| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Direct collision, find impulse magnitude |
| Difficulty | Moderate -0.8 This is a straightforward M1 momentum conservation problem requiring direct application of the momentum equation and impulse formula. Part (a) involves a single-step calculation using conservation of momentum with clearly defined masses and velocities. Part (b) is a direct application of the impulse-momentum theorem. No problem-solving insight or multi-step reasoning is required—just routine application of standard mechanics formulas. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| \(4m.2u - m.5u = -4m.\frac{1}{2}u + mv\) | M1 A1 | |
| \(3mu = -2mu + mv\) | ||
| \(v = 5u\), opposite direction | A1, A1 cso | (4) |
| Answer | Marks |
|---|---|
| \(I = 4m(\frac{1}{2}u - (-2u))\) OR \(I = m(5u - (-5u))\) | M1 A1 |
| \(= 10mu\) | \(= 10mu\) |
| A1 | (3) |
### Part (a)
$4m.2u - m.5u = -4m.\frac{1}{2}u + mv$ | M1 A1 |
$3mu = -2mu + mv$ | |
$v = 5u$, opposite direction | A1, A1 **cso** | (4)
### Part (b)
$I = 4m(\frac{1}{2}u - (-2u))$ OR $I = m(5u - (-5u))$ | M1 A1 |
$= 10mu$ | $= 10mu$ | |
| A1 | (3)
**Total for Question 1: 7**
---Two particles $P$ and $Q$ have masses $4m$ and $m$ respectively. The particles are moving towards each other on a smooth horizontal plane and collide directly. The speeds of $P$ and $Q$ immediately before the collision are $2u$ and $5u$ respectively. Immediately after the collision, the speed of $P$ is $\frac{1}{2}u$ and its direction of motion is reversed.
\begin{enumerate}[label=(\alph*)]
\item Find the speed and direction of motion of $Q$ after the collision. [4]
\item Find the magnitude of the impulse exerted on $P$ by $Q$ in the collision. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2013 Q1 [7]}}