Moderate -0.3 This is a straightforward application of the work-energy principle with clearly stated forces and distances. Students must resolve forces parallel to the slope, calculate net work done, and apply W = ΔKE twice. While it requires careful sign management for the two cases, the method is standard and all values are given explicitly, making it slightly easier than average.
A constant resistance to motion of magnitude 350 N acts on a car of mass 1250 kg. The engine of the car exerts a constant driving force of 1200 N. The car travels along a road inclined at an angle of \(\theta\) to the horizontal, where \(\sin \theta = 0.05\). Find the speed of the car when it has moved 100 m from rest in each of the following cases.
• The car is moving up the hill.
• The car is moving down the hill. [7]
Apply Newton’s second law for motion down the hill
[a = 1475/1250 = 1.18]
A1
Correct Newton’s law for motion down the hill
Up the hill: v 2 = 0 + 2 × 0.18 × 100
Answer
Marks
Guidance
Down the hill: v 2 = 0 + 2 × 1.18 × 100
M1
Use their a in the constant acceleration equations either to
find v going up or going down the hill
Answer
Marks
Guidance
Up the hill: v = 6 ms–1
A1
Down the hill: v = 15.4 ms–1
A1
Allow v = 2√59
Alternative method for question 4
Answer
Marks
Guidance
[1200 × 100 = 350 × 100 + 1250g × 100 × 0.05 + ½ × 1250 × v 2]
M1
Attempt the work-energy equation for motion up the hill
A1
Correct work-energy equation for motion up the hill
[1200 × 100 + 1250g × 100 × 0.05 = 350 × 100 + ½ × 1250 × v 2]
M1
Attempt work-energy equation for motion down the hill
A1
Correct work-energy equation for motion down the hill
M1
Attempt to solve either energy equation to find
either v going up the hill or v going down the hill
Answer
Marks
Guidance
Up the hill: v = 6 ms–1
A1
Down the hill: v = 15.4 ms–1
A1
Allow v = 2√59
7
Answer
Marks
Guidance
Question
Answer
Marks
Question 4:
4 | [1200 – 350 – 1250 × 10 × 0.05 = 1250a] | M1 | Apply Newton’s second law for motion up the hill
[a = 225/1250 = 0.18] | A1 | Correct Newton’s law for motion up the hill
[1200 – 350 + 1250 × 10 × 0.05 = 1250a] | M1 | Apply Newton’s second law for motion down the hill
[a = 1475/1250 = 1.18] | A1 | Correct Newton’s law for motion down the hill
Up the hill: v 2 = 0 + 2 × 0.18 × 100
Down the hill: v 2 = 0 + 2 × 1.18 × 100 | M1 | Use their a in the constant acceleration equations either to
find v going up or going down the hill
Up the hill: v = 6 ms–1 | A1
Down the hill: v = 15.4 ms–1 | A1 | Allow v = 2√59
Alternative method for question 4
[1200 × 100 = 350 × 100 + 1250g × 100 × 0.05 + ½ × 1250 × v 2] | M1 | Attempt the work-energy equation for motion up the hill
A1 | Correct work-energy equation for motion up the hill
[1200 × 100 + 1250g × 100 × 0.05 = 350 × 100 + ½ × 1250 × v 2] | M1 | Attempt work-energy equation for motion down the hill
A1 | Correct work-energy equation for motion down the hill
M1 | Attempt to solve either energy equation to find
either v going up the hill or v going down the hill
Up the hill: v = 6 ms–1 | A1
Down the hill: v = 15.4 ms–1 | A1 | Allow v = 2√59
7
Question | Answer | Marks | Guidance
A constant resistance to motion of magnitude 350 N acts on a car of mass 1250 kg. The engine of the car exerts a constant driving force of 1200 N. The car travels along a road inclined at an angle of $\theta$ to the horizontal, where $\sin \theta = 0.05$. Find the speed of the car when it has moved 100 m from rest in each of the following cases.
• The car is moving up the hill.
• The car is moving down the hill. [7]
\hfill \mbox{\textit{CAIE M1 2019 Q4 [7]}}