Standard +0.3 This is a straightforward work-energy problem requiring application of the work-energy principle with multiple forces. Students must account for kinetic energy change, potential energy change, work against resistance, and work by driving force—all standard M1 techniques with no conceptual surprises. The multi-step calculation and bookkeeping of energy terms makes it slightly above average difficulty.
A lorry of mass 12 000 kg moves up a straight hill of length 500 m, starting at the bottom with a speed of \(24 \text{ m s}^{-1}\) and reaching the top with a speed of \(16 \text{ m s}^{-1}\). The top of the hill is 25 m above the level of the bottom of the hill. The resistance to motion of the lorry is 7500 N. Find the driving force of the lorry. [6]
Cambridge International AS/A Level – May/June 2015
9709
43
WD by DF
= 3000000 – 1920000 + 7500×500
Driving force = 4830000÷500
Answer
Marks
Driving force is 9660 N
A1
M1
Answer
Marks
Guidance
A1
6
For using DF = WD by DF÷500
Alternative Method for 4
Answer
Marks
4
2 2
[16 = 24 + 2 × 500a]
-2
a = – 0.32 ms
Weight component down hill =
12000g × 25/500
25
DF – 7500 – 12000g ×
500
=12000 × (– 0.32)
Answer
Marks
Driving force is 9660 N
M1
A1
B1
M1
A1
Answer
Marks
Guidance
A1
6
2 2
For using v = u + 2as
For using Newton’s 2nd law
Question 4:
4 | 1
2 2
KE loss = × 12000(24 – 16 )
2
PE gain = 12000g × 25 | B1
B1
M1 | For using WD by DF
= PE gain – KE loss
+ WD against resistance
Page 6 | Mark Scheme | Syllabus | Paper
Cambridge International AS/A Level – May/June 2015 | 9709 | 43
WD by DF
= 3000000 – 1920000 + 7500×500
Driving force = 4830000÷500
Driving force is 9660 N | A1
M1
A1 | 6 | For using DF = WD by DF÷500
Alternative Method for 4
4 | 2 2
[16 = 24 + 2 × 500a]
-2
a = – 0.32 ms
Weight component down hill =
12000g × 25/500
25
DF – 7500 – 12000g ×
500
=12000 × (– 0.32)
Driving force is 9660 N | M1
A1
B1
M1
A1
A1 | 6 | 2 2
For using v = u + 2as
For using Newton’s 2nd law
A lorry of mass 12 000 kg moves up a straight hill of length 500 m, starting at the bottom with a speed of $24 \text{ m s}^{-1}$ and reaching the top with a speed of $16 \text{ m s}^{-1}$. The top of the hill is 25 m above the level of the bottom of the hill. The resistance to motion of the lorry is 7500 N. Find the driving force of the lorry. [6]
\hfill \mbox{\textit{CAIE M1 2015 Q4 [6]}}