Standard +0.3 This is a standard work-energy problem requiring conservation of energy on the smooth section AB, then work-energy theorem including friction on BC. It involves routine application of PE = mgh, KE = ½mv², and work done against friction = μR×distance. The multi-step nature and friction calculation make it slightly above average difficulty, but it follows a standard template with no novel insight required.
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The diagram shows a vertical cross-section \(A B C\) of a surface. The part of the surface containing \(A B\) is smooth and \(A\) is 2.5 m above the level of \(B\). The part of the surface containing \(B C\) is rough and is at \(45 ^ { \circ }\) to the horizontal. The distance \(B C\) is 4 m (see diagram). A particle \(P\) of mass 0.2 kg is released from rest at \(A\) and moves in contact with the curve \(A B\) and then with the straight line \(B C\). The coefficient of friction between \(P\) and the part of the surface containing \(B C\) is 0.4 . Find the speed with which \(P\) reaches \(C\).
\(\sqrt{2} - 0.4\sqrt{2} = 0.2a \rightarrow a = 3\sqrt{2}\text{ ms}^{-2}\) and \(V_C^2 = V_B^2 + 2 \times 3\sqrt{2} \times 4\)
M1, A1
For using Newton's 2nd law to find acceleration along \(BC\) and using \(v^2 = u^2 + 2as\) to find \(V_C\)
Speed at \(C\) is \(9.16\text{ ms}^{-1}\)
A1
# Question 4:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Frictional force $= 0.4 \times 2\cos45° = 0.4\sqrt{2}$ | M1, A1 | For using $R = 2\cos45°$ and $F = \mu R$ |
| KE gain $= \frac{1}{2} \times 0.2 \times V_C^2$ and PE loss $= 0.2 \times g \times (2.5 + 2\sqrt{2})$ | B1 | |
| | M1 | For using KE gain from $A$ to $C$ = PE loss from $A$ to $C$ – Work done by frictional force |
| $0.1V_C^2 = (5 + 4\sqrt{2}) - 0.4\sqrt{2} \times 4$ | A1 | |
| Speed at $C$ is $9.16\text{ ms}^{-1}$ | A1 | |
| | **Total: 6** | |
### First Alternative (last four marks):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2} \times 0.2 \times V_B^2 = 0.2 \times g \times 2.5 \rightarrow V_B^2 = 50$ | B1 | |
| $0.1(V_C^2 - V_B^2) = 0.2 \times g \times (4 \div \sqrt{2}) - 0.4\sqrt{2} \times 4$ | M1, A1 | For using KE gain from $B$ to $C$ = PE loss from $B$ to $C$ – Work done by frictional force |
| Speed at $C$ is $9.16\text{ ms}^{-1}$ | A1 | |
### Second Alternative (last four marks):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2} \times 0.2 \times V_B^2 = 0.2 \times g \times 2.5 \rightarrow V_B^2 = 50$ | B1 | |
| $\sqrt{2} - 0.4\sqrt{2} = 0.2a \rightarrow a = 3\sqrt{2}\text{ ms}^{-2}$ and $V_C^2 = V_B^2 + 2 \times 3\sqrt{2} \times 4$ | M1, A1 | For using Newton's 2nd law to find acceleration along $BC$ **and** using $v^2 = u^2 + 2as$ to find $V_C$ |
| Speed at $C$ is $9.16\text{ ms}^{-1}$ | A1 | |
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\includegraphics[max width=\textwidth, alt={}, center]{48f66bd5-33c1-4ce9-85f9-69faf10e871c-3_574_483_260_829}
The diagram shows a vertical cross-section $A B C$ of a surface. The part of the surface containing $A B$ is smooth and $A$ is 2.5 m above the level of $B$. The part of the surface containing $B C$ is rough and is at $45 ^ { \circ }$ to the horizontal. The distance $B C$ is 4 m (see diagram). A particle $P$ of mass 0.2 kg is released from rest at $A$ and moves in contact with the curve $A B$ and then with the straight line $B C$. The coefficient of friction between $P$ and the part of the surface containing $B C$ is 0.4 . Find the speed with which $P$ reaches $C$.
\hfill \mbox{\textit{CAIE M1 2015 Q4 [6]}}