| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Vertical drop and bounce |
| Difficulty | Standard +0.3 This is a standard M2 mechanics question involving routine application of SUVAT equations, coefficient of restitution formula, impulse-momentum theorem, and energy calculations. All steps follow textbook procedures with no novel problem-solving required, making it slightly easier than average for A-level. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.03f Impulse-momentum: relation6.03j Perfectly elastic/inelastic: collisions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v^2 = 5^2 + 2g(1.6)\) | B1 | Complete method to find \(v\ (= 7.507\ldots)\) |
| \(0.7 \times 7.507\ldots\ (= 5.255\ldots)\) | B1 | \(0.7 \times \text{cv}(v)\), but not \(\text{cv}(v) = 5\); may be seen in (ii) |
| \((0.7 \times \text{cv}(v))^2 = 2gh\) | M1 | Complete method to find \(h\), with final speed 0; allow \(\text{cv}(v) = 5\) for method |
| \(h = 1.41 \text{ m}\) | A1 | Exact 1.409 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| M1 | Change in momentum found, with relevant velocities ie \(\text{cv}(v)\) and \(0.7 \times \text{cv}(v)\) but not \(\text{cv}(v) = 5\) | |
| \(0.2(7.507\ldots)(0.7) - (-0.2)(7.507\ldots)\) | A1ft | This may be negative; ft on their \(v\) found in (i) |
| Impulse \(= 2.55 \text{ N s}\), upwards | A1 | \((2.5524\ldots)\) Must have direction also. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.2(9.8)(1.6) + \frac{1}{2}(0.2)(5^2) - 0.2(9.8)(\text{cv}(h))\) | M1 | Change in energy found, all energy terms needed and no extra terms. This may be negative |
| OR \(\frac{1}{2}(0.2)(7.507\ldots^2) - \frac{1}{2}(0.2)(0.7 \times 7.507\ldots)^2\) | A1ft | |
| Loss of energy \(= 2.87 \text{ J}\) | A1 | \((2.87436\) exact); art 2.87; allow \(-2.87\) |
## Question 5:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v^2 = 5^2 + 2g(1.6)$ | B1 | Complete method to find $v\ (= 7.507\ldots)$ |
| $0.7 \times 7.507\ldots\ (= 5.255\ldots)$ | B1 | $0.7 \times \text{cv}(v)$, **but not** $\text{cv}(v) = 5$; may be seen in (ii) |
| $(0.7 \times \text{cv}(v))^2 = 2gh$ | M1 | Complete method to find $h$, with final speed 0; **allow** $\text{cv}(v) = 5$ for method |
| $h = 1.41 \text{ m}$ | A1 | Exact 1.409 |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| | M1 | Change in momentum found, with relevant velocities ie $\text{cv}(v)$ and $0.7 \times \text{cv}(v)$ but not $\text{cv}(v) = 5$ |
| $0.2(7.507\ldots)(0.7) - (-0.2)(7.507\ldots)$ | A1ft | This may be negative; ft on their $v$ found in (i) |
| Impulse $= 2.55 \text{ N s}$, upwards | A1 | $(2.5524\ldots)$ Must have direction also. |
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.2(9.8)(1.6) + \frac{1}{2}(0.2)(5^2) - 0.2(9.8)(\text{cv}(h))$ | M1 | Change in energy found, all energy terms needed and no extra terms. This may be negative |
| OR $\frac{1}{2}(0.2)(7.507\ldots^2) - \frac{1}{2}(0.2)(0.7 \times 7.507\ldots)^2$ | A1ft | |
| Loss of energy $= 2.87 \text{ J}$ | A1 | $(2.87436$ exact); art 2.87; allow $-2.87$ |
---5 A small sphere of mass 0.2 kg is projected vertically downwards with a speed of $5 \mathrm {~ms} ^ { - 1 }$ from a height of 1.6 m above horizontal ground. It hits the ground and rebounds vertically upwards coming to instantaneous rest at a height of $h \mathrm {~m}$ above the ground. The coefficient of restitution between the sphere and the ground is 0.7 .\\
(i) Find $h$.\\
(ii) Find the magnitude and direction of the impulse exerted on the sphere by the ground.\\
(iii) Find the loss of energy of the sphere between the instant of projection and the instant it comes to instantaneous rest at height $h \mathrm {~m}$.
\hfill \mbox{\textit{OCR M2 2015 Q5 [10]}}