| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2004 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile passing through given point |
| Difficulty | Standard +0.3 This is a standard two-dimensional projectile motion problem requiring simultaneous equations from horizontal and vertical displacement formulas. While it involves multiple steps (setting up equations, solving for two unknowns, then finding speed and direction), the techniques are routine for M2 level with no novel insight required. Slightly above average difficulty due to the algebraic manipulation needed. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2 = VT\sin 35° - 5T^2\) or \(2 = 25\tan 35° - \dfrac{25^2 \times 10}{2V^2\cos^2 35°}\) | B1 | |
| \(25 = VT\cos 35°\) | B1 | |
| For obtaining \(V^2\) or \(T^2\) in \(AV^2 = B\) or \(CT^2 = D\) form where \(A,B,C,D\) are numerical | M1 | |
| \([(25\tan 35° - 2)\cos^2 35°]V^2 = 3125\) (aef) or \(5T^2 = 25\tan 35° - 2\) (aef) | ||
| \(V = 17.3\) or \(T = 1.76\) | A1 | |
| \(T = 1.76\) or \(V = 17.3\) (ft \(VT = 30.519\ldots\)) | B1 ft | 5 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| For using \(\dot{y} = V\sin 35° - gT\) (must be component of \(V\) for M1) | M1 | |
| \(\dot{y}_M\ (= 9.94 - 17.61 = -7.67) < 0 \to\) moving downwards | A1 ft | ft on \(V\) and \(T\) |
| For using \(v_M^2 = (V\cos 35°)^2 + \dot{y}_M^{\ 2}\) | M1 | |
| \(v_M^2 = ((14.20)^2 + (-7.67)^2)\) or for using the principle of conservation of energy \(\left(\tfrac{1}{2}m(v_M^2 - 17.3^2) = -mg \times 2\right)\) | ||
| \(v_M = 16.1\) ms\(^{-1}\) | A1 | 4 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Either: Compare 25 with \(\tfrac{1}{2}R\!\left(\tfrac{\frac{1}{2}v^2\sin 70°}{g}\right)\) | M1 | |
| \(25 > 14.1 \to\) moving downwards | A1 | |
| Or: Compare 1.76 with time to greatest height \(\left(\dfrac{V\sin 35°}{g}\right)\) | M1 | |
| \(1.76 > 0.994 \to\) moving downwards | A1 | |
| Or: \(\dfrac{dy}{dx} = \tan 35° - \dfrac{g \cdot 10}{V^2\cos^2 35°}\ (= -0.54)\) used | M1 | |
| As \(\tan\phi\) is negative \(\to\) moving downwards | A1 |
## Question 6:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2 = VT\sin 35° - 5T^2$ or $2 = 25\tan 35° - \dfrac{25^2 \times 10}{2V^2\cos^2 35°}$ | B1 | |
| $25 = VT\cos 35°$ | B1 | |
| For obtaining $V^2$ or $T^2$ in $AV^2 = B$ or $CT^2 = D$ form where $A,B,C,D$ are numerical | M1 | |
| $[(25\tan 35° - 2)\cos^2 35°]V^2 = 3125$ (aef) or $5T^2 = 25\tan 35° - 2$ (aef) | | |
| $V = 17.3$ or $T = 1.76$ | A1 | |
| $T = 1.76$ or $V = 17.3$ (ft $VT = 30.519\ldots$) | B1 ft | **5 marks total** |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| For using $\dot{y} = V\sin 35° - gT$ (must be component of $V$ for M1) | M1 | |
| $\dot{y}_M\ (= 9.94 - 17.61 = -7.67) < 0 \to$ moving downwards | A1 ft | ft on $V$ and $T$ |
| For using $v_M^2 = (V\cos 35°)^2 + \dot{y}_M^{\ 2}$ | M1 | |
| $v_M^2 = ((14.20)^2 + (-7.67)^2)$ or for using the principle of conservation of energy $\left(\tfrac{1}{2}m(v_M^2 - 17.3^2) = -mg \times 2\right)$ | | |
| $v_M = 16.1$ ms$^{-1}$ | A1 | **4 marks total** |
**Lines 1 and 2 Alternative Methods:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| **Either:** Compare 25 with $\tfrac{1}{2}R\!\left(\tfrac{\frac{1}{2}v^2\sin 70°}{g}\right)$ | M1 | |
| $25 > 14.1 \to$ moving downwards | A1 | |
| **Or:** Compare 1.76 with time to greatest height $\left(\dfrac{V\sin 35°}{g}\right)$ | M1 | |
| $1.76 > 0.994 \to$ moving downwards | A1 | |
| **Or:** $\dfrac{dy}{dx} = \tan 35° - \dfrac{g \cdot 10}{V^2\cos^2 35°}\ (= -0.54)$ used | M1 | |
| As $\tan\phi$ is negative $\to$ moving downwards | A1 | |
---6 A particle is projected from a point $O$ on horizontal ground. The velocity of projection has magnitude $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and direction upwards at $35 ^ { \circ }$ to the horizontal. The particle passes through the point $M$ at time $T$ seconds after the instant of projection. The point $M$ is 2 m above the ground and at a horizontal distance of 25 m from $O$.\\
(i) Find the values of $V$ and $T$.\\
(ii) Find the speed of the particle as it passes through $M$ and determine whether it is moving upwards or downwards.
\hfill \mbox{\textit{CAIE M2 2004 Q6 [9]}}