs only a magnitude, so it is scalar. The velocity includes the direction east, so velocity is the vector.

Tier 2 · Standard

1. 1. A scale drawing uses $1.0\,\text{cm}$ to represent $4.0\,\text{N}$. Describe the arrow that represents a force of $20\,\text{N}$ acting vertically downwards.[2 marks]

   Answer

   • The arrow is $5.0\,\text{cm}$ long.
   • Its arrowhead points vertically downwards.

   Method: The required length is $20\div4.0=5.0\,\text{cm}$. Draw that length vertically and put the arrowhead at the downward end to show the force direction.

Tier 3 · Hard

1. 1. A cyclist's speedometer reads $11\,\text{m/s}$ while the cyclist travels south-west. Give the cyclist's speed and velocity, and explain why these are different quantities even though their magnitudes match.[3 marks]

   Answer

   • Speed: $11\,\text{m/s}$.
   • Velocity: $11\,\text{m/s}$ south-west.
   • Speed is scalar, while velocity includes direction and is vector.

   Method: Keep only the magnitude for speed. Attach the stated direction to the same magnitude for velocity. The direction is what makes the velocity a different, vector quantity.

Tier 1 · Easy

1. 1. State whether friction and gravitational force are contact or non-contact forces.[1 mark]

   Answer

   • Friction is contact; gravitational force is non-contact.

   Method: Friction requires touching surfaces. Gravitational force acts between separated masses, so it does not require contact.

Tier 2 · Standard

1. 1. A magnet attracts an iron pin across a small gap. Describe the force interaction between the magnet and the pin.[2 marks]

   Answer

   • The magnetic force is non-contact.
   • The magnet pulls the pin and the pin pulls the magnet, with the forces directed towards each other.

   Method: The objects are separated, so the interaction is non-contact. The interaction acts on both objects: draw one arrow on the pin towards the magnet and one on the magnet towards the pin.

Tier 3 · Hard

1. 1. A parachutist is falling through the air while attached to an open parachute by cords. Name one non-contact force and two different contact forces acting in this situation, giving the object on which each named force acts.[3 marks]

   Answer

   • Gravitational force acts on the parachutist or parachute and is non-contact.
   • Air resistance acts on the parachute or parachutist and is contact.
   • Tension acts through the cords on the parachutist or parachute and is contact.

   Method: Identify gravity as acting without physical contact. Then identify collisions with air as air resistance and pulling by the cords as tension; both require contact with the object experiencing the force.

Tier 1 · Easy

1. 1. A $7.5\,\text{kg}$ bag is in a region where $g=9.8\,\text{N/kg}$. Calculate the bag's weight.[1 mark]

   Answer

   • $73.5\,\text{N}$

   Method: Use $W=mg$: $W=7.5\times9.8=73.5\,\text{N}$.

Tier 2 · Standard

1. 1. A sample has mass $2.4\,\text{kg}$. Its weight is $23.5\,\text{N}$ at location A and $3.8\,\text{N}$ at location B. Calculate the gravitational field strength at each location.[3 marks]

   Answer

   • At A, $g=9.79\,\text{N/kg}$.
   • At B, $g=1.58\,\text{N/kg}$.

   Method: Rearrange $W=mg$ to $g=W/m$. At A, $g=23.5\div2.4=9.79\,\text{N/kg}$. At B, $g=3.8\div2.4=1.58\,\text{N/kg}$.

Tier 3 · Hard

1. 1. An explorer has mass $68\,\text{kg}$. A newtonmeter would show $666\,\text{N}$ for the explorer on planet P and $177\,\text{N}$ on moon Q. Determine $g$ at P and Q, then explain what happens to the explorer's mass during the journey.[4 marks]

   Answer

   • At P, $g=9.79\,\text{N/kg}$.
   • At Q, $g=2.60\,\text{N/kg}$.
   • The mass remains $68\,\text{kg}$ because changing gravitational field strength changes weight, not mass.

   Method: Use $g=W/m$. For P, $666\div68=9.79\,\text{N/kg}$; for Q, $177\div68=2.60\,\text{N/kg}$. Mass is independent of location, so it stays $68\,\text{kg}$ while the weight changes.

