| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2018 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Difficulty | Standard +0.3 This is a standard two-part work-energy question requiring systematic application of SUVAT equations and energy conservation. Part (i) needs finding initial speed from free fall, then applying work-energy theorem. Part (ii) requires equation of motion with resistance force. All techniques are routine M1 content with clear problem structure and no novel insights required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks |
|---|---|
| 7(i) | At liquid surface, speed = 0 + g × 0.8 [= 8] |
| Answer | Marks | Guidance |
|---|---|---|
| 0.3g × ½ (0 + v) × 0.8 = ½ (0.3) v2 → v = 8 | B1 | Using constant acceleration equation v = u + at |
| Answer | Marks | Guidance |
|---|---|---|
| PE lost in water = 0.3g × 1.25 [ = 3.75] | B1 | |
| [½ × 0.3 × (82 – v2) + 0.3g × 1.25 = 1.2] | M1 | Using work-energy for downward motion in the tank |
| Answer | Marks |
|---|---|
| v = 9 m s–1 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Height above tank = ½ × g × 0.82 [= 3.2] | B1 | |
| Total PE loss = 0.3g × (3.2 + 1.25) [= 13.35] | B1 | |
| [0.3g × (3.2 + 1.25) = ½ × 0.3 × v2 + 1.2] | M1 | Work-energy equation for the total downward motion |
| v = 9 m s–1 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 7(ii) | [–0.3g – 1.8 = 0.3a] | M1 |
| a = –16 | A1 | |
| [1.25 = 7T + ½ × (–16) × T2] | M1 | Using constant acceleration equations to find the time, T, for the particle |
| Answer | Marks | Guidance |
|---|---|---|
| T = 0.25 (or 0.625, on the way down) | A1 | |
| [v at surface = 7 + (–16) × 0.25 = 3] | B1 | Using v = u + aT or equivalent to find v at surface |
| [0 = 3 – gt → t = 0.3] | M1 | Attempt to find the time, t, taken for the particle to travel from the surface |
| Answer | Marks | Guidance |
|---|---|---|
| Total time = T + t = 0.55 s | A1 | |
| Question | Answer | Marks |
| 7(ii) | Alternative method for Q7(ii) | |
| [–0.3g – 1.8 = 0.3a] | M1 | Using Newton’s 2nd law for the upward motion in the tank |
| a = –16 | A1 | |
| v2 = 72 + 2 × (–16) × 1.25 = 9 → v = 3 | B1 | Using constant acceleration equations to find v at the surface |
| Answer | Marks | Guidance |
|---|---|---|
| or 3 = 7 + (–16) × T | M1 | Using s = ½ (u + v) × T or v = u + aT to find the time, T, for the particle |
| Answer | Marks | Guidance |
|---|---|---|
| T = 0.25 | A1 | |
| [0 = 3 – gt → t = 0.3] | M1 | Attempt to find the time, t, taken for the particle to travel from the surface |
| Answer | Marks | Guidance |
|---|---|---|
| Total time = T + t = 0.55 s | A1 | |
| Question | Answer | Marks |
| 7(ii) | Second Alternative method for Q7(ii) | |
| [½ × 0.3 × (72 – v2) = 0.3g × 1.25 + 1.8 × 1.25] | M1 | Work-energy equation for motion from bottom to surface |
| A1 | Correct equation | |
| v = 3 | B1 | Find v at surface from rearrangement of work-energy |
| [1.25 = ½ (7 + 3) × T] | M1 | Using s = ½ (u + v) × T to find the time T, for the particle to travel from |
| Answer | Marks | Guidance |
|---|---|---|
| T = 0.25 | A1 | |
| [0 = 3 – 10t → t = 0.3] | M1 | Attempt to find the time, t, taken for the particle to travel from the surface |
| Answer | Marks |
|---|---|
| Total time = T + t = 0.55 s | A1 |
Question 7:
--- 7(i) ---
7(i) | At liquid surface, speed = 0 + g × 0.8 [= 8]
or
0.3g × ½ (0 + v) × 0.8 = ½ (0.3) v2 → v = 8 | B1 | Using constant acceleration equation v = u + at
or
PE loss = KE gain
