Standard +0.3 This is a standard mechanics problem requiring application of SUVAT equations and Newton's second law on an inclined plane with friction. Students must find acceleration from kinematics (s=2.25m, u=0, t=1.5s gives a), then resolve forces parallel to the plane (mg sin 30° - μmg cos 30° = ma) to find μ. It's slightly above average difficulty due to the two-stage approach and careful resolution of forces, but follows a well-practiced method with no conceptual surprises.
\(A\) and \(B\) are points on the same line of greatest slope of a rough plane inclined at \(30°\) to the horizontal. \(A\) is higher up the plane than \(B\) and the distance \(AB\) is \(2.25 \text{ m}\). A particle \(P\), of mass \(m \text{ kg}\), is released from rest at \(A\) and reaches \(B\) \(1.5 \text{ s}\) later. Find the coefficient of friction between \(P\) and the plane.
[6]
Question 3:
3 | 2
2.25 = ½ a(1.5 )
a = 2
o
R = mgcos30
o µ o
mgsin30 - mgcos30 = 2m
Coefficient of friction is 0.346 | M1
A1
B1
M1
A1 ft
A1 6 | 2
For using s = ½ at
For applying Newton’s second
µ
law (3 terms) and F = R
ft incorrect a or R or consistent
sin/cos mix
3 | 2
KE gain = ½ m3
o
R = mgcos30
µ o
2.25 mgcos30 =
o 2
mg(2.25sin30 ) – ½ m3
Coefficient of friction is 0.346 | M1
A1
B1
M1
A1ft
A1 6 | For using (0 + v)/2 = s/t to find
v and hence KE gain from ½
B
2
mv
B
µ
For using F = R and
2.25F = PE loss – KE gain
ft incorrect v or R or consistent
B
sin/cos mix
$A$ and $B$ are points on the same line of greatest slope of a rough plane inclined at $30°$ to the horizontal. $A$ is higher up the plane than $B$ and the distance $AB$ is $2.25 \text{ m}$. A particle $P$, of mass $m \text{ kg}$, is released from rest at $A$ and reaches $B$ $1.5 \text{ s}$ later. Find the coefficient of friction between $P$ and the plane.
[6]
\hfill \mbox{\textit{CAIE M1 2005 Q3 [6]}}