| Exam Board | Edexcel |
|---|---|
| Module | FM1 AS (Further Mechanics 1 AS) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Particle-wall perpendicular collision |
| Difficulty | Moderate -0.5 This is a straightforward two-part mechanics question requiring standard application of kinematic equations (v² = u² + 2as) to find speed before impact, then impulse-momentum theorem to find speed after impact, followed by a direct kinetic energy calculation. While it involves multiple steps, each step uses routine A-level mechanics formulas with no problem-solving insight required, making it slightly easier than average. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Speed just before impact: \(v^2 = u^2 + 2as = 2 \times 9.8 \times 3.6 (= 70.56)\) | M1 | Use suvat or energy to find speed before impact |
| \(v = 8.4\) (m s\(^{-1}\)) | A1 | Correct answer. Accept \(\sqrt{70.56}\), \(\sqrt{7.2g}\) |
| Use of \(I = mv - mu\): \(4.2 = 0.3(w-(-8.4))\) | M1 | Complete strategy to find \(w\); must use difference in velocities. Vigilant for sign fudges |
| Follow their 8.4 | A1ft | Correct unsimplified equation using their speed |
| \(w = 5.6\) (m s\(^{-1}\)) | A1 | Correct positive answer |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| KE lost \(= \frac{1}{2}m(v^2 - w^2)\) | M1 | Correct method using speeds immediately before and after impact |
| \(= \frac{0.3}{2}(8.4^2 - 5.6^2)\) — Follow their 8.4 and 5.6 | A1ft | Correct expression for their speeds. Accept subtraction either way round |
| \(= 5.88\) (J) | A1 | Correct solution only. Accept 5.9 |
| (3) |
## Question 1:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Speed just before impact: $v^2 = u^2 + 2as = 2 \times 9.8 \times 3.6 (= 70.56)$ | M1 | Use suvat or energy to find speed before impact |
| $v = 8.4$ (m s$^{-1}$) | A1 | Correct answer. Accept $\sqrt{70.56}$, $\sqrt{7.2g}$ |
| Use of $I = mv - mu$: $4.2 = 0.3(w-(-8.4))$ | M1 | Complete strategy to find $w$; must use difference in velocities. Vigilant for sign fudges |
| Follow their 8.4 | A1ft | Correct unsimplified equation using their speed |
| $w = 5.6$ (m s$^{-1}$) | A1 | Correct positive answer |
| **(5)** | | |
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| KE lost $= \frac{1}{2}m(v^2 - w^2)$ | M1 | Correct method using speeds immediately before and after impact |
| $= \frac{0.3}{2}(8.4^2 - 5.6^2)$ — Follow their 8.4 and 5.6 | A1ft | Correct expression for their speeds. Accept subtraction either way round |
| $= 5.88$ (J) | A1 | Correct solution only. Accept 5.9 |
| **(3)** | | |
---\begin{enumerate}
\item A small ball of mass 0.3 kg is released from rest from a point 3.6 m above horizontal ground. The ball falls freely under gravity, hits the ground and rebounds vertically upwards.
\end{enumerate}
In the first impact with the ground, the ball receives an impulse of magnitude 4.2 Ns . The ball is modelled as a particle.\\
(a) Find the speed of the ball immediately after it first hits the ground.\\
(b) Find the kinetic energy lost by the ball as a result of the impact with the ground.
\hfill \mbox{\textit{Edexcel FM1 AS 2018 Q1 [8]}}