9231 · 3.1
Motion of a projectile flashcards
Revision flashcards for Cambridge 9231 Motion of a projectile (syllabus 3.1). Flip, recall, then mark a real past-paper question.
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What is the key assumption made when modelling projectile motion?
Air resistance is negligible, and the acceleration due to gravity, $g$, is constant.
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What is the acceleration of a projectile in the horizontal direction?
$a_x = 0$. This means the horizontal component of velocity, $v_x$, is constant throughout the motion.
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What is the acceleration of a projectile in the vertical direction (taking upwards as positive)?
$a_y = -g$, where $g \approx 9.8 \, \text{m s}^{-2}$.
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What is the vertical component of velocity, $v_y$, at the projectile's maximum height?
$v_y = 0$. The projectile is momentarily at rest in the vertical direction as it changes from moving upwards to moving downwards.
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For a projectile launched and landing on level ground, what is its final vertical displacement, $s_y$?
$s_y = 0$. It returns to its initial vertical height.
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How do you find the time of flight for a projectile on level ground?
Set the vertical displacement $s_y = 0$ in the equation $s_y = u_y t + \frac{1}{2}a_y t^2$ and solve for $t$. The non-zero solution is the time of flight.
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How do you find the horizontal range, $R$, of a projectile?
First, find the total time of flight, $T$. Then, use the horizontal motion equation: $R = u_x T$.
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When analysing motion on a plane inclined at an angle $\alpha$ to the horizontal, what are the components of acceleration?
Parallel to the plane (downwards): $g \sin \alpha$. Perpendicular to the plane (into the plane): $g \cos \alpha$.
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What is a common mistake when dealing with projectiles on an inclined plane?
Forgetting to resolve the acceleration due to gravity ($g$) into components parallel and perpendicular to the plane. Using $-g$ for the 'vertical' acceleration is incorrect in this rotated coordinate system.
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How do you find the time of flight for a projectile on an inclined plane?
Set the displacement *perpendicular* to the plane to zero and solve for $t$ using the appropriate SUVAT equation with acceleration $a_\perp = -g \cos \alpha$.
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What is the 'range' on an inclined plane?
It is the distance from the point of projection to the point of landing, measured *along the surface of the plane*.