2020 A/L Physics Past Paper · Question 30 (English Medium)

Answer: (2) only at $t = 70\ \text{s}$

What the question asks

We must find: at what times the body's velocity $v$ returns to zero, after starting from rest.

Reading the options

The options are of two kinds: the times where the force is zero ($40$ s, $100$ s), the interval where the force is constant ($60$–$80$ s), and $70$ s, $120$ s.

The trap here is that zero force is not zero velocity.

Relevant theory

momentum = mass × velocity

Impulse = Impulsive force x Time of action of the force

The quantity on the left-hand side is called the impulse, the product of force and time, and is measured in N s. The quantity on the right-hand side is the change of momentum produced by this impulse.

Impulse = Change in momentum

Required equations

$\Delta I = F\,\Delta t \;(\text{one strip of area under the graph})$

$mv - 0 = \sum F\,\Delta t = \text{area}_{0 \to t}$

$v = 0 \iff \text{area}_{0 \to t} = 0$

Reasoning and calculation

Axes: horizontal axis time $t$ (s), vertical axis force $F$ (N).

Each small strip is $F \times \Delta t$ — a small impulse (N s).

Their sum, that is the area under the graph (negative below the axis), is the total momentum change so far.

$160$

$I_1 = \tfrac{1}{2} \times 40 \times 8$

$-80$

$I_2 = -\tfrac{1}{2} \times 20 \times 8$

$I_3 = -8 \times 10$

$0$

$I_1 + I_2 + I_3 = 160 - 80 - 80$

Since the body starts from rest ($u = 0$), at any instant $mv =$ the area so far.

So the velocity becomes zero not when the force is zero.

It becomes zero only when the area returns to zero.

$-8 \times 10$

$-\tfrac{1}{2} \times 20 \times 8$

$80$

$+\tfrac{1}{2} \times 20 \times 8$

$0 - 80 - 80 + 80$

After $70$ s, the total area falls to $-160$ N s by $100$ s.

It becomes $-80$ N s by $120$ s — it never reaches zero again.

So after $70$ s the body keeps moving in the opposite direction.

Checks

Unit check: area $\text{N} \times \text{s} = \text{kg m s}^{-1}$ — the unit of momentum ✓.

Dividing it by $m$ (kg) gives $\text{m s}^{-1}$.

$0.5 * 40 * F - 0.5 * 20 * F - F * (t - 60)$

Second way (symmetry): from $40$ s, the force line falls with slope $-0.4t$, reaching $-8$ N at $60$ s.

The negative area of $40$–$60$ s ($80$ N s) is half the area of $0$–$40$ s.

The remaining $80$ N s is cancelled by the $-8$ N force in $80 \div 8 = 10$ s — giving $60 + 10 = 70$ s.

Conclusion

Velocity $= \dfrac{\text{the } F\text{–}t \text{ area so far}}{m}$; that area returns to zero only at $t = 70$ s — the answer is (2).

Thinking the body stops at the instants the force is zero ($40$ s, $100$ s).

Zero force means only zero acceleration. The velocity is decided by the total impulse gained so far (the area under the graph).