| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/3 (Pre-U Mathematics Paper 3) |
| Year | 2018 |
| Session | June |
| Marks | 5 |
| Topic | Projectiles |
| Type | Speed at specific time or position |
| Difficulty | Moderate -0.8 This is a straightforward projectile motion question requiring only the application of standard kinematic equations to find horizontal and vertical velocity components at a given time, then combining them using Pythagoras and trigonometry. It's a routine calculation with no problem-solving insight required, making it easier than average but not trivial since it involves multiple steps and vector resolution. |
| Spec | 3.02e Two-dimensional constant acceleration: with vectors3.02i Projectile motion: constant acceleration model |
**Question 7**
Horizontal component is $8\cos40°$ **B1** (6.128355545)
Vertical component $= 8\sin40° - 10 \times 0.4$ **B1** (1.142300877)
Use Pythagoras or trigonometry **M1** Using *their* horizontal and vertical components
Speed $= 6.23\ \text{ms}^{-1}$ **A1**
Direction $= 10.6°$ above the horizontal **A1** May be clarified on a diagram, need either direction arrows on components or direction arrow on resultant7 A particle is projected with a speed of $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, at an angle of $40 ^ { \circ }$ above the horizontal. Find the speed and direction of motion of the particle 0.4 s after projection.
\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2018 Q7 [5]}}