| Exam Board | Edexcel |
|---|---|
| Module | FM1 AS (Further Mechanics 1 AS) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Projectile energy - basic KE/PE calculation |
| Difficulty | Standard +0.3 This is a straightforward application of conservation of energy and work-energy principle with standard mechanics setup. Part (a) requires basic energy conservation (KE + PE), part (b) is conceptual understanding of air resistance, and part (c) applies work-energy principle with constant resistance force. All steps are routine for FM1 students with no novel problem-solving required, making it slightly easier than average A-level difficulty. |
| Spec | 6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempt at use of conservation of energy principle | M1 | Correct no. of terms (two KE and one PE), dimensionally correct, condone sign errors. \(m\) does not need to be substituted, allow cancelled \(m\)'s. N.B. M0 if clearly using \(v^2 = u^2 + 2as\) for whole motion. M0 if they use \(0.5g \times (2.5 + 0.01d)\) |
| \(\frac{1}{2}m \times 25^2 - \frac{1}{2}mU^2 = mg \times 2.5\) | A1 | Correct equation in \(U\) only with at most one error |
| A1 | Correct equation in \(U\) only | |
| \(U = 24\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The value of \(U\) would be larger | B1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| WD against resistance \(= 2000 \times 0.01d\) | M1 | Use of work \(=\) force \(\times\) distance (allow if \(0.01\) is omitted) |
| Use of work-energy principle | M1 | Correct no. of terms (one KE, one PE, one Work Done), dimensionally correct. N.B. M0 if not using work-energy principle |
| \(2000 \times 0.01d = 0.5g \times 0.01d + \frac{1}{2} \times 0.5 \times 25^2\) | A1 | Correct equation in \(d\) only with at most one error (omission of \(0.01\) twice is one error) |
| A1 | Correct equation in \(d\) only | |
| \(d = 7.83\) (3 sf). N.B. If PE term omitted, \(d = 7.8152\), scores max M1M0A0A0A0 | A1 | cao |
## Question 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempt at use of conservation of energy principle | M1 | Correct no. of terms (two KE and one PE), dimensionally correct, condone sign errors. $m$ does not need to be substituted, allow cancelled $m$'s. N.B. M0 if clearly using $v^2 = u^2 + 2as$ for whole motion. M0 if they use $0.5g \times (2.5 + 0.01d)$ |
| $\frac{1}{2}m \times 25^2 - \frac{1}{2}mU^2 = mg \times 2.5$ | A1 | Correct equation in $U$ only with at most one error |
| | A1 | Correct equation in $U$ only |
| $U = 24$ | A1 | cao |
**(4 marks)**
---
## Question 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| The value of $U$ would be larger | B1 | cao |
**(1 mark)**
---
## Question 3(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| WD against resistance $= 2000 \times 0.01d$ | M1 | Use of work $=$ force $\times$ distance (allow if $0.01$ is omitted) |
| Use of work-energy principle | M1 | Correct no. of terms (one KE, one PE, one Work Done), dimensionally correct. N.B. M0 if not using work-energy principle |
| $2000 \times 0.01d = 0.5g \times 0.01d + \frac{1}{2} \times 0.5 \times 25^2$ | A1 | Correct equation in $d$ only with at most one error (omission of $0.01$ twice is one error) |
| | A1 | Correct equation in $d$ only |
| $d = 7.83$ (3 sf). N.B. If PE term omitted, $d = 7.8152$, scores max M1M0A0A0A0 | A1 | cao |
**(5 marks)**
---\begin{enumerate}
\item A stone of mass 0.5 kg is projected vertically upwards with a speed $U \mathrm {~ms} ^ { - 1 }$ from a point $A$. The point $A$ is 2.5 m above horizontal ground.
\end{enumerate}
The speed of the stone as it hits the ground is $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$\\
The motion of the stone from the instant it is projected from $A$ until the instant it hits the ground is modelled as that of a particle moving freely under gravity.\\
(a) Use the model and the principle of conservation of mechanical energy to find the value of $U$.
In reality, the stone will be subject to air resistance as it moves from $A$ to the ground.\\
(b) State how this would affect your answer to part (a).
The ground is soft and the stone sinks a vertical distance $d \mathrm {~cm}$ into the ground. The resistive force exerted on the stone by the ground is modelled as a constant force of magnitude 2000 N and the stone is modelled as a particle.\\
(c) Use the model and the work-energy principle to find the value of $d$, giving your answer to 3 significant figures.
\hfill \mbox{\textit{Edexcel FM1 AS 2023 Q3 [10]}}