2018 A/L Physics Past Paper · Question 13 (English Medium)

Average force on the ant

An ant of mass 2×10−62 \times 10^{-6} kg (2 milligrams), which is stationary at the edge of a horizontal smooth strip is removed in 0.2 s by blowing with mouth. The direction of blowing is horizontal as shown by the arrows in the figure. If the ant is thrown out in the direction of the blowing with a horizontal velocity of 0.5 m s⁻¹, the average force exerted on the ant by the blow is

  1. (1) 5×10−65 \times 10^{-6} N
  2. (2) 1×10−51 \times 10^{-5} N
  3. (3) 2×10−52 \times 10^{-5} N
  4. (4) 1×10−31 \times 10^{-3} N
  5. (5) 5×10−35 \times 10^{-3} N
Answer and explanation

Answer: (1) 5×10−65 \times 10^{-6} N

Given information

We know:

Mass of the ant: m=2×10−6m = 2 \times 10^{-6} kg

At the start it is still: u=0u = 0

Speed at the end: v=0.5 m s−1v = 0.5\ \text{m s}^{-1}

Time of the blow: t=0.2t = 0.2 s

The strip is smooth. So there is no friction.

What the question asks

We want: the average force FF of the blow on the ant.

Relevant theory

The rate of change of moments of an object is proportional to the unbalanced (resultant) force which acts on the object.

Required equations

F=ΔpΔt=m(v−u)tF = \frac{\Delta p}{\Delta t} = \frac{m(v - u)}{t}

Applying it to this question

F=2×10−6×(0.5−0)0.2F = \frac{2\times10^{-6}\times(0.5 - 0)}{0.2}

Reasoning and calculation

1×10−61\times10^{-6}

Δp=2×10−6×0.5\Delta p = 2\times10^{-6}\times0.5

F=1×10−60.2F = \dfrac{1\times10^{-6}}{0.2}

F=12×10−6+1F = \dfrac{1}{2}\times10^{-6+1}

5×10−65\times10^{-6}

FF

Checks

m∗v/tm * v / t

Check the unit.

kg×m s−1s=kg m s−2=N\frac{\text{kg} \times \text{m s}^{-1}}{\text{s}} = \text{kg m s}^{-2} = \text{N}

That is the unit of force. Good.

Does the size make sense?

The ant's weight is mg=2×10−5mg = 2 \times 10^{-5} N.

The blow is 5×10−65 \times 10^{-6} N. That is smaller than its weight.

That fits a gentle blow.

Conclusion

F=5×10−6 NF = 5\times10^{-6}\ \text{N}

F=mvt=2×10−6×0.50.2=5×10−6 NF = \frac{mv}{t} = \frac{2 \times 10^{-6} \times 0.5}{0.2} = 5 \times 10^{-6}\ \text{N}

So the answer is option (1).

Explore

The ant sits still at the edge of the strip.

For 0.20.2 s the air from the mouth pushes it, and it goes faster and faster.

When the blowing stops, the ant leaves the strip at 0.5 m s−10.5\ \text{m s}^{-1}.

Its momentum changed by mvmv; that change divided by 0.20.2 s is the force: 5×10−65 \times 10^{-6} N.

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