A leisure-park ride consists of a carriage that moves along a railed track.

Question 19

(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]

Reference: Past Exam Paper – March 2018 Paper 22 Q2

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 p_Y – momentum at X p_X}

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