2020 A/L Physics Past Paper · Question 42 (English Medium)

Answer: (5) Figure (5)

What the question asks

We must find: the graph of the vertical component of velocity $v$ against time $t$, from the jump until the person reaches the ground.

Relevant theory

$\therefore F = ma$

F - Resultant force in newtons

The british scientist George Stokes (1819 — 1903 A. D), considering a spherical body

moving in a liquid presented the relationship between the magnitude of the viscous

frictional force 'F’ and the terminal velocity 'v’ attained by the body by the following

equation.

The resultant force on the sphere becomes zero.

The sphere now begins to move with a uniform velocity.

Required equations

$mg - R = ma$

$R = R(v),\quad R(0) = 0$

$R(v_T) = mg \;\Rightarrow\; a = 0$

Reasoning and calculation

Axes: horizontal axis time $t$, vertical axis the vertical component of velocity $v$ (downward positive).

The slope of the graph is the acceleration $a = g - \dfrac{R}{m}$.

Starting condition: at $t = 0$, $v = 0$, so the resistance $R = 0$ and $a = g$ — the graph starts from the origin with the steepest slope.

Sections: the behaviour changes at two moments — when the parachute opens, and when the ground is reached.

Three parts: before opening, after opening, at the ground.

Before opening

As the speed grows the resistance grows; $a = g - \dfrac{R}{m}$ keeps falling.

The speed keeps rising, but more and more slowly: a downward-curving rise, towards a large terminal speed.

After opening

The resistance is suddenly large: $R > mg$, $a$ is negative, large at first — the speed falls rapidly.

As the speed falls the resistance falls too and the drop slows: an upward-curving fall, down to a new, small terminal speed.

Small terminal speed, then the ground

$R = mg$: $a = 0$ — a horizontal line; on reaching the ground the speed drops to zero at once — a vertical fall.

Boundary values: at the start $v = 0$; at the moment of opening the speed is at its largest (continuous, with no jump); just before the ground, a small constant $v_2 > 0$; at the ground, $v = 0$.

Slope, curvature, sign: before opening the slope is positive and keeps falling (a downward curve).

After opening the slope is negative and its magnitude keeps falling (an upward curve); at the terminal speed the slope is zero.

$v$ is always positive (downward).

Reading the options

Comparing now with the choices: a downward-curving rise, an upward-curving fall, a short horizontal part, a vertical drop at the end — Figure (5).

Why the other options fail

A symmetric arch; no terminal speed.

Lines that look like constant acceleration.

The rise curves upward.

Conclusion

As the resistance grows with the speed the rise curves downward; once opened, resistance exceeds the weight and the speed falls to a small terminal speed; it is zero at the ground — the answer is (5).

Assuming the person rises or stops on opening the parachute.

The speed falls after opening, but still downward — it settles at a small terminal speed where resistance equals weight.