Three points $$\(A, B\)$$ and $$\(C\)$$ lie on a line of greatest slope of a plane inclined at an angle of $$\(30^{\circ}\)$$ to the horizontal, with $$\(A B=1 \mathrm{~m}\)$$ and $$\(B C=1 \mathrm{~m}\)$$, as shown in the diagram. A particle of mass $$\(0.2 \mathrm{~kg}\)$$ is released from rest at $$\(A\)$$ and slides down the plane. The part of the plane from $$\(A\)$$ to $$\(B\)$$ is smooth. The part of the plane from $$\(B\)$$ to $$\(C\)$$ is rough, with coefficient of friction $$\(\mu\)$$ between the plane and the particle. Given that $$\(\mu=\frac{1}{2} \sqrt{3}\)$$, find the speed of the particle at $$\(C\)$$. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Exam No：9709_w20_qp_41 Year：2020 Question No：7(a)
Answer:
$
0.2 \times 10 \times 0.5=\frac{1}{2} \times 0.2 \times v_{B}^{2}
$
$
v_{B}^{2}=10
$
Alternative method for the first 3 marks
$
0.2 \times 10 \times \sin 30=0.2 a, a=5
$
$
v_{B}^{2}=0^{2}+2 \times 5 \times 1
$
\(v_{B}^{2}=10\)
THEN, either this method for the next 5 marks
$
R=0.2 \times 10 \times \cos 30=\sqrt{3}
$
\(F=\frac{\sqrt{3}}{2} \times 0.2 \times \frac{\sqrt{3}}{2} \times 10=1.5\)
PE loss \(=0.2 \times 10 \times 0.5=1\)
WD against \(F=1.5 \times 1\)
$
\frac{1}{2} 0.2 \times 10+0.2 \times 10 \times 0.5=1.5 \times 1+\frac{1}{2} 0.2 v_{C}^{2}
$
\( v_{c}=\sqrt{5}=2.24 \mathrm{~ms}^{-1} \)
OR, this method for the next 5 marks
\(R=0.2 \times 10 \times \cos 30=\sqrt{3}\)
\(F=\frac{\sqrt{3}}{2} \times 0.2 \times \frac{\sqrt{3}}{2} \times 10=1.5\)
\(0.2 \times 10 \sin 30-1.5=0.2 a \quad a=-2.5\)
\(v_{c}^{2}=10+2 \times-2.5 \times 1\)
\(v_{c}=\sqrt{5}=2.24 \mathrm{~ms}^{-1}\)
Alternative method for question \(7(a)\)
PE loss \(=0.2 \times 10 \times 2 \sin 30=2\)
\(\mathrm{KE}\) gain \(=\frac{1}{2} \times 0.2 \times v_{C}^{2}\)
Both PE loss and KE gain correct
\(R=0.2 \times 10 \times \cos 30=\sqrt{ } 3\)
\(F=\frac{\sqrt{3}}{2} \times 0.2 \times \frac{\sqrt{3}}{2} \times 10=1.5\)
WD against \(F=1.5 \times 1\)
\(0.2 \times 10 \times 1=1.5 \times 1+\frac{1}{2} \times 0.2 \times v_{C}^{2}\)
\(v_{c}=\sqrt{5}=2.24 \mathrm{~ms}^{-1}\)
0.2 \times 10 \times 0.5=\frac{1}{2} \times 0.2 \times v_{B}^{2}
$
$
v_{B}^{2}=10
$
Alternative method for the first 3 marks
$
0.2 \times 10 \times \sin 30=0.2 a, a=5
$
$
v_{B}^{2}=0^{2}+2 \times 5 \times 1
$
\(v_{B}^{2}=10\)
THEN, either this method for the next 5 marks
$
R=0.2 \times 10 \times \cos 30=\sqrt{3}
$
\(F=\frac{\sqrt{3}}{2} \times 0.2 \times \frac{\sqrt{3}}{2} \times 10=1.5\)
PE loss \(=0.2 \times 10 \times 0.5=1\)
WD against \(F=1.5 \times 1\)
$
\frac{1}{2} 0.2 \times 10+0.2 \times 10 \times 0.5=1.5 \times 1+\frac{1}{2} 0.2 v_{C}^{2}
$
\( v_{c}=\sqrt{5}=2.24 \mathrm{~ms}^{-1} \)
OR, this method for the next 5 marks
\(R=0.2 \times 10 \times \cos 30=\sqrt{3}\)
\(F=\frac{\sqrt{3}}{2} \times 0.2 \times \frac{\sqrt{3}}{2} \times 10=1.5\)
\(0.2 \times 10 \sin 30-1.5=0.2 a \quad a=-2.5\)
\(v_{c}^{2}=10+2 \times-2.5 \times 1\)
\(v_{c}=\sqrt{5}=2.24 \mathrm{~ms}^{-1}\)
Alternative method for question \(7(a)\)
PE loss \(=0.2 \times 10 \times 2 \sin 30=2\)
\(\mathrm{KE}\) gain \(=\frac{1}{2} \times 0.2 \times v_{C}^{2}\)
Both PE loss and KE gain correct
\(R=0.2 \times 10 \times \cos 30=\sqrt{ } 3\)
\(F=\frac{\sqrt{3}}{2} \times 0.2 \times \frac{\sqrt{3}}{2} \times 10=1.5\)
WD against \(F=1.5 \times 1\)
\(0.2 \times 10 \times 1=1.5 \times 1+\frac{1}{2} \times 0.2 \times v_{C}^{2}\)
\(v_{c}=\sqrt{5}=2.24 \mathrm{~ms}^{-1}\)
Knowledge points:
4.1.4 understand that a contact force between two surfaces can be represented by two components, the normal component and the frictional component
4.1.6 understand the concepts of limiting friction and limiting equilibrium, recall the definition of coefficient of friction, and use the relationship F = nR or F G nR, as appropriate
4.4.1 apply Newton’s laws of motion to the linear motion of a particle of constant mass moving under the action of constant forces, which may include friction, tension in an inextensible string and thrust in a connecting rod If any other forces resisting motion are to be considered (e.g. air resistance) this will be indicated in the question.
4.4.3 solve simple problems which may be modelled as the motion of a particle moving vertically or on an inclined plane with constant acceleration Including, for example, motion of a particle on a rough plane where the acceleration while moving up the plane is different from the acceleration while moving down the plane.
4.5.2 understand the concepts of gravitational potential energy and kinetic energy, and use appropriate formulae
Solution:
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