Tier 1 · Easy

1. 1. Two horizontal forces act on a crate: $16\,\text{N}$ east and $9\,\text{N}$ west. Calculate the resultant force.[1 mark]

   Answer

   • $7\,\text{N}$ east

   Method: The forces oppose, so subtract their magnitudes: $16-9=7\,\text{N}$. The larger force acts east, so the resultant is east.

Tier 2 · Standard

1. 1. A model boat is pulled forwards by $42\,\text{N}$. Water resistance is $27\,\text{N}$ and air resistance is $6\,\text{N}$, both backwards. Determine the resultant force on the boat.[2 marks]

   Answer

   • $9\,\text{N}$ forwards

   Method: The total backward force is $27+6=33\,\text{N}$. Therefore the resultant is $42-33=9\,\text{N}$ forwards.

Tier 3 · Hard

1. 1. Forces on a rail cart are $85\,\text{N}$ east, $34\,\text{N}$ west and $19\,\text{N}$ west. Calculate the resultant, then state the additional single force needed to make the forces balanced.[3 marks]

   Answer

   • The resultant is $32\,\text{N}$ east.
   • An additional force of $32\,\text{N}$ west is needed.

   Method: Combine the westward forces: $34+19=53\,\text{N}$. The resultant is $85-53=32\,\text{N}$ east. A balancing force must be equal and opposite, so it is $32\,\text{N}$ west.

Tier 1 · Easy

1. 1. A horizontal force of $35\,\text{N}$ moves a box $4.0\,\text{m}$ horizontally. Calculate the work done by the force.[1 mark]

   Answer

   • $140\,\text{J}$

   Method: The movement is along the force's line of action, so $W=Fs=35\times4.0=140\,\text{J}$.

Tier 2 · Standard

1. 1. A winch lifts a load vertically through $6.5\,\text{m}$ using a constant upward force of $480\,\text{N}$. Calculate the work done and identify the energy store that increases.[3 marks]

   Answer

   • $3.12\times10^3\,\text{J}$
   • The gravitational potential energy store of the load increases.

   Method: The displacement is along the upward force, so $W=480\times6.5=3120\,\text{J}$. Lifting transfers energy to the load's gravitational potential energy store.

Tier 3 · Hard

1. 1. A powered trolley moves $12\,\text{m}$ along a level floor. Its motor provides a forward force of $180\,\text{N}$ while friction is $140\,\text{N}$. Calculate the work done by the motor, the energy transferred thermally by friction and the remaining energy transferred to the trolley's kinetic energy store.[5 marks]

   Answer

   • Motor work: $2160\,\text{J}$.
   • Thermal transfer: $1680\,\text{J}$.
   • Increase in kinetic energy store: $480\,\text{J}$.

   Method: The motor does $W=180\times12=2160\,\text{J}$. Work against friction is $140\times12=1680\,\text{J}$ and is transferred thermally. The remainder is $2160-1680=480\,\text{J}$, transferred to the kinetic energy store.

Tier 1 · Easy

1. 1. A spring's length changes from $0.18\,\text{m}$ to $0.23\,\text{m}$. Calculate its extension.[1 mark]

   Answer

   • $0.050\,\text{m}$

   Method: Extension is stretched length minus original length: $e=0.23-0.18=0.050\,\text{m}$.

Tier 2 · Standard

1. 1. Within its linear region, a spring extends by $0.075\,\text{m}$ when a force of $18\,\text{N}$ is applied. Calculate the spring constant and predict the extension produced by $12\,\text{N}$.[3 marks]

   Answer

   • $k=240\,\text{N/m}$
   • Extension at $12\,\text{N}$ is $0.050\,\text{m}$.

   Method: Rearrange $F=ke$ to $k=F/e=18\div0.075=240\,\text{N/m}$. Still within the linear region, $e=F/k=12\div240=0.050\,\text{m}$.

Tier 3 · Hard

1. 1. A spring of constant $240\,\text{N/m}$ is stretched by $0.18\,\text{m}$ without exceeding its limit of proportionality. Calculate the applied force and the elastic potential energy stored. State what observation after unloading would show that the spring had instead been inelastically deformed.[5 marks]

   Answer

   • Applied force: $43.2\,\text{N}$.
   • Elastic potential energy: $3.89\,\text{J}$.
   • Inelastic deformation would be shown if the spring did not return to its original length.

