Moderate -0.3 This is a straightforward work-energy problem requiring application of the work-energy principle with one resistance force. Students must recognize that total work = change in KE + work against resistance, then solve a simple equation for m. Standard M1 content with clear structure and no conceptual subtlety.
An athlete has mass \(m\) kg. The athlete runs along a horizontal road against a constant resistance force of magnitude 24 N. The total work done by the athlete in increasing his speed from 5 ms\(^{-1}\) to 6 ms\(^{-1}\) while running a distance of 50 metres is 1541 J.
Find the value of \(m\). [4]
Attempt at work energy equation with 4 relevant terms;
dimensionally correct. Allow sign errors.
1
m(6−5)2
M0 for .
2
341
m= =62
Answer
Marks
5.5
A1
SC for constant acceleration method (question only gives total work done, and does not suggest constant force)
Answer
Marks
Guidance
a=0.11
SCB1
From 62 =52 +2a50 SOI.
m=62
SCB1
1541
From −24=m0.11 or 30.82−24=m0.11.
50
4
Answer
Marks
Guidance
Question
Answer
Marks
Question 1:
1 | 1
KE = m52
before
2
1
KE = m62
after
2 | B1 | For either correct.
1
m(6−5)2
Do not allow .
2
1
Note: Difference = m11.
2
WD against resistance
= 5024 =1200 | B1 | (m−24)50.
Do not allow if errors such as e.g.
1 m ( 62 −52) +5024=1541
2
5.5m=1541−1200 | M1 | Attempt at work energy equation with 4 relevant terms;
dimensionally correct. Allow sign errors.
1
m(6−5)2
M0 for .
2
341
m= =62
5.5 | A1
SC for constant acceleration method (question only gives total work done, and does not suggest constant force)
a=0.11 | SCB1 | From 62 =52 +2a50 SOI.
m=62 | SCB1 | 1541
From −24=m0.11 or 30.82−24=m0.11.
50
4
Question | Answer | Marks | Guidance
An athlete has mass $m$ kg. The athlete runs along a horizontal road against a constant resistance force of magnitude 24 N. The total work done by the athlete in increasing his speed from 5 ms$^{-1}$ to 6 ms$^{-1}$ while running a distance of 50 metres is 1541 J.
Find the value of $m$. [4]
\hfill \mbox{\textit{CAIE M1 2024 Q1 [4]}}