| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2001 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Collision with friction after impact |
| Difficulty | Standard +0.3 This is a standard M1 collision problem combining conservation of momentum with friction. Part (a) requires setting up two simultaneous equations (momentum conservation and the given speed relationship), part (b) is direct impulse calculation, and part (c) applies work-energy theorem with friction. All techniques are routine for M1 with no novel insight required, making it slightly easier than average. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.03b Conservation of momentum: 1D two particles6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Before: \(A(\underset{0.6}{\rightarrow}4.5)\) \((\underset{0.2}{\leftarrow}3)B\); Momentum conserved on system | M1 | |
| After: \(\frac{v}{2} \rightarrow \quad v \rightarrow\) | ||
| \(0.6 \times 4.5 - 0.2 \times 3 = 0.6 \times \frac{v}{2} + 0.2 \times v\) | A1, M1, A1 | |
| Solving for \(v \rightarrow V = 4.2\) | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Change in momentum of A or B attempted | M1 | |
| \(0.2(3+4.2)\) or \(0.6(4.5-2.1) \rightarrow 1.44\) units NS | A1 f.t. B1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(R = mg\) | B1 | |
| \(\mu R = \mu mg =\) retarding force or deceleration \(\mu g\) | M1 | |
| \(v^2 = u^2 + 2as\) applied: \(0 = 4.2^2 - 2\mu g \times 2\) | M1 A1 f.t. | |
| \(\mu = \frac{4.2^2}{4g} = 0.45\) | DM1, A1 | (6) DM1 depends on M1+1 |
## Question 5:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Before: $A(\underset{0.6}{\rightarrow}4.5)$ $(\underset{0.2}{\leftarrow}3)B$; Momentum conserved on system | M1 | |
| After: $\frac{v}{2} \rightarrow \quad v \rightarrow$ | | |
| $0.6 \times 4.5 - 0.2 \times 3 = 0.6 \times \frac{v}{2} + 0.2 \times v$ | A1, M1, A1 | |
| Solving for $v \rightarrow V = 4.2$ | | (4) |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Change in momentum of A or B attempted | M1 | |
| $0.2(3+4.2)$ or $0.6(4.5-2.1) \rightarrow 1.44$ units NS | A1 f.t. B1 | (3) |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $R = mg$ | B1 | |
| $\mu R = \mu mg =$ retarding force or deceleration $\mu g$ | M1 | |
| $v^2 = u^2 + 2as$ applied: $0 = 4.2^2 - 2\mu g \times 2$ | M1 A1 f.t. | |
| $\mu = \frac{4.2^2}{4g} = 0.45$ | DM1, A1 | (6) DM1 depends on M1+1 |
---5. Two small balls $A$ and $B$ have masses 0.6 kg and 0.2 kg respectively. They are moving towards each other in opposite directions on a horizontal table when they collide directly. Immediately before the collision, the speed of $A$ is $4.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the speed of $B$ is $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Immediately after the collision, $A$ and $B$ move in the same direction and the speed of $B$ is twice the speed of $A$.
By modelling the balls as particles, find
\begin{enumerate}[label=(\alph*)]
\item the speed of $B$ immediately after the collision,
\item the magnitude of the impulse exerted on $B$ in the collision, stating the units in which your answer is given.
The table is rough. After the collision, $B$ moves a distance of 2 m on the table before coming to rest.
\item Find the coefficient of friction between $B$ and the table.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2001 Q5 [13]}}