   Method: Use $F=ke=240\times0.18=43.2\,\text{N}$. Then $E_e=\frac12ke^2=0.5\times240\times0.18^2=3.888\,\text{J}$, which rounds to $3.89\,\text{J}$. A permanent extension after removing the force indicates inelastic deformation.

Tier 1 · Easy

1. 1. A force of $28\,\text{N}$ acts perpendicular to a handle $0.35\,\text{m}$ from its pivot. Calculate the moment.[1 mark]

   Answer

   • $9.8\,\text{N m}$

   Method: Use $M=Fd$: $M=28\times0.35=9.8\,\text{N m}$.

Tier 2 · Standard

1. 1. A seesaw is balanced. A child of weight $360\,\text{N}$ sits $1.5\,\text{m}$ to the left of the pivot. Calculate how far to the right of the pivot a child of weight $450\,\text{N}$ must sit.[3 marks]

   Answer

   • $1.2\,\text{m}$

   Method: Balance clockwise and anticlockwise moments: $360\times1.5=450d$. Therefore $d=540\div450=1.2\,\text{m}$.

Tier 3 · Hard

1. 1. A uniform $4.0\,\text{m}$ beam of weight $220\,\text{N}$ is supported at its left end and at a point $3.2\,\text{m}$ from the left end. A $300\,\text{N}$ load is placed at the right end. Calculate the upward force from the support at $3.2\,\text{m}$.[4 marks]

   Answer

   • $513\,\text{N}$ upward

   Method: The beam's weight acts at its centre, $2.0\,\text{m}$ from the left end. Taking moments about the left support gives $R\times3.2=(220\times2.0)+(300\times4.0)=1640\,\text{N m}$. Hence $R=1640\div3.2=512.5\,\text{N}$, or $513\,\text{N}$ upward.

Tier 1 · Easy

1. 1. A gas pushes normally on a hatch with force $240\,\text{N}$. The hatch area is $0.080\,\text{m}^2$. Calculate the pressure.[1 mark]

   Answer

   • $3000\,\text{Pa}$

   Method: Use $p=F/A$: $p=240\div0.080=3000\,\text{Pa}$.

Tier 2 · Standard

1. 1. A liquid exerts a normal force of $54\,\text{N}$ on a sensor of area $30\,\text{cm}^2$. Calculate the pressure on the sensor.[3 marks]

   Answer

   • $1.8\times10^4\,\text{Pa}$

   Method: Convert the area: $30\,\text{cm}^2=30\times10^{-4}=0.0030\,\text{m}^2$. Then $p=54\div0.0030=18000\,\text{Pa}$.

Tier 3 · Hard

1. 1. A sealed chamber produces the same normal force of $1260\,\text{N}$ on either of two removable panels. Panel X measures $0.30\,\text{m}$ by $0.20\,\text{m}$ and panel Y measures $0.45\,\text{m}$ by $0.35\,\text{m}$. Calculate the pressure on each panel and determine which panel experiences the lower pressure.[5 marks]

   Answer

   • Panel X: $2.10\times10^4\,\text{Pa}$.
   • Panel Y: $8.00\times10^3\,\text{Pa}$.
   • Panel Y experiences the lower pressure.

   Method: Panel X has area $0.30\times0.20=0.060\,\text{m}^2$, so $p_X=1260\div0.060=21000\,\text{Pa}$. Panel Y has area $0.45\times0.35=0.1575\,\text{m}^2$, so $p_Y=1260\div0.1575=8000\,\text{Pa}$. The same force spread over Y's larger area gives the lower pressure.

Tier 1 · Easy

1. 1. Oil has density $820\,\text{kg/m}^3$. Calculate the pressure due to a $1.5\,\text{m}$ column of the oil when $g=9.8\,\text{N/kg}$.[2 marks]

   Answer

   • $1.21\times10^4\,\text{Pa}$

   Method: Use $p=h\rho g=1.5\times820\times9.8=12054\,\text{Pa}$, which is $1.21\times10^4\,\text{Pa}$ to three significant figures.

