CAIE A-Level Mathematics A2 4.4.3 Constant Acceleration Questions
Practise combining dynamics, friction and kinematics to solve multi-stage mechanics problems.
- Syllabus
- 2028–2030
- Course
- Mathematics 9709
- Level
- A2
Practise combining dynamics, friction and kinematics to solve multi-stage mechanics problems.
A straight slope of length 60 m is inclined at an angle of 12∘ to the horizontal. A bobsled starts at the top of the slope with a speed of 5 ms−1. The bobsled slides directly down the slope.
It is given that there is no resistance to the bobsled's motion.
Find its speed when it reaches the bottom of the slope.
a=2.08[2.07911…]
From mgsin12=ma.
Marking guidance:
Allow exact (e.g. a=gsin12 ).
v2=52+2a×60
For use of v2=u2+2as with u=5 and s=60.
Allow sign errors but a must be either gsin12 or gcos12 only.
Speed =16.6 ms−1[16.567861…]
AWRT 16.6
Alternative Method for Q5(a)
For attempt at work energy equation
3 terms, dimensionally correct.
Allow sign errors;
allow sin/cos mix on PE term - condone m missing
from all terms. Must be a weight component.
21mv2=21m×52+mg×60sin12
Correct equation.
(for reference: 60sin12=12.4747… )
Speed =16.6 ms−1[16.567861…]
AWRT 16.6
It is given instead that the coefficient of friction between the bobsled and the slope is 0.03 .
Find the time that it takes for the bobsled to reach the bottom of the slope.
R=mgcos12
Resolving correctly perpendicular to the plane.
mgsin12−F=ma
*M1
Use of Newton's second law, correct number of terms;
allow sign errors;
Marking guidance:
allow sin/cos mix (must be a weight component).
For use of F=0.03 R to get equation in a (and m ) only
Where R is a component of weight only (dimensionally correct but allow sin/cosmix).
[⇒a=1.79[1.78567….]]60=5t+21at2 and solve for t
Dependent on previous two M marks.
For use of s=ut+21at2 with s=60, u=5 and their a or other complete method to find positive value(s) of t.
Time =5.86 s[5.86260…]
AWRT 5.86 (from using a=1.79 or better).
AWRT 5.85 (from using a=1.8 ).
AWRT 5.87 (from correct working).
Alternative method for Question 5(b): Using energy
R=mgcos12
Resolving correctly perpendicular to the plane.
21mv2−21m×52=60×mgsin12−60×F
(*M1)
Use of work-energy principle, correctly number of relevant terms;
allow sign errors;
allow sin/cos mix on PE term (must be a weight component).
For use of F=0.03 R to get equation in v (and m ) only
Where R is a component of weight only (dimensionally correct but allow sin/cos mix)
[⇒v=15.5[15.46870….]].60=21(5+v)t and solve for t
Dependent on previous two M marks.
Use of s=21(u+v)t with s=60, u=5 and their v or other complete method to find positive value(s) of t.
Time =5.86 s[5.86260…]