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GCSE / IGCSE Physics – Forces and Motion

Speed, acceleration, motion graphs, Newton’s laws, momentum, moments.

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12 questions  ·  64 marks  ·  answers included  ·  GCSE/IGCSE
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This worksheet covers:

  • Speed and acceleration
  • Distance-time and velocity-time graphs
  • Resultant forces and Newton’s laws
  • Momentum and collisions
  • Stopping distances
  • Hooke’s law
  • Moments

Calculator allowed; use $g = 9.81\,\text{N kg}^{-1}$ (some boards use $10$); allow about 60 minutes.

Questions

Section A – Skills

Show equations, substitution and units.
1

A cyclist travels $1500$ m in $120$ s.(a) Calculate the average speed of the cyclist.(b) The cyclist then accelerates uniformly from $12\,\text{m s}^{-1}$ to $30\,\text{m s}^{-1}$ in $6.0$ s. Calculate the acceleration.

Show answer
(a) $v = \dfrac{1500}{120} = 12.5\,\text{m s}^{-1}$(b) $a = \dfrac{30 – 12}{6.0} = 3.0\,\text{m s}^{-2}$
[4]
2

A resultant force of $450$ N acts on a car of mass $900$ kg.(a) Calculate the acceleration of the car.(b) State Newton’s first law of motion.

Show answer
(a) $a = \dfrac{450}{900} = 0.5\,\text{m s}^{-2}$(b) An object remains at rest or moves at constant velocity unless acted on by a resultant force
[3]
3

Two forces act on a box at right angles: $30$ N to the east and $40$ N to the north.(a) Calculate the magnitude of the resultant force.(b) Describe how you would find its direction using a scale drawing.

Show answer
(a) $\sqrt{30^2 + 40^2} = 50$ N(b) Draw the two vectors tip-to-tail to scale, draw the resultant from start to finish, measure the angle with a protractor (about $53^\circ$ from east towards north)
[4]
4

A spring has a spring constant of $25\,\text{N m}^{-1}$ and is stretched by $12$ cm within its limit of proportionality.(a) Calculate the force applied.(b) Calculate the elastic potential energy stored.

Show answer
(a) $F = kx = 25 \times 0.12 = 3.0$ N(b) $E = \frac{1}{2}kx^2 = \frac{1}{2}(25)(0.12)^2 = 0.18$ J
[4]
5

A car of mass $1200$ kg travels at $15\,\text{m s}^{-1}$.(a) Calculate its momentum.(b) Calculate its kinetic energy.

Show answer
(a) $p = mv = 18\,000\,\text{kg m s}^{-1}$(b) $E_k = \frac{1}{2}(1200)(15)^2 = 135\,000$ J $= 135$ kJ
[4]

Section B – Exam-style

Longer structured questions.
6

A train starts from rest and accelerates uniformly to $16\,\text{m s}^{-1}$ in $8$ s. It then travels at $16\,\text{m s}^{-1}$ for $12$ s before decelerating uniformly to rest in $4$ s.(a) Sketch the velocity-time graph, labelling the axes and key values.(b) Calculate the initial acceleration.(c) Calculate the total distance travelled.

Show answer
(a) Straight line from $(0, 0)$ to $(8, 16)$, horizontal to $(20, 16)$, straight line down to $(24, 0)$(b) $2.0\,\text{m s}^{-2}$(c) Area $= 64 + 192 + 32 = 288$ m
[6]
7

A skydiver of mass $70$ kg has just left an aircraft. At one instant the air resistance acting on her is $300$ N.(a) Calculate her weight.(b) Calculate her acceleration at this instant.(c) Explain, in terms of forces, why she eventually reaches terminal velocity.

Show answer
(a) $W = 70 \times 9.81 = 687$ N(b) Resultant $= 687 – 300 = 387$ N; $a = \dfrac{387}{70} = 5.5\,\text{m s}^{-2}$(c) As speed increases, air resistance increases until it equals the weight; resultant force is zero so acceleration is zero and velocity is constant
[6]
8

A car of mass $1200$ kg moving at $15\,\text{m s}^{-1}$ collides with a stationary van of mass $800$ kg. The two vehicles move off together.(a) State the principle of conservation of momentum.(b) Calculate the common velocity after the collision.(c) Explain why crumple zones reduce injuries in such a collision.

Show answer
(a) Total momentum before a collision equals total momentum after, provided no external forces act(b) $1200 \times 15 = 2000v$, so $v = 9.0\,\text{m s}^{-1}$(c) They increase the time over which momentum changes, reducing the rate of change of momentum and so the force on occupants
[6]
9

A driver travelling at $20\,\text{m s}^{-1}$ sees a hazard. Her reaction time is $0.7$ s. The car has mass $1000$ kg and the brakes provide a force of $6000$ N.(a) Calculate the thinking distance.(b) Calculate the deceleration and hence the braking distance.(c) State two factors that would increase the braking distance.

Show answer
(a) $20 \times 0.7 = 14$ m(b) $a = 6.0\,\text{m s}^{-2}$; $s = \dfrac{v^2}{2a} = \dfrac{400}{12} = 33.3$ m(c) Any two: wet or icy road, worn tyres or brakes, greater speed, greater mass
[7]
10

A uniform seesaw is pivoted at its centre. A child of mass $30$ kg sits $1.5$ m from the pivot on the left.(a) Calculate the moment of the child’s weight about the pivot.(b) An adult of mass $45$ kg sits on the right. Calculate how far from the pivot the adult must sit for the seesaw to balance.

Show answer
(a) $30 \times 9.81 \times 1.5 = 441$ N m(b) $441 = 45 \times 9.81 \times d$, so $d = 1.0$ m
[5]

Section C – Challenge

Extended problems.
11

A tennis ball of mass $0.16$ kg is struck by a racket and leaves at $40\,\text{m s}^{-1}$, having been at rest. The ball is in contact with the racket for $0.020$ s.(a) Calculate the change in momentum of the ball.(b) Calculate the average force exerted by the racket.(c) Using Newton’s third law, state the force the ball exerts on the racket.(d) A car accelerates from rest to $25\,\text{m s}^{-1}$ over a distance of $100$ m. Use $v^2 = u^2 + 2as$ to find its acceleration.

Show answer
(a) $0.16 \times 40 = 6.4\,\text{kg m s}^{-1}$(b) $F = \dfrac{6.4}{0.020} = 320$ N(c) $320$ N on the racket, in the opposite direction(d) $a = \dfrac{25^2}{2 \times 100} = 3.1\,\text{m s}^{-2}$ ($3.125$)
[8]
12

A uniform beam of length $4.0$ m and weight $200$ N rests horizontally on two supports, one at the left end and one at the right end. A load of $500$ N is placed $3.0$ m from the left end.(a) Taking moments about the left support, calculate the upward force from the right support.(b) Calculate the upward force from the left support.(c) Explain what would happen if the load were moved to $3.9$ m from the left end.

Show answer
(a) $4.0R = 200 \times 2.0 + 500 \times 3.0 = 1900$, so $R = 475$ N(b) $700 – 475 = 225$ N(c) Right support force rises to $(400 + 1950)/4 = 587.5$ N and left falls to $112.5$ N; beam still balanced but nearly all the load is carried by the right support
[7]

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