Tier 2 · Standard

1. 1. Two pressure sensors are $0.40\,\text{m}$ and $2.10\,\text{m}$ below the surface of a liquid of density $960\,\text{kg/m}^3$. Calculate the pressure difference when $g=9.8\,\text{N/kg}$.[3 marks]

   Answer

   • $1.60\times10^4\,\text{Pa}$

   Method: The depth difference is $2.10-0.40=1.70\,\text{m}$. Hence $\Delta p=\Delta h\rho g=1.70\times960\times9.8=15993.6\,\text{Pa}$, or $1.60\times10^4\,\text{Pa}$.

Tier 3 · Hard

1. 1. A fully submerged cuboid is $0.30\,\text{m}$ high and has horizontal top and bottom areas of $0.012\,\text{m}^2$. It is in water of density $1000\,\text{kg/m}^3$, where $g=9.8\,\text{N/kg}$. Calculate the pressure difference between its bottom and top, use this to calculate the upthrust, and predict its initial motion if its weight is $30\,\text{N}$.[5 marks]

   Answer

   • Pressure difference: $2940\,\text{Pa}$.
   • Upthrust: $35.3\,\text{N}$.
   • It initially accelerates upwards because upthrust exceeds weight.

   Method: The depth difference is the cuboid height, so $\Delta p=h\rho g=0.30\times1000\times9.8=2940\,\text{Pa}$. The force difference is $F=\Delta pA=2940\times0.012=35.28\,\text{N}$. Since $35.28\,\text{N}>30\,\text{N}$, the resultant is upward and the cuboid initially accelerates upward.

Tier 1 · Easy

1. 1. State what microscopic event produces atmospheric pressure on a window.[1 mark]

   Answer

   • Air molecules collide with the window surface.

   Method: Use the particle model: moving air molecules repeatedly strike the surface, and their collisions create pressure.

Tier 2 · Standard

1. 1. Explain why a barometer records a lower atmospheric pressure at the top of a tall mountain than at sea level.[3 marks]

   Answer

   • There are fewer air molecules above the mountain-top instrument.
   • The air is less dense at the greater altitude.
   • There are fewer molecular collisions with the instrument, so the pressure is lower.

   Method: Link greater altitude to a smaller amount and weight of air above the surface. This means lower air density and fewer collisions per unit area, giving lower pressure.

Tier 3 · Hard

1. 1. At high altitude, the pressure inside a sealed case is $101\,\text{kPa}$ and the atmospheric pressure outside is $75\,\text{kPa}$. A flat lid has area $0.012\,\text{m}^2$. Calculate the resultant force on the lid and state its direction.[4 marks]

   Answer

   • $312\,\text{N}$ outwards

   Method: The pressure difference is $101-75=26\,\text{kPa}=26000\,\text{Pa}$. Using $F=\Delta pA$ gives $F=26000\times0.012=312\,\text{N}$. The larger pressure is inside, so the resultant force is outwards.

Tier 1 · Easy

1. 1. A walker travels $14\,\text{m}$ north and then $5\,\text{m}$ south. Determine the distance and displacement.[2 marks]

   Answer

   • Distance $19\,\text{m}$; displacement $9\,\text{m}$ north.

   Method: Distance adds both path lengths: $14+5=19\,\text{m}$. Taking north as positive, displacement is $14-5=9\,\text{m}$ north.

Tier 2 · Standard

1. 1. A robot moves $6.0\,\text{m}$ east and then $8.0\,\text{m}$ north. Calculate its distance and the magnitude and direction of its displacement.[4 marks]

   Answer

   • Distance $14.0\,\text{m}$.
   • Displacement $10.0\,\text{m}$ at $53.1^\circ$ north of east.

   Method: The path length is $6.0+8.0=14.0\,\text{m}$. The straight-line displacement is $\sqrt{6.0^2+8.0^2}=10.0\,\text{m}$. Its direction is $\tan^{-1}(8.0/6.0)=53.1^\circ$ north of east.

Tier 3 · Hard

1. 1. A survey drone flies $600\,\text{m}$ east, $250\,\text{m}$ west and then $120\,\text{m}$ north. Calculate the total distance and the magnitude and direction of its displacement from launch.[5 marks]

   Answer

   • Distance $970\,\text{m}$.
   • Displacement $370\,\text{m}$ at $18.9^\circ$ north of east.

