| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2020 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Multiple sequential collisions |
| Difficulty | Standard +0.3 This is a standard two-part mechanics question requiring routine application of SUVAT equations, conservation of momentum, and friction calculations. Part (a) involves finding speed down an incline then applying momentum conservation—straightforward bookwork. Part (b) adds friction and coalescence but follows standard procedures. The multi-step nature and 10 total marks place it slightly above average, but no novel insight or complex problem-solving is required. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form3.03f Weight: W=mg3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model6.02i Conservation of energy: mechanical energy principle6.03b Conservation of momentum: 1D two particles6.03i Coefficient of restitution: e6.03j Perfectly elastic/inelastic: collisions |
| Answer | Marks |
|---|---|
| 7(a) | 0.3gsin 30 = 0.3a (a = 5) |
| (M1 for applying Newton’s second law parallel to the plane) | M1 |
| v2 = 0 + 2 × 2.5 × a | M1 |
| v = 5 | A1 |
| 0.3 × 5 + 0 = 0.3 × 2 + 0.2 w | M1 |
| Velocity of Q = 4.5 ms–1 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 7(b) | 0.3 × z + 0 = 0.5 × 1.2 | M1 |
| Velocity of P before collision z = 2 | A1 | |
| Friction force on P after reaches horizontal plane F = μ × 0.3 g | B1 |
| Answer | Marks |
|---|---|
| 2 2 | M1 |
| Coefficient μ = 0.7 | A1 |
| Answer | Marks |
|---|---|
| 0.3 × z + 0 = 0.5 × 1.2 | M1 |
| Velocity of P before collision z = 2 | A1 |
| Friction force on P after reaches horizontal plane F = μ × 0.3 g | B1 |
| a = (52 – 22) / (2 × 1.5) = 7, F = 0.3 × 7 | M1 |
| Coefficient μ = 0.7 | A1 |
Question 7:
--- 7(a) ---
7(a) | 0.3gsin 30 = 0.3a (a = 5)
(M1 for applying Newton’s second law parallel to the plane) | M1
v2 = 0 + 2 × 2.5 × a | M1
v = 5 | A1
0.3 × 5 + 0 = 0.3 × 2 + 0.2 w | M1
Velocity of Q = 4.5 ms–1 | A1
5
Question | Answer | Marks
--- 7(b) ---
7(b) | 0.3 × z + 0 = 0.5 × 1.2 | M1
Velocity of P before collision z = 2 | A1
Friction force on P after reaches horizontal plane F = μ × 0.3 g | B1
1 1
μ × 0.3g × 1.5 = × 0.3 × 52 – × 0.3 × 22
2 2 | M1
Coefficient μ = 0.7 | A1
Alternative method for question 7(b)
0.3 × z + 0 = 0.5 × 1.2 | M1
Velocity of P before collision z = 2 | A1
Friction force on P after reaches horizontal plane F = μ × 0.3 g | B1
a = (52 – 22) / (2 × 1.5) = 7, F = 0.3 × 7 | M1
Coefficient μ = 0.7 | A1
5A particle $P$ of mass $0.3\text{ kg}$, lying on a smooth plane inclined at $30°$ to the horizontal, is released from rest. $P$ slides down the plane for a distance of $2.5\text{ m}$ and then reaches a horizontal plane. There is no change in speed when $P$ reaches the horizontal plane. A particle $Q$ of mass $0.2\text{ kg}$ lies at rest on the horizontal plane $1.5\text{ m}$ from the end of the inclined plane (see diagram). $P$ collides directly with $Q$.
\includegraphics{figure_7}
\begin{enumerate}[label=(\alph*)]
\item It is given that the horizontal plane is smooth and that, after the collision, $P$ continues moving in the same direction, with speed $2\text{ m s}^{-1}$.
Find the speed of $Q$ after the collision. [5]
\item It is given instead that the horizontal plane is rough and that when $P$ and $Q$ collide, they coalesce and move with speed $1.2\text{ m s}^{-1}$.
Find the coefficient of friction between $P$ and the horizontal plane. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2020 Q7 [10]}}