Hint
Choose and retain a positive direction; distinguish signed quantities from magnitudes. Use total mass and all three friction forces; cancel internal tensions.
Worked solution
(a)
Use total mass and all three friction forces; cancel internal tensions.
The full-system pull minus μg times total mass leaves resultant 10 N, giving \(\displaystyle a= \frac{1m}{s^{2}}\).
Answer: \(1\) \(\displaystyle \frac{m}{s^{2}}\)
(b)
Isolate both trailing particles when calculating the front string tension.
AB pulls the B+C subsystem against its friction, giving \(\displaystyle T_{\mathrm{AB}}=(M_{2}+M_{3})(a+\mu g)=\)\(8\) N.
Answer: \(8\) N
(c)
Their speed at the break is at=\(\displaystyle \frac{2m}{s}\). With the front pull gone, the remaining pair decelerates at \(\displaystyle \mu g=\)\(1\) \(\displaystyle \frac{m}{s^{2}}\). The extra stopping time is \(2\) s.
Answer: \(2\) s.
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit complete valid alternative methods and consequential reasoning. An unexplained result does not earn a method mark.
(a) · 2 marks
- 1 mark: Use total mass and all three friction forces; cancel internal tensions.
- 1 mark: Give the correct requested result, retaining signs, units and any domain restrictions.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Equal acceleration does not make two separate string tensions equal.
(b) · 2 marks
- 1 mark: Isolate both trailing particles when calculating the front string tension.
- 1 mark: Give the correct requested result, retaining signs, units and any domain restrictions.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Choose the correct total mass for each isolated subsystem.
(c) · 3 marks
- 1 mark: Transfer the common pre-break velocity to the new B+C stage.
- 1 mark: Use the new external friction resultant and stop at the first zero velocity.
- 1 mark: Give the complete requested result, using the required units, precision and boundary inclusion.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Equal acceleration does not make two separate string tensions equal.
Total: 7 marks.