   Method: Distance is $600+250+120=970\,\text{m}$. The net components are $350\,\text{m}$ east and $120\,\text{m}$ north. The magnitude is $\sqrt{350^2+120^2}=370\,\text{m}$ and the direction is $\tan^{-1}(120/350)=18.9^\circ$ north of east.

Tier 1 · Easy

1. 1. A toy car travels $24\,\text{m}$ in $6.0\,\text{s}$ at constant speed. Calculate its speed.[1 mark]

   Answer

   • $4.0\,\text{m/s}$

   Method: Use $v=s/t$: $v=24\div6.0=4.0\,\text{m/s}$.

Tier 2 · Standard

1. 1. A runner covers $300\,\text{m}$ in $48\,\text{s}$, rests for $12\,\text{s}$ and then covers another $200\,\text{m}$ in $32\,\text{s}$. Calculate the average speed for the whole interval.[3 marks]

   Answer

   • $5.43\,\text{m/s}$

   Method: Total distance is $300+200=500\,\text{m}$. Total time includes the rest: $48+12+32=92\,\text{s}$. Average speed is $500\div92=5.43\,\text{m/s}$.

Tier 3 · Hard

1. 1. A delivery drone travels $1.8\,\text{km}$ to a site in $2\,\text{min}\ 20\,\text{s}$. It waits for $30\,\text{s}$, then returns along the same route at a constant $15\,\text{m/s}$. Calculate its average speed for the complete trip from departure to return.[5 marks]

   Answer

   • $12.4\,\text{m/s}$

   Method: Convert $1.8\,\text{km}$ to $1800\,\text{m}$ and the outward time to $140\,\text{s}$. The return time is $1800\div15=120\,\text{s}$. Total distance is $3600\,\text{m}$ and total time is $140+30+120=290\,\text{s}$, so average speed is $3600\div290=12.4\,\text{m/s}$.

Tier 1 · Easy

1. 1. A train moves north at a speed of $18\,\text{m/s}$. State its velocity.[1 mark]

   Answer

   • $18\,\text{m/s}$ north

   Method: Velocity is speed with direction, so attach the stated direction north to the magnitude $18\,\text{m/s}$.

Tier 2 · Standard

1. 1. A cart moves $50\,\text{m}$ east and then $20\,\text{m}$ west in a total time of $10\,\text{s}$. Calculate its average velocity.[3 marks]

   Answer

   • $3.0\,\text{m/s}$ east

   Method: The displacement is $50-20=30\,\text{m}$ east. Average velocity is displacement divided by time: $30\div10=3.0\,\text{m/s}$ east.

Tier 3 · Hard

1. 1. A rescue vehicle travels $90\,\text{m}$ east and then $120\,\text{m}$ north in $30\,\text{s}$. Calculate the magnitude and direction of its average velocity, and compare this with its average speed.[5 marks]

   Answer

   • Average velocity $5.0\,\text{m/s}$ at $53.1^\circ$ north of east.
   • Average speed $7.0\,\text{m/s}$, which is greater because it uses distance rather than displacement.

   Method: Displacement is $\sqrt{90^2+120^2}=150\,\text{m}$ at $\tan^{-1}(120/90)=53.1^\circ$ north of east. Average velocity is $150\div30=5.0\,\text{m/s}$ in that direction. Distance is $90+120=210\,\text{m}$, so average speed is $210\div30=7.0\,\text{m/s}$.

Tier 1 · Easy

1. 1. A runner travels $156\,\text{m}$ in $48\,\text{s}$ at constant speed. Calculate the gradient of the distance–time graph.[2 marks]

   Answer

   • $3.25\,\text{m s}^{-1}$

   Method: The graph gradient is speed. Use $v=\Delta s/\Delta t=156/48=3.25\,\text{m s}^{-1}$.

Tier 2 · Standard

1. 1. A distance–time graph is a straight line from $(0\,\text{s},0\,\text{m})$ to $(24\,\text{s},120\,\text{m})$. It is then horizontal for $8\,\text{s}$. State what happens during the horizontal section and calculate the speed during the first section.[3 marks]

   Answer

   • The object is stationary for $8\,\text{s}$.
   • $5.0\,\text{m s}^{-1}$

   Method: A horizontal line has zero gradient, so the distance does not change and the object is stationary. For the first section, $v=120/24=5.0\,\text{m s}^{-1}$.

