| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Two projectiles meeting - 2D flight |
| Difficulty | Challenging +1.8 This is a challenging M2 projectile collision problem requiring multiple sophisticated steps: finding collision point using the constraint that both particles travel horizontally simultaneously, analyzing an oblique collision with coefficient of restitution, tracking post-collision motion with a rebound, and finally calculating position at a specific angle. The problem integrates projectile motion, impulse-momentum, and collision theory across three interconnected parts, requiring sustained problem-solving over 17 marks with non-trivial geometric and algebraic manipulation. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model6.03b Conservation of momentum: 1D two particles6.03e Impulse: by a force6.03f Impulse-momentum: relation6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(32\left(\frac{3}{5}\right)\right)^2 - 2(9.8)h = 0\) | M1 | Use of \(v^2 = u^2 + 2as\) with \(v = 0\) (or \((20(24/25))^2 - 2(9.8)h = 0\)) |
| \(h = 18.8 \text{ m}\) | A1 | \(h = 18.808163...\) or \(\frac{4608}{245}\) |
| \(t = \frac{32\left(\frac{3}{5}\right)}{9.8} (= 1.96)\) | B1 | \(1.9591836...\), \(\frac{96}{49}\) |
| \(AB = \left(32\left(\frac{4}{5}\right) + 20\left(\frac{7}{25}\right)\right)t = 61.1 \text{ m}\) | B1 | \(61.126530...\) or \(\frac{14976}{245}\) |
| [4] | ||
| OR | ||
| For last two marks may use range formula | B1 | For correct expression for finding half the range for one particle (50.155... or 10.971...) |
| \(61.1\text{m}\) | B1 | \(61.126530...\) or \(\frac{14976}{245}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Speed of \(P\) at impact \(\sqrt{2(9.8)\left(\frac{4608}{245}\right)}\) | B1 ft | cv(h) from (i); (19.2) |
| Speed of \(P\) after impact \(\sqrt{2(9.8)(5)}\) | B1 | \(7\sqrt{2} = 9.899494937\) |
| \(I = 3\left(7\sqrt{2} - \left(-\frac{96}{5}\right)\right)\) | M1 | Attempt at Impulse = change in momentum |
| \(I = 87.3 \text{ N s, upwards}\) | A1 [4] | Must be positive; 87.298484... ;must have direction stated. |
| Answer | Marks | Guidance |
|---|---|---|
| M1* | Attempt at use of coefficient of restitution, component of 32 and 20 needed; | |
| \(0 - v_B = -\frac{1}{2}\left(32\left(\frac{4}{5}\right) - \left(-20\left(\frac{7}{25}\right)\right)\right)\) | ||
| \(v_B = 15.6\) | A1 | \(\frac{78}{5}\) |
| Vertical component of speed \(v_B \tan 25\) | B1* | 7.2743399467 (soi); |
| \((v_B \tan 25)^2 = 2(9.8)y\) | M1* | Use of \(v^2 = u^2 + 2as\) vertically with \(u = 0\) to find vertical displacement with their vertical speed component |
| \(y = 2.70\) | A1 | 2.6998412... |
| \(t = \frac{v_B \tan 25}{9.8}\) | B1* | \(t = 0.74228565...\) |
| \(x = v_B t\) | B1* | \(x = 11.5796562...\) |
| \(QC = \sqrt{x^2 + y^2}\) | M1 dep* | |
| \(QC = 11.9 \text{ m}\) | A1 [9] | 11.890230... |
## (i)
$\left(32\left(\frac{3}{5}\right)\right)^2 - 2(9.8)h = 0$ | M1 | Use of $v^2 = u^2 + 2as$ with $v = 0$ (or $(20(24/25))^2 - 2(9.8)h = 0$)
$h = 18.8 \text{ m}$ | A1 | $h = 18.808163...$ or $\frac{4608}{245}$
$t = \frac{32\left(\frac{3}{5}\right)}{9.8} (= 1.96)$ | B1 | $1.9591836...$, $\frac{96}{49}$
$AB = \left(32\left(\frac{4}{5}\right) + 20\left(\frac{7}{25}\right)\right)t = 61.1 \text{ m}$ | B1 | $61.126530...$ or $\frac{14976}{245}$
[4] | |
OR | |
For last two marks may use range formula | B1 | For correct expression for finding half the range for one particle (50.155... or 10.971...)
$61.1\text{m}$ | B1 | $61.126530...$ or $\frac{14976}{245}$
## (ii)
Speed of $P$ at impact $\sqrt{2(9.8)\left(\frac{4608}{245}\right)}$ | B1 ft | cv(h) from (i); (19.2)
Speed of $P$ after impact $\sqrt{2(9.8)(5)}$ | B1 | $7\sqrt{2} = 9.899494937$
$I = 3\left(7\sqrt{2} - \left(-\frac{96}{5}\right)\right)$ | M1 | Attempt at Impulse = change in momentum
$I = 87.3 \text{ N s, upwards}$ | A1 [4] | Must be positive; 87.298484... ;must have direction stated.
## (iii)
| M1* | Attempt at use of coefficient of restitution, component of 32 and 20 needed;
$0 - v_B = -\frac{1}{2}\left(32\left(\frac{4}{5}\right) - \left(-20\left(\frac{7}{25}\right)\right)\right)$ | |
$v_B = 15.6$ | A1 | $\frac{78}{5}$
Vertical component of speed $v_B \tan 25$ | B1* | 7.2743399467 (soi);
$(v_B \tan 25)^2 = 2(9.8)y$ | M1* | Use of $v^2 = u^2 + 2as$ vertically with $u = 0$ to find vertical displacement with their vertical speed component
$y = 2.70$ | A1 | 2.6998412...
$t = \frac{v_B \tan 25}{9.8}$ | B1* | $t = 0.74228565...$
$x = v_B t$ | B1* | $x = 11.5796562...$
$QC = \sqrt{x^2 + y^2}$ | M1 dep* |
$QC = 11.9 \text{ m}$ | A1 [9] | 11.890230...
Allow wrong $v_B$ for max M1 A0 B1 M1 A0 B1 M1 A0
If $v_B$ "found" not using restitution then M0 A0 B1 M1 A0 B1 B1 M1 A0 max
---A particle $P$ is projected with speed $32 \text{ m s}^{-1}$ at an angle of elevation $\alpha$, where $\sin \alpha = \frac{3}{4}$, from a point $A$ on horizontal ground. At the same instant a particle $Q$ is projected with speed $20 \text{ m s}^{-1}$ at an angle of elevation $\beta$, where $\sin \beta = \frac{24}{25}$, from a point $B$ on the same horizontal ground. The particles move freely under gravity in the same vertical plane and collide with each other at the point $C$ at the instant when they are travelling horizontally (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Calculate the height of $C$ above the ground and the distance $AB$. [4]
\end{enumerate}
Immediately after the collision $P$ falls vertically. $P$ hits the ground and rebounds vertically upwards, coming to instantaneous rest at a height 5 m above the ground.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Given that the mass of $P$ is 3 kg, find the magnitude and direction of the impulse exerted on $P$ by the ground. [4]
\end{enumerate}
The coefficient of restitution between the two particles is $\frac{1}{2}$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Find the distance of $Q$ from $C$ at the instant when $Q$ is travelling in a direction of $25°$ below the horizontal. [9]
\end{enumerate}
\hfill \mbox{\textit{OCR M2 2016 Q7 [17]}}