Hint

Choose and retain a positive direction; distinguish signed quantities from magnitudes. Add gravity and friction along the plane during upward sliding.

Worked solution

(a)

The normal reaction is Mg cos α. Both Mg sin α and μMg cos α act down the plane, so \(Ma=-Mg\sin\alpha-\mu Mg\cos\alpha\) gives \(\displaystyle a=\)\(\displaystyle - \frac{34}{5}\) \(\displaystyle \frac{m}{s^{2}}\).

Answer: \(\displaystyle - \frac{34}{5}\) \(\displaystyle \frac{m}{s^{2}}\).

(b)

Use the upward-stage deceleration and zero velocity at the first stop.

Using 0=\(\displaystyle U^{2}\)−2bs gives \(\displaystyle \frac{605}{68}\) m.

Answer: \(\displaystyle \frac{605}{68}\) m

(c)

To remain still would require friction Mg sin α. Its cap is μMg cos α, which is smaller because \(\displaystyle \frac{3}{5}\) > \(\displaystyle \frac{2}{25}\). It slides back, with friction now uphill. Downward Newton balance gives \(\displaystyle a=g(\sin \alpha -\mu \cos \alpha )=\)\(\displaystyle \frac{26}{5}\) \(\displaystyle \frac{m}{s^{2}}\).

Answer: It cannot remain at rest; downward acceleration is \(\displaystyle \frac{26}{5}\) \(\displaystyle \frac{m}{s^{2}}\).

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) · 3 marks

• 1 mark: Draw gravity and friction acting down the plane during upward sliding.
• 1 mark: Resolve along the plane with the stated positive direction and use Newton’s second law.
• 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: Friction reverses when sliding reverses.

(b) · 2 marks

• 1 mark: Use the upward-stage deceleration and zero velocity at the first stop.
• 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: Check whether the particle can stay at rest before assuming a return slide.

(c) · 3 marks

• 1 mark: Compare the friction required for rest with its limiting value and interpret the inequality.
• 1 mark: Reverse friction for the downward stage and apply the new along-plane force equation.
• 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: Friction reverses when sliding reverses.

Total: 8 marks.