Date | May 2011 | Marks available | 3 | Reference code | 11M.2.SL.TZ1.3 |
Level | Standard level | Paper | Paper 2 | Time zone | Time zone 1 |
Command term | Determine | Question number | 3 | Adapted from | N/A |
Question
This question is about circular motion.
A ball of mass 0.25 kg is attached to a string and is made to rotate with constant speed v along a horizontal circle of radius r = 0.33m. The string is attached to the ceiling and makes an angle of 30° with the vertical.
(i) On the diagram above, draw and label arrows to represent the forces on the ball in the position shown.
(ii) State and explain whether the ball is in equilibrium.
Determine the speed of rotation of the ball.
Markscheme
(i) [1] each for correct arrow and (any reasonable) labelling;
Award [1 max] for arrows in correct direction but not starting at the ball.
(ii) no;
because the two forces on the ball can never cancel out / there is a net force on
the ball / the ball moves in a circle / the ball has acceleration/it is changing
direction;
Award [0] for correct answer with no or wrong argument.
\(T\left( { = \frac{{mg}}{{\cos {{30}^ \circ }}}} \right) = 2.832{\rm{N}}\);
\(\frac{{m{v^2}}}{r} = T\sin {30^ \circ }\);
\(v = \left( {\sqrt {\frac{{Tr\sin {{30}^ \circ }}}{m}} = \sqrt {\frac{{2.832 \times 0.33 \times \sin {{30}^ \circ }}}{{0.25}}} } \right) = 1.4{\rm{m}}{{\rm{s}}^{ - 1}}\);
or
\(T\cos {30^ \circ } = mg\);
\(T\sin {30^ \circ } = \frac{{m{v^2}}}{r}\);
\(v = \left( {\sqrt {gr\tan {{30}^ \circ }} = \sqrt {9.81 \times 0.33 \times \tan {{30}^ \circ }} } \right) = 1.4{\rm{m}}{{\rm{s}}^{ - 1}}\);