| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2020 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Direct collision, find final speed |
| Difficulty | Moderate -0.5 This is a straightforward M1 momentum conservation problem with clearly defined masses and velocities. Part (a) requires setting up and solving simultaneous equations from momentum conservation and the given speed relationship. Parts (b) and (c) are direct applications of impulse formula and Newton's third law. While it requires careful algebraic manipulation, it follows a standard template with no conceptual surprises or novel problem-solving required. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| CLM: \(4mu + 2mu = mv + 2m \times 4v\) | M1 A1 | M1: Correct number of terms, dimensionally correct, condone sign errors. Allow even if both assumed moving at same speed after collision. A1: Correct equation, allow cancelled \(m\)'s or consistent extra \(g\)'s |
| \(4v = \frac{8u}{3}\) (2.7\(u\) or better) | A1 | Correct answer, must be positive as it's a speed, single term |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\pm m(v - 4u)\) OR \(\pm 2m(4v - u)\) | M1 A1ft | M1: Dimensionally correct impulse-momentum equation (M0 if \(g\) included), correct terms, condone sign errors, must be difference of momenta using EITHER \(m\) and \(4u\) with \(v_P\) OR \(2m\) and \(u\) with \(v_Q\). A1ft: Correct expression in terms of \(m\) and \(u\), follow their \(v_P\) or \(v_Q\). A0ft if both assumed same speed after collision |
| \(\frac{10mu}{3}\) (3.3\(mu\) or better) | A1 | cao, must be positive as it's a magnitude |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Opposite to the direction of motion | B1 | Any clear equivalent |
| (1) |
# Question 1:
## Part (a)
**Diagram:** $P(m)$ moving at $4u$, $Q(2m)$ moving at $u$, after collision $P$ moves at $v$, $Q$ moves at $4v$
| Answer/Working | Marks | Guidance |
|---|---|---|
| CLM: $4mu + 2mu = mv + 2m \times 4v$ | M1 A1 | M1: Correct number of terms, dimensionally correct, condone sign errors. Allow even if both assumed moving at same speed after collision. A1: Correct equation, allow cancelled $m$'s or consistent extra $g$'s |
| $4v = \frac{8u}{3}$ (2.7$u$ or better) | A1 | Correct answer, must be positive as it's a speed, single term |
| | **(3)** | |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\pm m(v - 4u)$ **OR** $\pm 2m(4v - u)$ | M1 A1ft | M1: Dimensionally correct impulse-momentum equation (M0 if $g$ included), correct terms, condone sign errors, must be difference of momenta using EITHER $m$ and $4u$ with $v_P$ OR $2m$ and $u$ with $v_Q$. A1ft: Correct expression in terms of $m$ and $u$, follow their $v_P$ or $v_Q$. A0ft if both assumed same speed after collision |
| $\frac{10mu}{3}$ (3.3$mu$ or better) | A1 | cao, must be positive as it's a magnitude |
| | **(3)** | |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Opposite to the direction of motion | B1 | Any clear equivalent |
| | **(1)** | |
**Total: (7)**\begin{enumerate}
\item Two particles, $P$ and $Q$, with masses $m$ and $2 m$ respectively, are moving in the same direction along the same straight line when they collide directly. Immediately before they collide, $P$ is moving with speed $4 u$ and $Q$ is moving with speed $u$. Immediately after they collide, both particles are moving in the same direction and the speed of $Q$ is four times the speed of $P$.\\
(a) Find the speed of $Q$ immediately after the collision.\\
(b) Find the magnitude of the impulse exerted by $Q$ on $P$ in the collision.\\
(c) State clearly the direction of this impulse.\\
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2020 Q1 [7]}}