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Date May 2013 Marks available 3 Reference code 13M.2.SL.TZ1.2
Level Standard level Paper Paper 2 Time zone Time zone 1
Command term Calculate Question number 2 Adapted from N/A

Question

This question is about circular motion.

The diagram shows a car moving at a constant speed over a curved bridge. At the position shown, the top surface of the bridge has a radius of curvature of 50 m.

Explain why the car is accelerating even though it is moving with a constant speed.

[2]
a.

On the diagram, draw and label the vertical forces acting on the car in the position shown.

[2]
b.

Calculate the maximum speed at which the car will stay in contact with the bridge.

[3]
c.

Markscheme

direction changing;

velocity changing so accelerating; 

a.

weight/gravitational force/mg/w/Fw/Fg and reaction/normal reaction/perpendicular contact force/N/R/FN/FR both labelled; (do not allow “gravity” for “weight”.)

weight between wheels (in box) from centre of mass and reactions at both wheels / single reaction acting along same line of action as the weight;

Judge by eye. Look for reasonably vertical lines with weight force longer than (sum of) reaction(s). Extra forces (eg centripetal force) loses the second mark.

 

b.

\(g = \frac{{{v^2}}}{r}\);
\(v = \sqrt {50 \times 9.8} \);

22(ms-1);

Allow [3] for a bald correct answer.

c.

Examiners report

Most candidates failed to state that acceleration is the rate of change of velocity and that as velocity is a vector it has both magnitude and direction. With there being a change in direction the car accelerates. Many erroneously talked about there being a change of direction of the acceleration – the direction is always centripetal.

a.

 Few marked in reaction acting at each wheel and the weight acting from the centre of gravity. The weight needed to be larger than the combined reaction to give a resultant centripetal force (this is shown by the relative length of the lines). Most candidates were unconcerned about the point of application of the forces and often added spurious horizontal and/or centripetal forces. Centripetal forces, being the resultant of the other force, should not be marked in on free body diagrams like this. 

b.

The majority of candidates made a good attempt at calculating the maximum speed by equating the weight to the centripetal force (that is, in the limit there is no reaction force).

c.

Syllabus sections

Core » Topic 6: Circular motion and gravitation » 6.1 – Circular motion
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