Standard +0.3 This is a standard connected particles problem using energy methods with straightforward setup: identify initial/final PE and KE, apply conservation of energy, and solve for speed. The calculation involves basic trigonometry (sin θ = 3/5 given) and simple algebra. While it requires understanding of the energy method and careful bookkeeping of PE changes for both particles, it's a routine textbook exercise with no novel insight required.
\includegraphics{figure_4}
Two particles \(A\) and \(B\), of masses \(0.8\text{ kg}\) and \(1.6\text{ kg}\) respectively, are connected by a light inextensible string. Particle \(A\) is placed on a smooth plane inclined at an angle \(\theta\) to the horizontal, where \(\sin \theta = \frac{3}{5}\). The string passes over a small smooth pulley \(P\) fixed at the top of the plane, and \(B\) hangs freely (see diagram). The section \(AP\) of the string is parallel to a line of greatest slope of the plane. The particles are released from rest with both sections of the string taut. Use an energy method to find the speed of the particles after each particle has moved a distance of \(0.5\text{ m}\), assuming that \(A\) has not yet reached the pulley. [6]
[Gain in PE = 0.8 g × 0.5 × sinθ ] or [Loss in PE = 1.6 g × 0.5]
Answer
Marks
Guidance
A B
M1
For PE change of either particle (irrespective of sign)
Loss in PE = 1.6 g × 0.5 – 0.8 g × 0.5 × 0.6
A1
Change of PE
[1.2v2 =8−2.4]
M1
Energy equation originating from 4 terms
Speed is 2.16 (m s–1)
A1
Total:
6
SC for using Newton II equations and v2 = u2+ 2as (max 2/6)
[16 – T = 1.6a and T – 8sinθ = 0.8a] → a = 4.67 (ms–2) B1
14
[v2 = 2 × × 0.5] → speed is 2.16 (ms–1) B1
3
Alternative method 1 for Question 4
[ 1× 0.8 × v2] or [0.8 g × 0.5 × sinθ ]
Answer
Marks
Guidance
2
M1
For KE gain or PE gain of particle A
1× 0.8 × v2 + 0.8 g × 0.5 × 0.6
Answer
Marks
Guidance
2
A1
Total energy gain for particle A
[16 – T =1.6a and T – 8sinθ = 0.8a → T = ....] 8.53
M1
Forms and solves Newton II equations to find tension T
128
WD = × 0.5
T
Answer
Marks
Guidance
15
A1
Finds WD
Tension
[ 1 × 0.8 × v2 + 0.8 g × 0.5 × 0.6 = 128 × 0.5]
Answer
Marks
Guidance
2 15
M1
Energy equation (3 terms)
Question
Answer
Marks
4
Speed is 2.16 (m s–1)
A1
Total:
6
Alternative method 2 for Question 4
[ 1× 1.6 × v2] or [1.6 g × 0.5]
Answer
Marks
Guidance
2
M1
For KE gain or PE loss of particle B
1.6 g × 0.5 – 1 × 1.6 × v2
Answer
Marks
Guidance
2
A1
Energy change for particle B
[16 – T = 1.6a and T – 8sinθ = 0.8a → T = ....] 8.53
M1
Forms and solves Newton II equations to find tension T
WD = 128 × 0.5
T
Answer
Marks
Guidance
15
A1
Finds WD
Tension
1.6 g × 0.5 – 1 × 1.6 × v2 = 128× 0.5]
Answer
Marks
Guidance
2 15
M1
Energy equation (3 terms)
Speed is 2.16 (m s–1)
A1
Total:
6
Question
Answer
Marks
Question 4:
4 | 1 1
[ × 0.8 × v2] or [ × 1.6 × v2]
2 2 | M1 | For KE of either particle
Gain in KE = 1×0.8×v2+ 1×1.6×v2
2 2 | A1 | Total KE
[Gain in PE = 0.8 g × 0.5 × sinθ ] or [Loss in PE = 1.6 g × 0.5]
A B | M1 | For PE change of either particle (irrespective of sign)
Loss in PE = 1.6 g × 0.5 – 0.8 g × 0.5 × 0.6 | A1 | Change of PE
[1.2v2 =8−2.4] | M1 | Energy equation originating from 4 terms
Speed is 2.16 (m s–1) | A1
Total: | 6
SC for using Newton II equations and v2 = u2+ 2as (max 2/6)
[16 – T = 1.6a and T – 8sinθ = 0.8a] → a = 4.67 (ms–2) B1
14
[v2 = 2 × × 0.5] → speed is 2.16 (ms–1) B1
3
Alternative method 1 for Question 4
[ 1× 0.8 × v2] or [0.8 g × 0.5 × sinθ ]
2 | M1 | For KE gain or PE gain of particle A
1× 0.8 × v2 + 0.8 g × 0.5 × 0.6
2 | A1 | Total energy gain for particle A
[16 – T =1.6a and T – 8sinθ = 0.8a → T = ....] 8.53 | M1 | Forms and solves Newton II equations to find tension T
128
WD = × 0.5
T
15 | A1 | Finds WD
Tension
[ 1 × 0.8 × v2 + 0.8 g × 0.5 × 0.6 = 128 × 0.5]
2 15 | M1 | Energy equation (3 terms)
Question | Answer | Marks | Guidance
4 | Speed is 2.16 (m s–1) | A1
Total: | 6
Alternative method 2 for Question 4
[ 1× 1.6 × v2] or [1.6 g × 0.5]
2 | M1 | For KE gain or PE loss of particle B
1.6 g × 0.5 – 1 × 1.6 × v2
2 | A1 | Energy change for particle B
[16 – T = 1.6a and T – 8sinθ = 0.8a → T = ....] 8.53 | M1 | Forms and solves Newton II equations to find tension T
WD = 128 × 0.5
T
15 | A1 | Finds WD
Tension
1.6 g × 0.5 – 1 × 1.6 × v2 = 128× 0.5]
2 15 | M1 | Energy equation (3 terms)
Speed is 2.16 (m s–1) | A1
Total: | 6
Question | Answer | Marks | Guidance
\includegraphics{figure_4}
Two particles $A$ and $B$, of masses $0.8\text{ kg}$ and $1.6\text{ kg}$ respectively, are connected by a light inextensible string. Particle $A$ is placed on a smooth plane inclined at an angle $\theta$ to the horizontal, where $\sin \theta = \frac{3}{5}$. The string passes over a small smooth pulley $P$ fixed at the top of the plane, and $B$ hangs freely (see diagram). The section $AP$ of the string is parallel to a line of greatest slope of the plane. The particles are released from rest with both sections of the string taut. Use an energy method to find the speed of the particles after each particle has moved a distance of $0.5\text{ m}$, assuming that $A$ has not yet reached the pulley. [6]
\hfill \mbox{\textit{CAIE M1 2018 Q4 [6]}}