Tier 3 · Hard

1. 1. A robot moves along a straight track. Its distance from the start is $0\,\text{m}$ at $0\,\text{s}$, $54\,\text{m}$ at $12\,\text{s}$, $150\,\text{m}$ at $28\,\text{s}$ and $150\,\text{m}$ at $40\,\text{s}$. The graph joins these points with straight lines. Calculate the speed in each time interval and the average speed over all $40\,\text{s}$.[5 marks]

   Answer

   • $4.5\,\text{m s}^{-1}$ from $0$ to $12\,\text{s}$
   • $6.0\,\text{m s}^{-1}$ from $12$ to $28\,\text{s}$
   • $0\,\text{m s}^{-1}$ from $28$ to $40\,\text{s}$
   • Average speed $=3.75\,\text{m s}^{-1}$

   Method: Use the gradient of each section. The speeds are $54/12=4.5\,\text{m s}^{-1}$, $(150-54)/(28-12)=96/16=6.0\,\text{m s}^{-1}$ and $(150-150)/(40-28)=0\,\text{m s}^{-1}$. The total distance is $150\,\text{m}$, so the average speed is $150/40=3.75\,\text{m s}^{-1}$.

Tier 1 · Easy

1. 1. A scooter increases its velocity from $3.0\,\text{m s}^{-1}$ to $15.0\,\text{m s}^{-1}$ in $8.0\,\text{s}$. Calculate its average acceleration.[2 marks]

   Answer

   • $1.5\,\text{m s}^{-2}$

   Method: Use $a=(v-u)/t=(15.0-3.0)/8.0=12.0/8.0=1.5\,\text{m s}^{-2}$.

Tier 2 · Standard

1. 1. A velocity–time graph rises uniformly from $0$ to $12\,\text{m s}^{-1}$ in $4.0\,\text{s}$, stays horizontal for $6.0\,\text{s}$, then falls uniformly to $0$ in $3.0\,\text{s}$. Determine the acceleration in each section.[4 marks]

   Answer

   • $+3.0\,\text{m s}^{-2}$
   • $0\,\text{m s}^{-2}$
   • $-4.0\,\text{m s}^{-2}$

   Method: Acceleration is the graph gradient. The first section gives $(12-0)/4.0=+3.0\,\text{m s}^{-2}$. The horizontal section has zero gradient. The final section gives $(0-12)/3.0=-4.0\,\text{m s}^{-2}$.

Tier 3 · Hard

1. 1. A train accelerates uniformly from $8.0\,\text{m s}^{-1}$ to $20.0\,\text{m s}^{-1}$ while travelling $168\,\text{m}$. Calculate its acceleration and the time taken.[5 marks]

   Answer

   • $1.0\,\text{m s}^{-2}$
   • $12\,\text{s}$

   Method: Use $v^2-u^2=2as$: $20.0^2-8.0^2=2a(168)$. Thus $336=336a$, so $a=1.0\,\text{m s}^{-2}$. Then $a=(v-u)/t$, so $t=(20.0-8.0)/1.0=12\,\text{s}$.

Tier 1 · Easy

1. 1. A boat moves at constant velocity. Its propeller provides a forward force of $680\,\text{N}$. Determine the total resistive force.[2 marks]

   Answer

   • $680\,\text{N}$ opposite to the motion

   Method: Constant velocity means zero resultant force. The resistive force must therefore be equal in magnitude and opposite in direction to the $680\,\text{N}$ propeller force.

Tier 2 · Standard

1. 1. A van travels in a straight line at a steady $18\,\text{m s}^{-1}$. The engine force is $920\,\text{N}$. Explain the motion in terms of the forces and state the resultant force.[3 marks]

   Answer

   • The resistive forces total $920\,\text{N}$ backwards.
   • The resultant force is $0\,\text{N}$, so the van continues at constant velocity.

   Method: Steady speed in a straight line means constant velocity. By Newton's First Law the resultant force is zero, so the backward resistive forces balance the $920\,\text{N}$ engine force.