PE lost in water = 0.3g × 1.25 [ = 3.75] | B1
[½ × 0.3 × (82 – v2) + 0.3g × 1.25 = 1.2] | M1 | Using work-energy for downward motion in the tank
PE loss + KE loss = Work done against resistance
v = 9 m s–1 | A1
Alternative method for Q7(i)
Height above tank = ½ × g × 0.82 [= 3.2] | B1
Total PE loss = 0.3g × (3.2 + 1.25) [= 13.35] | B1
[0.3g × (3.2 + 1.25) = ½ × 0.3 × v2 + 1.2] | M1 | Work-energy equation for the total downward motion
v = 9 m s–1 | A1
4
Question | Answer | Marks | Guidance
--- 7(ii) ---
7(ii) | [–0.3g – 1.8 = 0.3a] | M1 | Using Newton’s 2nd law for the upward motion in the tank
a = –16 | A1
[1.25 = 7T + ½ × (–16) × T2] | M1 | Using constant acceleration equations to find the time, T, for the particle
to travel from the bottom to the surface of the liquid
T = 0.25 (or 0.625, on the way down) | A1
[v at surface = 7 + (–16) × 0.25 = 3] | B1 | Using v = u + aT or equivalent to find v at surface
[0 = 3 – gt → t = 0.3] | M1 | Attempt to find the time, t, taken for the particle to travel from the surface
to reach maximum height using their v ≠ 7
Total time = T + t = 0.55 s | A1
Question | Answer | Marks | Guidance
7(ii) | Alternative method for Q7(ii)
[–0.3g – 1.8 = 0.3a] | M1 | Using Newton’s 2nd law for the upward motion in the tank
a = –16 | A1
v2 = 72 + 2 × (–16) × 1.25 = 9 → v = 3 | B1 | Using constant acceleration equations to find v at the surface
1.25 = ½ (7 + 3) × T
or 3 = 7 + (–16) × T | M1 | Using s = ½ (u + v) × T or v = u + aT to find the time, T, for the particle
to travel from the bottom to the surface of the liquid
T = 0.25 | A1
[0 = 3 – gt → t = 0.3] | M1 | Attempt to find the time, t, taken for the particle to travel from the surface
to reach maximum height using their v ≠ 7
Total time = T + t = 0.55 s | A1
Question | Answer | Marks | Guidance
7(ii) | Second Alternative method for Q7(ii)
[½ × 0.3 × (72 – v2) = 0.3g × 1.25 + 1.8 × 1.25] | M1 | Work-energy equation for motion from bottom to surface
A1 | Correct equation
v = 3 | B1 | Find v at surface from rearrangement of work-energy
[1.25 = ½ (7 + 3) × T] | M1 | Using s = ½ (u + v) × T to find the time T, for the particle to travel from
the bottom to the surface of the liquid
T = 0.25 | A1
[0 = 3 – 10t → t = 0.3] | M1 | Attempt to find the time, t, taken for the particle to travel from the surface
to reach maximum height using their v ≠ 7
Total time = T + t = 0.55 s | A1
7A particle of mass $0.3$ kg is released from rest above a tank containing water. The particle falls vertically, taking $0.8$ s to reach the water surface. There is no instantaneous change of speed when the particle enters the water. The depth of water in the tank is $1.25$ m. The water exerts a force on the particle resisting its motion. The work done against this resistance force from the instant that the particle enters the water until it reaches the bottom of the tank is $1.2$ J.
\begin{enumerate}[label=(\roman*)]
\item Use an energy method to find the speed of the particle when it reaches the bottom of the tank. [4]
\end{enumerate}
When the particle reaches the bottom of the tank, it bounces back vertically upwards with initial speed $7$ m s$^{-1}$. As the particle rises through the water, it experiences a constant resistance force of $1.8$ N. The particle comes to instantaneous rest $t$ seconds after it bounces on the bottom of the tank.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the value of $t$. [7]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2018 Q7 [11]}}