User interface language: English | Español

Date November 2011 Marks available 2 Reference code 11N.2.SL.TZ0.4
Level Standard level Paper Paper 2 Time zone Time zone 0
Command term Calculate Question number 4 Adapted from N/A

Question

This question is in two parts. Part 1 is about forces. Part 2 is about internal energy.

Part 1 Forces

A railway engine is travelling along a horizontal track at a constant velocity.

On the diagram above, draw labelled arrows to represent the vertical forces that act on the railway engine.

[3]
a.

Explain, with reference to Newton’s laws of motion, why the velocity of the railway engine is constant.

[2]
b.

The constant horizontal velocity of the railway engine is 16 ms–1. A total horizontal resistive force of 76 kN acts on the railway engine.

Calculate the useful power output of the railway engine.

[2]
c.

The power driving the railway engine is switched off. The railway engine stops, from its speed of 16 ms–1, without braking in a distance of 1.1 km. A student hypothesizes that the horizontal resistive force is constant.

Based on this hypothesis, calculate the mass of the railway engine.

[2]
d.

Another hypothesis is that the horizontal force in (c) consists of two components. One component is a constant frictional force of 19 kN. The other component is a resistive force F that varies with speed v where F is proportional to v3.

(i) State the value of the magnitude of F when the railway engine is travelling at 16 ms–1.

(ii) Determine the total horizontal resistive force when the railway engine is travelling at 8.0 ms–1.

[5]
e.

On its journey, the railway engine now travels around a curved track at constant speed. Explain whether or not the railway engine is accelerating.

[3]
f.

Markscheme

The shaded box shows the acceptable range of position for W/mg.
single downward arrow labelled W/weight or mg/gravity force; (do not allow gravity)
two upward arrows labelled reaction/contact forces; (do not allow for only one arrow seen)
arrow positions as shown in diagram;

a.

horizontal forces have resultant of zero; (must describe or imply horizontal force)
valid statement linked to theory (e.g. Newton 1/Newton 2/conservation of momentum)
explaining why zero force results in constant velocity/zero acceleration;

b.

power =16×76000;
1.2 MW;

c.

acceleration\( = \frac{{{{16}^2}}}{{2 \times 1100}}\left( { = 0.116} \right)\);
\(m = \left( {\frac{{7.6 \times {{10}^4}}}{{0.116}} = } \right)6.5 \times {10^5}{\rm{kg}}\);
Award [2] for a bald correct answer.

or

use of Fs=\(\frac{1}{2}m{v^2}\);
\(m = \left( {\frac{{2 \times 7.6 \times {{10}^4} \times 1100}}{{{{16}^2}}} = } \right)6.5 \times {10^5}{\rm{kg}}\);
Award [2] for a bald correct answer.

d.

(i) 57 kN;

(ii) \({F_8} = \frac{{{F_{16}}}}{{{2^3}}}\);
F8=7.1(kN);
total force =19+7.1(kN);
=26 kN;
Award [4] for a bald correct answer.

or

\(k = \left( {\frac{{57 \times {{10}^3}}}{{{{16}^3}}}} \right) = 13.91\);
F8=(13.91×83)=7.1(kN);
total force=19+7.1(kN);
=26 kN;
Award [4] for a bald correct answer.

e.

direction of engine is constantly changing;
velocity is speed + direction / velocity is a vector;
engine is accelerating as velocity is changing;
Award [0] for a bald correct answer.

or

centripetal force required to maintain circular motion;
quotes Newton 1/Newton 2;
so engine is accelerating as a force acts;
Award [0] for a bald correct answer.

f.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.
[N/A]
e.
[N/A]
f.

Syllabus sections

Core » Topic 2: Mechanics » 2.1 – Motion
Show 111 related questions

View options