| Exam Board | Edexcel |
|---|---|
| Module | FM1 AS (Further Mechanics 1 AS) |
| Year | 2020 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Particle-wall perpendicular collision |
| Difficulty | Standard +0.3 This is a straightforward Further Mechanics question requiring standard application of work-energy principle with constant resistance, followed by a collision with given coefficient of restitution. Part (c) requires conceptual understanding of energy loss in collisions. While it's FM1 content (inherently harder than single maths), the techniques are routine and well-practiced, making it slightly easier than average for an A-level question overall. |
| Spec | 6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Scheme | Marks | Guidance |
| \(\frac{1}{2}mgH\) | B1 | Work done against resistance (allow -ve). Can be implied by use of \(\frac{3}{2}mgH\) (work done against resistance + work done against weight) |
| \(\frac{1}{2}m(8gH - v^2)\) | B1 | KE loss (allow -ve) |
| Apply the work-energy principle | M1 | Correct no. of terms, dimensionally correct. Condone sign errors. |
| \(\frac{1}{2}mgH = \frac{1}{2}m(8gH - v^2) - mgH\) | A1 | Correct unsimplified equation |
| \(v = \sqrt{5gH}\) | A1 | Correct answer (any equivalent but must be in terms of \(g\) and \(H\)). Accept \(2.2\sqrt{gH}\) or better |
| Answer | Marks | Guidance |
|---|---|---|
| Scheme | Marks | Guidance |
| Use NLR to find rebound speed: \(\frac{1}{2}\sqrt{5gH}\) | M1 | Use of NLR |
| Apply the work-energy principle or \(suvat\) with \(a = \frac{1}{2}g\) | M1 | Correct no. of terms, dimensionally correct |
| \(\frac{1}{2}mgH = mgH - \frac{1}{2}m(v_1^2 - \frac{1}{4} \times 5gH)\) or \((v_1)^2 = \frac{5gH}{4} + 2 \times \frac{g}{2} \times H\) | A1ft | Correct equation with at most one error ft on their answer to (a). M1A1ft is available to a candidate who has not scored the first M1 |
| \(v_1 = \frac{3}{2}\sqrt{gH}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Scheme | Marks | Guidance |
| Since \(e < 1\), ball loses energy in its collision with the ceiling. | B1 | Clear explanation. Need to identify that the loss of KE occurs in the impact with the ceiling. Do not insist on seeing \(e < 1\) or equivalent. If they include incorrect additional statements then B0 |
| Scheme | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}mgH$ | B1 | Work done against resistance (allow -ve). Can be implied by use of $\frac{3}{2}mgH$ (work done against resistance + work done against weight) |
| $\frac{1}{2}m(8gH - v^2)$ | B1 | KE loss (allow -ve) |
| Apply the work-energy principle | M1 | Correct no. of terms, dimensionally correct. Condone sign errors. |
| $\frac{1}{2}mgH = \frac{1}{2}m(8gH - v^2) - mgH$ | A1 | Correct unsimplified equation |
| $v = \sqrt{5gH}$ | A1 | Correct answer (any equivalent but must be in terms of $g$ and $H$). Accept $2.2\sqrt{gH}$ or better |
**Question 4b:**
| Scheme | Marks | Guidance |
|--------|-------|----------|
| Use NLR to find rebound speed: $\frac{1}{2}\sqrt{5gH}$ | M1 | Use of NLR |
| Apply the work-energy principle or $suvat$ with $a = \frac{1}{2}g$ | M1 | Correct no. of terms, dimensionally correct |
| $\frac{1}{2}mgH = mgH - \frac{1}{2}m(v_1^2 - \frac{1}{4} \times 5gH)$ or $(v_1)^2 = \frac{5gH}{4} + 2 \times \frac{g}{2} \times H$ | A1ft | Correct equation with at most one error ft on their answer to (a). M1A1ft is available to a candidate who has not scored the first M1 |
| $v_1 = \frac{3}{2}\sqrt{gH}$ | A1 | |
**Question 4c:**
| Scheme | Marks | Guidance |
|--------|-------|----------|
| Since $e < 1$, ball loses energy in its collision with the ceiling. | B1 | Clear explanation. Need to identify that the loss of KE occurs in the impact with the ceiling. Do not insist on seeing $e < 1$ or equivalent. If they include incorrect additional statements then B0 |\begin{enumerate}
\item A small ball, of mass $m$, is thrown vertically upwards with speed $\sqrt { 8 g H }$ from a point $O$ on a smooth horizontal floor. The ball moves towards a smooth horizontal ceiling that is a vertical distance $H$ above $O$. The coefficient of restitution between the ball and the ceiling is $\frac { 1 } { 2 }$\\
In a model of the motion of the ball, it is assumed that the ball, as it moves up or down, is subject to air resistance of constant magnitude $\frac { 1 } { 2 } \mathrm { mg }$.\\
Using this model,\\
(a) use the work-energy principle to find, in terms of $g$ and $H$, the speed of the ball immediately before it strikes the ceiling,\\
(b) find, in terms of $g$ and $H$, the speed of the ball immediately before it strikes the floor at $O$ for the first time.
\end{enumerate}
In a simplified model of the motion of the ball, it is assumed that the ball, as it moves up or down, is subject to no air resistance.
Using this simplified model,\\
(c) explain, without any detailed calculation, why the speed of the ball, immediately before it strikes the floor at $O$ for the first time, would still be less than $\sqrt { 8 g H }$
\hfill \mbox{\textit{Edexcel FM1 AS 2020 Q4 [11]}}