| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2018 |
| Session | October |
| Marks | 6 |
| 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 standard application of conservation of momentum and impulse-momentum theorem. Part (a) involves simple algebraic manipulation with given values, and part (b) is direct recall of impulse = change in momentum. Both parts are routine textbook exercises with no problem-solving insight required, making it easier than average A-level questions. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| VILU SIHI NI III M I ION OC | VIIV 5141 NI 311814 ION OC | VI4V SIHI NI JIIYM ION OC |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0.8 \times 4 - 2 \times 2 = 2v - 0.8 \times 2.5\) | M1A1 | CLM, correct no. of terms, dim correct, condone extra \(g\)'s, correct pairings of mass and velocity |
| \(v = 0.6 \text{ m s}^{-1}\) | A1 (3) | Must be positive |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(I = 0.8(4 + 2.5) = 5.2\) Ns | M1A1,B1 (3) | Impulse-momentum equation, correct velocities, condone sign errors. M0 if \(g\) included |
| OR: \(I = 2(0.6 + 2) = 5.2\) Ns | M1A1,B1 [6] | B1 for Ns or kg m s\(^{-1}\). M0A0B1 possible |
# Question 1:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0.8 \times 4 - 2 \times 2 = 2v - 0.8 \times 2.5$ | M1A1 | CLM, correct no. of terms, dim correct, condone extra $g$'s, correct pairings of mass and velocity |
| $v = 0.6 \text{ m s}^{-1}$ | A1 (3) | Must be positive |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $I = 0.8(4 + 2.5) = 5.2$ Ns | M1A1,B1 (3) | Impulse-momentum equation, correct velocities, condone sign errors. M0 if $g$ included |
| **OR:** $I = 2(0.6 + 2) = 5.2$ Ns | M1A1,B1 [6] | B1 for Ns or kg m s$^{-1}$. M0A0B1 possible |
---\begin{enumerate}
\item A particle $P$ of mass 0.8 kg is moving along a straight horizontal line on a smooth hoizontal surface with speed $4 \mathrm {~ms} ^ { - 1 }$. A second particle $Q$ of mass 2 kg is moving, in the opposite direction to $P$, along the same straight line with speed $2 \mathrm {~ms} ^ { - 1 }$. The particles collide directly. Immediately after the collision the direction of motion of each particle is reversed and the speed of $P$ is $2.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(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, stating the units of your answer.\\
\end{enumerate}
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VILU SIHI NI III M I ION OC & VIIV 5141 NI 311814 ION OC & VI4V SIHI NI JIIYM ION OC \\
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Figure 1
A non-uniform plank $A B$ has weight 60 N and length 5 m . The plank rests horizontally in equilibrium on two smooth supports at $A$ and $C$, where $A C = 3 \mathrm {~m}$, as shown in Figure 1. A parcel of weight 12 N is placed on the plank at $B$ and the plank remains horizontal and in equilibrium. The magnitude of the reaction of the support at $A$ on the plank is half the magnitude of the reaction of the support at $C$ on the plank.
By modelling the plank as a non-uniform rod and the parcel as a particle,\\
(a) find the distance of the centre of mass of the plank from $A$.\\
(b) State briefly how you have used the modelling assumption\\
(i) that the parcel is a particle,\\
(ii) that the plank is a rod.\\
\hfill \mbox{\textit{Edexcel M1 2018 Q1 [6]}}