| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2018 |
| Session | March |
| Marks | 9 |
| Topic | Oblique and successive collisions |
| Type | Particle bouncing on inclined plane |
| Difficulty | Standard +0.8 This is a 2D collision problem requiring resolution of velocities in normal and tangential directions, application of coefficient of restitution, incorporation of frictional impulse, and energy loss calculation. While the techniques are standard for Further Mechanics, the combination of restitution, friction impulse, and the need to carefully track both components makes this moderately challenging, requiring more steps and care than typical A-level questions but not requiring novel insight. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.03i Coefficient of restitution: e6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| Components of initial vel: \(6.5 \cos 30° \downarrow\), \(6.5 \sin 30° \rightarrow\) | M1 | For correct use of either component |
| Upwards component of final vel: \(v_V = \frac{4}{5} \times 6.5 \cos 30°\) | M1 | |
| \(-2 = 2.5v_H - 2.5 \times 6.5 \sin 30°\) | M1 | Use of impulse-momentum; must involve horizontal components only |
| \(v_H = 2.45\) | A1 | |
| \(v = \sqrt{v_V^2 + v_H^2}\), \(\tan \theta = \frac{v_H}{v_V}\), oe | M1 | For use of either trig or Pythagoras |
| Velocity has magnitude 4.48 m s\(^{-1}\) at 33.1° to the normal | A1 [7] | Condone 33.2° |
| Answer | Marks | Guidance |
|---|---|---|
| Percentage loss is \(\frac{\frac{1}{2} \times 2.5 \times (6.5^2 - 4.482^2)}{\frac{1}{2} \times 2.5 \times 6.5^2} \times 100\) | M1 | Factors of \(\frac{1}{2} \times 6.5\) may be omitted |
| = 52.5 | A1 [2] | Condone answers 52 and 53 |
## (i)
Components of initial vel: $6.5 \cos 30° \downarrow$, $6.5 \sin 30° \rightarrow$ | M1 | For correct use of either component
Upwards component of final vel: $v_V = \frac{4}{5} \times 6.5 \cos 30°$ | M1 |
$-2 = 2.5v_H - 2.5 \times 6.5 \sin 30°$ | M1 | Use of impulse-momentum; must involve horizontal components only
$v_H = 2.45$ | A1 |
$v = \sqrt{v_V^2 + v_H^2}$, $\tan \theta = \frac{v_H}{v_V}$, oe | M1 | For use of either trig or Pythagoras
Velocity has magnitude 4.48 m s$^{-1}$ at 33.1° to the normal | A1 [7] | Condone 33.2°
## (ii)
Percentage loss is $\frac{\frac{1}{2} \times 2.5 \times (6.5^2 - 4.482^2)}{\frac{1}{2} \times 2.5 \times 6.5^2} \times 100$ | M1 | Factors of $\frac{1}{2} \times 6.5$ may be omitted
= 52.5 | A1 [2] | Condone answers 52 and 53
---6 A particle $P$ of mass 2.5 kg strikes a rough horizontal plane. Immediately before $P$ strikes the plane it has a speed of $6.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and its direction of motion makes an angle of $30 ^ { \circ }$ with the normal to the plane at the point of impact. The impact may be assumed to occur instantaneously. The coefficient of restitution between $P$ and the plane is $\frac { 2 } { 3 }$. The friction causes a horizontal impulse of magnitude 2 Ns to be applied to $P$ in the plane in which it is moving.\\
(i) Calculate the velocity of $P$ immediately after the impact with the plane.\\
(ii) $\quad P$ loses about $x \%$ of its kinetic energy as a result of the impact. Find the value of $x$.
\hfill \mbox{\textit{OCR Further Mechanics 2018 Q6 [9]}}