Motion in One & Two Dimensions (OCR A Level Physics)

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Katie M

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Motion in One & Two Dimensions

  • If a constant force acts upon an object, then it will experience a resultant acceleration, determined using F = ma
  • This motion can be investigated in one or two-dimensional planes, such as along the ground or on a slope
    • One-dimensional planes involve just up and down or left and right (on the ground)
    • Two-dimensional planes involve both up and down and left and right (on a slope)

  • On a slope, it is often simpler to resolve the forces into parallel and perpendicular components, rather than horizontally and vertically:
    • The weight, mg of the object acts vertically down
    • The frictional force, F between the slope and the object acts along the plane of the slope, in the direction opposing the motion
    • The normal reaction force, R acts perpendicular to the plane of contact between the object and the slope

Parallel and Perpendicular to the slope, downloadable AS & A Level Physics revision notes

The normal reaction force R, weight W and friction F on a block moving down a slope

Worked example

The diagram below shows the forces acting on a water drop on the windscreen of a stationary car.Windscreen Worked Example, downloadable AS & A Level Physics revision notesThe windscreen makes an angle of 30° to the horizontal. The weight of the water drop is 8.0 × 10–5 N.The normal contact force on the water drop is R. There is also a force F acting on the water drop as shown. The water drop is stationary.Determine:

(a) The component of the weight of the water-drop perpendicular to the windscreen

(b) The component of the weight of the water-drop parallel to the windscreen

(c) The magnitude of R

(d) The magnitude of F

Windscreen Worked Example Ans, downloadable AS & A Level Physics revision notes

Part (a)

    • Perpendicular component = W cos 30°
    • Perpendicular component = (8.0 × 10–5) cos 30° = 6.9 × 10–5 N

Part (b)

    • Parallel component = W sin 30°
    • Parallel component = (8.0 × 10–5) sin 30° = 4.0 × 10–5 N

Part (c)

    • R is equal to the perpendicular component of the weight
    • Perpendicular forces must be equal and opposite

R = 6.9 × 10–5 N

Part (d)

    • F is equal to the parallel component of the weight
    • Parallel forces must be equal and opposite

F = 4.0 × 10–5 N

Worked example

The diagram below shows a rider on a sledge sliding down an icy slope. The frictional forces acting on the sledge and the rider are negligible. The normal contact force N and the total weight W of the sledge and rider are shown.Icy Slope Worked Example, downloadable AS & A Level Physics revision notesThe acceleration of the sledge and rider down the slope is 2.0 m s–2.Determine the angle made by the slope to the horizontal.

Step 1: Write down the known quantities

    • Acceleration, a = 2.0 m s–2
    • Weight, W = mg
    • Component of weight parallel the slope, F = W sin θ = mg sin θ

Icy Slope Worked Example Ans, downloadable AS & A Level Physics revision notes

Step 2: Write down the equation relating force and acceleration

F = ma

Step 3: Substitute in the component of weight down the slope

mg sin θ = ma

Step 4: Rearrange for sin θ and calculate the angle

a = g sin θ

sin θ = 2.0 / 9.81

θ = 12°

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Katie M

Author: Katie M

Expertise: Physics

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.