Q#19 (Past Exam Paper – March 2018 Paper 22 Q2)
(a) Explain what is meant by
(i) work done, [1]
(ii) kinetic energy. [1]
(b) A leisure-park ride consists of a carriage that moves along a railed track. Part of the track lies in a vertical plane and follows an arc XY of a circle of radius 13 m, as shown in Fig. 2.1.
Fig. 2.1
The mass of the carriage is 580 kg. At point X, the carriage has velocity 22 m s-1 in a horizontal direction. The velocity of the carriage then decreases to 12 m s-1 in a vertical direction at point Y.
(i) For the carriage moving from X to Y
1. show that the decrease in kinetic energy is 9.9 × 104 J, [2]
2. calculate the gain in gravitational potential energy. [2]
(ii) Show that the length of the track from X to Y is 20 m. [1]
(iii) Use your answers in (b)(i) and (b)(ii) to calculate the average resistive force acting on the carriage as it moves from X to Y. [2]
(iv) Describe the change in the direction of the linear momentum of the carriage as it moves from X to Y. [1]
(v) Determine the magnitude of the change in linear momentum when the carriage moves from X to Y. [3]
[Total: 13]
Solution:
(a)
(i) Work done is defined as the product of force and the distance moved in the direction of the force.
(ii) Kinetic energy is the energy (of a mass/body) due to motion / speed / velocity
(b)
(i)
1.
{Since the speed changes, the carriage would have different kinetic energies at X and Y. The decrease in KE can be calculated as follows:
Decrease in KE = KE at X – KE at Y}
E = ½ mv2
(Δ)E = ½ × 580 × (222 – 122) = 9.9 × 104 J
2.
{When moving from X to Y, the carriage gains a vertical height of 13 m.}
(Δ)E = mg(Δ)h
ΔE = 580 × 9.81 × 13
ΔE = 7.4 × 104 J
(ii)
{The arc XY forms a quarter of the circumference a circle.
Arc XY = circumference / 4 = 2πr / 4}
length = (2π×13) / 4 or (π×26) / 4 or (π×13) / 2 = 20 m
(iii)
{The loss in KE was found to be 9.9 × 104 J while the gain in GPE was calculated to be 7.4 × 104 J.
This difference implies that some work has been done against resistance forces.
Work done against resistive force = (9.9 × 104 – 7.4 × 104) J
Also, work done = force × distance
The distance moved is along the arc XY (= 20 m).}
Average resistive force = (9.9 × 104 – 7.4 × 104) / 20
Average resistive force = 1300 N
(iv) From horizontal/right {at X} to vertical / up {at Y} or by 90°
(v)
{Momentum is a vector.
Here, the momentum vector has changed from horizontal (at X) to vertical {at Y}. We cannot have the change by simple calculation. We need to find the resultant momentum vector just like we find the resultant of other vectors.
Change in momentum = final momentum – initial momentum
Change in momentum = momentum at Y pY – momentum at X pX}
p = mv
or (580 × 22) or (580 × 12)
{The magnitude of the change in momentum can be found using Pythagoras’ theorem.}
Δp = [ (580×12)2 + (580×22)2 ]0.5
Δp = 1.5 × 104 N s
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