| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2002 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Collision with friction after impact |
| Difficulty | Moderate -0.8 This is a straightforward M1 collision problem requiring only standard application of conservation of momentum (part a is a 'show that'), followed by basic equations of motion with constant force/deceleration. All three parts use routine textbook methods with no problem-solving insight required, making it easier than average but not trivial due to the multi-part structure and numerical work involved. |
| Spec | 6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles6.03i Coefficient of restitution: e6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(1500 \times 10 + 2500 \times 5 = 1500 \times 4 + 2500 \times v\) | M1 A1 | |
| \(v = 8.6 \text{ m s}^{-1}\) | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(P: 1500a = -500 \Rightarrow a = -\frac{1}{3} \text{ m s}^{-2}\) | M1 | |
| \(0^2 = 4^2 - 2 \times \frac{1}{3} \times s \Rightarrow s = 24\) m | M1 A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(P: 0 = 4 - \frac{1}{3}t \Rightarrow t = 12\) s | M1 | |
| \(Q: s = 8.6 \times 12 = 103.2\) m | M1 A1 | |
| Distance apart \(= 103.2 - 24 = 79.2\) m | M1 A1 | (5) |
## Question 6:
### Part (a):
| Answer/Working | Marks | Notes |
|---|---|---|
| $1500 \times 10 + 2500 \times 5 = 1500 \times 4 + 2500 \times v$ | M1 A1 | |
| $v = 8.6 \text{ m s}^{-1}$ | A1 | **(3)** |
### Part (b):
| Answer/Working | Marks | Notes |
|---|---|---|
| $P: 1500a = -500 \Rightarrow a = -\frac{1}{3} \text{ m s}^{-2}$ | M1 | |
| $0^2 = 4^2 - 2 \times \frac{1}{3} \times s \Rightarrow s = 24$ m | M1 A1 | **(3)** |
### Part (c):
| Answer/Working | Marks | Notes |
|---|---|---|
| $P: 0 = 4 - \frac{1}{3}t \Rightarrow t = 12$ s | M1 | |
| $Q: s = 8.6 \times 12 = 103.2$ m | M1 A1 | |
| Distance apart $= 103.2 - 24 = 79.2$ m | M1 A1 | **(5)** |
---6. A railway truck $P$ of mass 1500 kg is moving on a straight horizontal track. The truck $P$ collides with a truck $Q$ of 2500 kg at a point $A$. Immediately before the collision, $P$ and $Q$ are moving in the same direction with speeds $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ respectively. Immediately after the collision, the direction of motion of $P$ is unchanged and its speed is $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. By modelling the trucks as particles,
\begin{enumerate}[label=(\alph*)]
\item show that the speed of $Q$ immediately after the collision is $8.6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
After the collision at $A$, the truck $P$ is acted upon by a constant braking force of magnitude 500 N . The truck $P$ comes to rest at the point $B$.
\item Find the distance $A B$.
After the collision $Q$ continues to move with constant speed $8.6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\item Find the distance between $P$ and $Q$ at the instant when $P$ comes to rest.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2002 Q6 [11]}}