Tier 3 · Hard

1. 1. A lift moves upward. The motor force is $6200\,\text{N}$ upward, its weight is $5800\,\text{N}$ and friction is $400\,\text{N}$ downward. Describe its motion. The motor force then falls to $5000\,\text{N}$ while the lift is still moving upward. Calculate the new resultant force and describe the change in motion.[5 marks]

   Answer

   • Initially the lift moves upward at constant velocity.
   • The new resultant force is $1200\,\text{N}$ downward.
   • The upward-moving lift decelerates.

   Method: Initially, the downward forces total $5800+400=6200\,\text{N}$, balancing the motor force, so the lift's velocity is constant. After the change, the downward resultant is $5800+400-5000=1200\,\text{N}$. Because this resultant opposes the upward velocity, the lift slows down.

Tier 1 · Easy

1. 1. A resultant force accelerates a $12\,\text{kg}$ object at $1.5\,\text{m s}^{-2}$. Calculate the resultant force.[2 marks]

   Answer

   • $18\,\text{N}$

   Method: Use $F=ma=12\times1.5=18\,\text{N}$.

Tier 2 · Standard

1. 1. A student uses a wheeled cart, a pulley and slotted masses to investigate how force affects acceleration while total mass stays constant. Describe how the student should change the force, measure the acceleration and improve the reliability of the results. State the expected relationship.[5 marks]

   Answer

   • Move slotted masses from the cart to the hanging mass so the pulling force changes but the total moving mass stays constant.
   • Use light gates or a motion sensor and a data logger to measure acceleration.
   • Repeat each force setting and calculate a mean, identifying anomalies.
   • Acceleration should be directly proportional to resultant force for constant mass.

   Method: Keep the combined mass of cart plus hanging masses unchanged. Transfer one slotted mass at a time from the cart to the hanger, increasing the driving force without changing total mass. Release the cart from the same position and obtain acceleration with light gates or a motion sensor. Repeat each setting, check anomalies and calculate a mean. A graph of acceleration against resultant force should be a straight line through the origin, showing $a\propto F$ when mass is constant.

Tier 3 · Hard

1. 1. A $1250\,\text{kg}$ car increases its velocity from $8.0\,\text{m s}^{-1}$ to $26.0\,\text{m s}^{-1}$ in $9.0\,\text{s}$. The resistive forces total $700\,\text{N}$. Calculate the acceleration, the resultant force and the engine force.[5 marks]

   Answer

   • $2.0\,\text{m s}^{-2}$
   • $2500\,\text{N}$ resultant
   • $3200\,\text{N}$ engine force

   Method: The acceleration is $a=(26.0-8.0)/9.0=2.0\,\text{m s}^{-2}$. Hence the resultant force is $F=ma=1250\times2.0=2500\,\text{N}$. The engine must supply the resultant plus the resistance: $2500+700=3200\,\text{N}$.

Tier 1 · Easy

1. 1. A hammer exerts a downward force on a nail. State the corresponding Newton's Third Law force.[2 marks]

   Answer

   • The nail exerts an equal-magnitude upward force on the hammer.

   Method: Reverse both the objects and the direction: hammer on nail downward pairs with nail on hammer upward. The two forces have equal magnitude.

Tier 2 · Standard

1. 1. A book rests on a table. The Earth pulls the book downward. Identify the Newton's Third Law partner to this force and explain why it is not the table's upward force on the book.[3 marks]

   Answer

   • The book pulls the Earth upward with an equal gravitational force.
   • The table force acts on the book, whereas the third-law partner acts on the Earth.

   Method: The interaction is gravitational and the two objects are Earth and book. Earth on book pairs with book on Earth. The normal contact force is a different interaction; it balances the weight on the book but is not its third-law partner.

Tier 3 · Hard

1. 1. A propeller pushes water backward with a force of $4.2\,\text{kN}$. The water resistance on the boat is $1.8\,\text{kN}$ backward and the boat's mass is $1200\,\text{kg}$. State the third-law force exerted by the water because of the propeller interaction, then calculate the boat's acceleration.[5 marks]

   Answer

   • The water pushes the propeller forward with $4.2\,\text{kN}$.
   • $2.0\,\text{m s}^{-2}$ forward

   Method: By Newton's Third Law, the water exerts $4.2\,\text{kN}$ forward on the propeller. The resultant force on the boat is $4.2-1.8=2.4\,\text{kN}=2400\,\text{N}$. Therefore $a=F/m=2400/1200=2.0\,\text{m s}^{-2}$ forward.

