Paper 4 Projectile Motion Answers

Q1 Horizontal range investigation

1. Squaring gives

$$x^2=u^2\frac{2h}{g}=\frac{2u^2}{g}h.$$

Thus $x^2$ is proportional to $h$, with gradient $2u^2/g$. [2] 2. If gradient is $G$, then $G=2u^2/g$, so

$$u=\sqrt{\frac{Gg}{2}}.$$

[2] 3. Use carbon paper or a paper target to mark the first impact point; repeat several launches at each height and average the measured range; measure from the point vertically below the launch point to the centre of the impact marks. [2] 4. It keeps $u$ constant so that changes in $x$ are due to changes in $h$. [1] 5. Keep the landing area clear, prevent the ball rolling into walkways, or wear eye protection if the launcher is spring-loaded. [1]

Q2 Video analysis of a projectile

1. Dividing by $t$ gives $y/t=u_y-\frac{1}{2}gt$ [1]. This has the form $Y=c+mt$ [1], with vertical intercept $u_y$ and gradient $-g/2$ [1].

$t/\mathrm{s}$	0.10	0.20	0.30	0.40	0.50
$(y/t)/\mathrm{m}{\mathrm{s}}^{-1}$	2.44	1.95	1.46	0.970	0.480

Correct values [1]; sensible precision and heading unit [1].

3. Award axes/units [1], sensible scales [1], accurate plots [2], and one straight best-fit line [1].

4. The gradient is approximately $-4.90\mathrm{m}{\mathrm{s}}^{-2}$ [1]. Since gradient $=-g/2$, $g=-2\times \text{gradient}\approx 9.80\mathrm{m}{\mathrm{s}}^{-2}$ [2]. Apply follow-through from the graph.
5. Accept an experiment-specific source with effect, for example: camera not perpendicular causes position-dependent parallax; an incorrect calibration length scales all $y$ values and hence the gradient; an incorrect frame rate scales all time values. [2]