Questions on scalars and vectors

Question 1

For each of these quantities, state whether it is a scalar or a vector: distance, velocity, energy, momentum, temperature, force, speed, displacement.

Accepted Answer

Scalars: distance, energy, temperature, speed (2). Vectors: velocity, momentum, force, displacement (2). One mark for each two correct.

Question 2

State the difference between a scalar quantity and a vector quantity, giving one example of each.

Accepted Answer

A scalar has magnitude only; a vector has magnitude and direction (1). Example scalar: speed (or energy, distance); example vector: velocity (or force) (1).

Question 3

Two forces act on a box along the same straight line: 9.0 N to the right and 4.0 N to the left. Calculate the magnitude of the resultant force and state its direction.

Accepted Answer

Along one line vectors add with sign: 9.0 − 4.0 (1) 5.0 N, to the right (1)

Question 4

State the condition for three coplanar forces acting at a point to be in equilibrium.

Accepted Answer

The resultant force (the vector sum of the three forces) is zero (1).

Question 5

A survey drone flies 8.0 km due north, then 6.0 km due south. Calculate the total distance it travels and its final displacement.

Accepted Answer

Distance = 8.0 + 6.0 = 14.0 km (1). Displacement = 8.0 − 6.0 = 2.0 km due north (1).

Question 6

Two forces of 7.0 N act on the same object. State the angle between the two forces that gives a resultant of (a) zero, (b) 14 N.

Accepted Answer

(a) 180° (the forces are opposite and cancel) (1). (b) 0° (the forces are in the same direction and add to 14 N) (1).

Question 7

State how the vector A − B is constructed from two coplanar vectors A and B, and state what this construction gives when it is applied to the initial and final velocities of a moving object.

Accepted Answer

Reverse B to give −B, a vector of the same magnitude pointing the opposite way, and add it to A tip to tail: A − B = A + (−B) (1). Applied to velocities it gives the change in velocity, Δv = v − u, with both velocities treated as vectors (1).

Question 8

Forces of 3.0 N and 4.0 N act on a particle at right angles to each other. Calculate the magnitude of the resultant force and the angle it makes with the 3.0 N force.

Accepted Answer

R = √(3.0² + 4.0²) (1) R = 5.0 N (1) tan θ = 4.0/3.0 (1) θ = 53° from the 3.0 N force (1)

Question 9

A force of 12.0 N acts at 30° above the horizontal. Calculate the horizontal and vertical components of the force.

Accepted Answer

Horizontal component = 12.0 cos 30°, vertical = 12.0 sin 30° (1) Horizontal: 10.4 N (1) Vertical: 6.0 N (1)

Question 10

An aircraft climbs with a velocity of 60 m s⁻¹ at 25° above the horizontal. Calculate the horizontal and vertical components of its velocity.

Accepted Answer

Horizontal component = 60 cos 25°, vertical = 60 sin 25° (1) Horizontal: 54.4 m s⁻¹ (1) Vertical: 25.4 m s⁻¹ (1)

Question 11

A walker travels 3.0 km due east, then 4.0 km due north. Calculate (a) the total distance travelled, (b) the magnitude of the displacement, and (c) the bearing of the displacement.

Accepted Answer

(a) Distance is the path length: 3.0 + 4.0 = 7.0 km (1) (b) √(3.0² + 4.0²) = 5.0 km (1) (c) tan θ = 3.0/4.0 east of north (1) Bearing 37° (1)

Question 12

A crate of weight 20 N rests on a slope inclined at 35° to the horizontal. Calculate the components of the weight parallel to and perpendicular to the slope. Show that your two components are consistent with the original weight.

Accepted Answer

Parallel: 20 sin 35° = 11.5 N (1) Perpendicular: 20 cos 35° = 16.4 N (1) Check: √(11.5² + 16.4²) (1) = 20 N, the original weight (1)

Question 13

A swimmer heads directly across a river at 1.4 m s⁻¹ relative to the water. The river is 42 m wide and the current flows at 0.50 m s⁻¹ parallel to the banks. Calculate (a) the magnitude of the swimmer's resultant velocity, (b) the time taken to cross, and (c) the distance downstream at which the swimmer reaches the far bank.

Accepted Answer

(a) R = √(1.4² + 0.50²) (1) = 1.5 m s⁻¹ (1). (b) The crossing depends only on the component across the river: t = 42/1.4 = 30 s (1). (c) x = 0.50 × 30 = 15 m downstream (1).

