| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Collision with two possible outcomes |
| Difficulty | Standard +0.3 This is a standard M1 momentum/impulse question requiring straightforward application of impulse-momentum theorem for part (a) and conservation of momentum for part (b). The 'two possible outcomes' refers to Q's direction after collision, making it slightly more interesting than basic exercises but still routine problem-solving with no novel insight required. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03e Impulse: by a force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(4.2 = 0.5(v - -4)\) | M1 | Impulse/momentum equation. Must be using \(I = \pm(mv - mu)\). Inclusion of \(g\) is M0 |
| Correct unsimplified equation | A1 | |
| \(v = 4.4\text{ ms}^{-1}\) | A1 | Must be positive - the question asks for the speed |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2 - 2m = -\dfrac{1}{2}v \pm m\) | M1 | Conservation of momentum. No missing/additional terms. Condone sign errors. Dimensionally correct. Follow their \(v\). Condone common factor of \(g\) throughout |
| Correct equation for one solution. Follow their \(v\) | A1ft | |
| Correct unsimplified equation(s) for both possible solutions. Follow their \(v\) | A1ft | |
| \(m = 1.4\) or \(4.2\) | A1 | Need both |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(4.2 = m(\pm1 - -2)\) | M1 | Impulse on \(Q\). Dimensionally correct. Condone sign errors |
| Correct equation for one solution | A1 | |
| Correct unsimplified equation for both possible solutions | A1 | |
| \(m = 1.4\) or \(4.2\) | A1 | Need both |
## Question 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4.2 = 0.5(v - -4)$ | M1 | Impulse/momentum equation. Must be using $I = \pm(mv - mu)$. Inclusion of $g$ is M0 |
| Correct unsimplified equation | A1 | |
| $v = 4.4\text{ ms}^{-1}$ | A1 | Must be positive - the question asks for the speed |
## Question 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2 - 2m = -\dfrac{1}{2}v \pm m$ | M1 | Conservation of momentum. No missing/additional terms. Condone sign errors. Dimensionally correct. Follow their $v$. Condone common factor of $g$ throughout |
| Correct equation for one solution. Follow their $v$ | A1ft | |
| Correct unsimplified equation(s) for both possible solutions. Follow their $v$ | A1ft | |
| $m = 1.4$ or $4.2$ | A1 | Need both |
**OR:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4.2 = m(\pm1 - -2)$ | M1 | Impulse on $Q$. Dimensionally correct. Condone sign errors |
| Correct equation for one solution | A1 | |
| Correct unsimplified equation for both possible solutions | A1 | |
| $m = 1.4$ or $4.2$ | A1 | Need both |
---3. Two particles, $P$ and $Q$, have masses 0.5 kg and $m \mathrm {~kg}$ respectively. They are moving in opposite directions towards each other along the same straight line on a smooth horizontal plane and collide directly. Immediately before the collision the speed of $P$ is $4 \mathrm {~ms} ^ { - 1 }$ and the speed of $Q$ is $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The magnitude of the impulse exerted on $P$ by $Q$ in the collision is 4.2 N s . As a result of the collision the direction of motion of $P$ is reversed.
\begin{enumerate}[label=(\alph*)]
\item Find the speed of $P$ immediately after the collision.
The speed of $Q$ immediately after the collision is $1 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\item Find the two possible values of $m$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2017 Q3 [7]}}