Tier 1 · Easy

1. 1. A driver's thinking distance is $12\,\text{m}$ and the braking distance is $28\,\text{m}$. Calculate the stopping distance.[1 mark]

   Answer

   • $40\,\text{m}$

   Method: Add the two components: $s_{\text{stop}}=12+28=40\,\text{m}$.

Tier 2 · Standard

1. 1. A car travels at $15\,\text{m s}^{-1}$. The driver's reaction time is $0.64\,\text{s}$ and the braking distance is $19\,\text{m}$. Calculate the thinking distance and the stopping distance.[3 marks]

   Answer

   • Thinking distance $=9.6\,\text{m}$
   • Stopping distance $=28.6\,\text{m}$

   Method: During the reaction time, $s=vt=15\times0.64=9.6\,\text{m}$. Therefore the stopping distance is $9.6+19=28.6\,\text{m}$.

Tier 3 · Hard

1. 1. A delivery van has a reaction time of $0.75\,\text{s}$. At $20\,\text{m s}^{-1}$ its braking distance is $32\,\text{m}$; at $25\,\text{m s}^{-1}$ its braking distance is $50\,\text{m}$. Calculate the stopping distance at each speed and the increase in stopping distance.[5 marks]

   Answer

   • $47\,\text{m}$ at $20\,\text{m s}^{-1}$
   • $68.75\,\text{m}$ at $25\,\text{m s}^{-1}$
   • Increase $=21.75\,\text{m}$

   Method: At $20\,\text{m s}^{-1}$, the thinking distance is $20\times0.75=15\,\text{m}$, so the stopping distance is $15+32=47\,\text{m}$. At $25\,\text{m s}^{-1}$, the thinking distance is $18.75\,\text{m}$, so the stopping distance is $18.75+50=68.75\,\text{m}$. The increase is $68.75-47=21.75\,\text{m}$.

Tier 1 · Easy

1. 1. State the typical range of human reaction times and give one factor that can increase a driver's reaction time.[2 marks]

   Answer

   • $0.2\,\text{s}$ to $0.9\,\text{s}$
   • One of tiredness, alcohol, some drugs or distraction.

   Method: Recall the AQA typical range, then name one accepted factor that delays the driver's response.

Tier 2 · Standard

1. 1. A driver travels at $22\,\text{m s}^{-1}$ and reacts in $0.48\,\text{s}$. Calculate the thinking distance.[2 marks]

   Answer

   • $10.56\,\text{m}$

   Method: The vehicle continues at its original speed during the reaction time, so $s=vt=22\times0.48=10.56\,\text{m}$.

Tier 3 · Hard

1. 1. A student measures reaction time four times without a distraction and obtains $0.31$, $0.29$, $0.30$ and $0.89\,\text{s}$. With a distraction the results are $0.44$, $0.47$, $0.45$ and $0.46\,\text{s}$. Identify the anomalous result, calculate suitable mean times and evaluate the effect of the distraction.[5 marks]

   Answer

   • $0.89\,\text{s}$ is anomalous.
   • Mean without distraction $=0.30\,\text{s}$ after excluding the anomaly.
   • Mean with distraction $=0.455\,\text{s}$.
   • The distraction increased reaction time by about $0.155\,\text{s}$.

   Method: The $0.89\,\text{s}$ value is far from the other no-distraction repeats, so exclude it with justification. The remaining mean is $(0.31+0.29+0.30)/3=0.30\,\text{s}$. The distracted mean is $(0.44+0.47+0.45+0.46)/4=0.455\,\text{s}$. The increase is $0.455-0.300=0.155\,\text{s}$, supporting the conclusion that the distraction slowed the response.

Tier 1 · Easy

1. 1. Give one example of poor vehicle condition that increases braking distance and explain why it does so.[2 marks]

   Answer

   • For example, worn tyres reduce friction with the road, reducing the braking force and increasing braking distance.

   Method: Use either tyres or brakes, then link the poor condition to reduced effective friction or braking force and therefore a longer braking distance.

Tier 2 · Standard