Question 14

A shop sign of weight 24 N hangs at rest from two chains. Each chain makes an angle of 55° with the horizontal. Calculate the tension in each chain.

Accepted Answer

Resolving vertically: 2T sin 55° = 24 (1). T = 24/(2 sin 55°) (1) = 14.6 N (1).

Question 15

A barge is pulled along a canal by two cables, one from each bank. Each cable is pulled with a force of 850 N at 25° to the canal's direction, one on each side. Explain why the resultant force on the barge is directed along the canal, and calculate its magnitude.

Accepted Answer

The components perpendicular to the canal are equal and opposite, so they cancel; only the components along the canal remain (1). R = 2 × 850 × cos 25° (1) = 1541 N ≈ 1.5 kN (1).

Question 16

A cyclist is travelling due east at 8.0 m s⁻¹. She rounds a bend and, without changing speed, is then travelling due north at 8.0 m s⁻¹. Calculate the magnitude of her change in velocity and state its direction.

Accepted Answer

The change is a vector subtraction, Δv = v − u = v + (−u) (1). Reversing the initial velocity gives 8.0 m s⁻¹ due west, to be added to 8.0 m s⁻¹ due north (1). The two are perpendicular, so |Δv| = √(8.0² + 8.0²) = 11.3 m s⁻¹ (1), directed north-west, a bearing of 315° (1).

Question 17

Two forces act on a particle: 6.0 N due east and 8.0 N due north. A third force keeps the particle in equilibrium. Determine the magnitude of the third force and its direction as a bearing.

Accepted Answer

The third force balances the resultant of the other two (1) Magnitude = √(6.0² + 8.0²) (1) = 10 N (1) Directed opposite the resultant: bearing 217° (1)

Question 18

Forces of 5.0 N and 8.0 N act on a point with an angle of 60° between their directions. By resolving into components (or otherwise), calculate the magnitude of the resultant and the angle it makes with the 8.0 N force.

Accepted Answer

Along the 8.0 N direction: 8.0 + 5.0 cos 60° = 10.5 N (1) Perpendicular: 5.0 sin 60° = 4.33 N (1) R = √(10.5² + 4.33²) (1) R = 11.4 N (1) At 22° to the 8.0 N force (1)

Question 19

Three forces of 5.0 N, 12 N and 13 N act on a particle which remains in equilibrium. Explain, using a vector triangle, how this is possible, and state the angle between the 5.0 N and 12 N forces.

Accepted Answer

Drawn head to tail the three forces close a triangle, so their vector sum is zero and equilibrium is possible (1). Since 5² + 12² = 13², the triangle is right-angled (1): the angle between the 5.0 N and 12 N forces is 90° (1).

Question 20

A cable-car cabin of weight 1800 N hangs at rest from the midpoint of a support cable. Each half of the cable makes an angle of 12° with the horizontal. The cable manufacturer quotes a maximum safe tension of 4.5 kN. Deduce whether the tension in the cable is within the safe limit.

Accepted Answer

Resolving vertically: 2T sin 12° = 1800 (1). T = 1800/(2 sin 12°) = 4329 N ≈ 4.3 kN (1). Comparison: 4.3 kN is less than the 4.5 kN limit (1), so the tension is safe, though with a margin of only about 171 N; a smaller sag angle would exceed the limit (1).

Question 21

An orienteer runs 4.0 km due east, then 3.0 km on a bearing of 030°. Determine the magnitude of her resultant displacement and its bearing.

Accepted Answer

East component: 4.0 + 3.0 sin 30° = 5.5 km (1); north component: 3.0 cos 30° = 2.60 km (1). Magnitude = √(5.5² + 2.60²) = 6.1 km (1). Bearing: tan θ = 5.5/2.60 east of north, giving 65°, bearing 065° (1).

Question 22

A crate of weight 620 N is hauled up a ramp at constant speed by a rope parallel to the ramp surface. The crate runs on rollers, so assume frictional forces are negligible. (a) Show that when the ramp is at 20° to the horizontal the tension in the rope is about 210 N. (b) The rope's maximum safe tension is 260 N. Determine the greatest ramp angle at which the crate can be hauled at constant speed.

Accepted Answer

(a) Constant speed means equilibrium along the slope, so the tension balances the weight component: T = W sin θ (1). T = 620 sin 20° = 212 N ≈ 210 N (1). (b) At the limit 260 = 620 sin θ (1), so sin θ = 0.419 (1), giving θ = 25° (1).