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3.3.4Forces and Motion

Motion in One and Two Dimensions

5 min read 8 parts Lesson 17 of 66 1 quiz Virtual lab

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Lesson summary

3.3.4 Motion in One and Two Dimensions — OCR A Level Physics A notes: Apply Newton's second law and vector resolution to motion in a line and to projectile motion in two dimensions.The resultant force on a body is equal to the rate of change of its momentum.

  • Horizontal
  • Worked example 1
  • Worked example 2
  • Worked example 3
  • Time of flight
  • Vertical: constant acceleration g downwards. Using up as positive, vy = u sin θ − gt and y = u sin θ t − ½gt2

Section 1 of 8Newton's second law

The resultant force on a body is equal to the rate of change of its momentum.

Newton's second law
For constant mass this gives F = ma, where F is the resultant force in N, m the mass in kg and a the acceleration in m s−2, with the acceleration in the same direction as the resultant force.
When two forces act on a body along a line, you add them with signs: Fnet = Fforward − Fbackward.
A body with zero resultant force moves with constant velocity, which may be zero.

Section 2 of 8One-dimensional motion

In one dimension, choose a positive direction, find the resultant force, then use a = F / m together with the equations of motion v = u + at, s = ut + ½at2, v2 = u2 + 2as.

What the symbols mean: s = displacement (m) · u = initial velocity (m s−1) · a = acceleration (m s−2) · t = time (s)
One-dimensional motion
For a lift or a vehicle going up or down a slope the same method works: add the forces along the line of motion.
For a slope, resolve the weight into components along and perpendicular to the plane, giving acceleration a = g sin θ for a frictionless slope.

Section 3 of 8Resolving vectors

A vector at angle θ to the horizontal has components: horizontal = V cos θ and vertical = V sin θ.

Resolving vectors
The two components in perpendicular directions act independently of each other.
Their combination is found by Pythagoras: V = √(Vx2 + Vy2), with direction tan θ = Vy / Vx.
This independence of perpendicular motions is the foundation of two-dimensional dynamics.

Section 4 of 8Projectile motion

A projectile moves under gravity alone (air resistance ignored), so the only force is weight acting vertically down. Therefore:

Horizontal
Horizontal no force, so no acceleration. The horizontal velocity is constant, ux = u cos θ, and x = uxt.
Vertical
Vertical constant acceleration g downwards. Using up as positive, vy = u sin θ − gt and y = u sin θ t − ½gt2.

The path is a parabola.

Projectile motion
For a launch from level ground the time of flight is T = 2u sin θ / g, the maximum height is H = u2 sin2θ / 2g and the range is R = u2 sin 2θ / g, greatest at 45° if air resistance is negligible.
A projectile launched horizontally from height h has initial vertical velocity zero; the time to land is t = √(2h/g), independent of its horizontal speed.
Two balls, one dropped and one thrown horizontally from the same height, therefore land at the same time.

At the top of the path the vertical velocity is zero but the horizontal velocity is not, so the speed at the top is u cos θ and the acceleration is still g.

Section 5 of 8Air resistance and real motion

Drag reduces the horizontal speed and shortens the range, and the path becomes asymmetric: the descent is steeper than the ascent. The optimum angle for maximum range falls below 45°. Real projectiles such as a javelin or a badminton shuttlecock need numerical modelling, but the idealised equations give a good first estimate for dense, slow objects.

Check results with limitsthe speed on landing from a launch height should exceed the launch speed, and the horizontal range should increase if the launch speed increases at fixed angle.

Air resistance and real motion
Where a question gives the landing height different from the launch height, solve the vertical equation for t (a quadratic) and take the positive, physically meaningful root.
Then use x = u cos θ × t for the horizontal distance.
Always state the sign convention for the vertical axis at the beginning.

Section 6 of 8Worked examples

  1. A ball is kicked from level ground at 20 m s−1 at 35° above the horizontal. Find the time of flight and the range. (g = 9.81 m s−2)

  2. A stone is thrown horizontally at 6.0 m s−1 from a cliff 45 m high. Find the time to reach the sea, the horizontal distance and the impact speed.

  3. A 1000 kg car on a frictionless 8.0° slope is released from rest. Find its acceleration and speed after 5.0 s.

Section 7 of 8Key terms

TermMeaning
Resultant forceThe single force that has the same effect as all the forces acting on a body.
ComponentThe effect of a vector in a chosen direction, V cos θ or V sin θ.
ProjectileAn object moving under gravity alone after launch.
Time of flightTotal time the projectile is in the air.
RangeHorizontal distance travelled by a projectile.
ParabolaThe curved path of a projectile in a uniform field without drag.

Section 8 of 8Exam technique

1
Giving a projectile a horizontal acceleration. Without drag, ax = 0.
2
Using the full launch speed in the vertical equations instead of the component u sin θ.
3
Thinking a heavier projectile lands sooner. Time of fall does not depend on mass.
4
Setting acceleration to zero at the top of the flight.
5
Using range = u²sin2θ/g for a launch from a height; it applies only to level ground.
Try it in the virtual labNewton’s second law
Physics 0625 & 9702 SUVAT Mastery Graph Interpretation

Constant Acceleration Equations (SUVAT) & Slopes

v = u + at  |  s = ut + ½at2  |  v2 = u2 + 2as  |  s = ½(u + v)t

  • Displacement-Time Graph: Gradient = Velocity (curve = accelerating).
  • Velocity-Time Graph: Gradient = Acceleration; Area under curve = Distance travelled / Displacement.
  • Free Fall in vacuum: a = g ≈ 9.81 m/s2 downwards (air resistance negligible).
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