2018 A/L Physics Past Paper · Question 14 (English Medium)

Answer: (1) $\dfrac{v_0^2}{2\mu g}$

What the question asks

We want: how far the object slides before it stops.

Only friction slows it down.

Give the answer using $m$, $v_0$, $\mu$ and $g$.

Relevant theory

The friction acting between two surfaces as they slide over one another is called dynamic friction.

Required equations

$F = \mu R,\quad R = mg$

$F = ma$

$v^2 = u^2 + 2as$

What the letters mean:

$R$ is the push of the ground on the object. On flat ground $R = mg$.

$a$ is the acceleration. It is negative, because the object slows down.

Starting speed: $u = v_0$

Speed at the end: $v = 0$

$s$ is the distance we want.

Reasoning and calculation

$\mu mg = m|a|$

$a = -\mu g$

$0 = v_0^2 - 2\mu g\,s$

$s = \frac{v_0^2}{2\mu g}$

Reading the options

We got $s = \dfrac{v_0^2}{2\mu g}$. That is option (1).

Options (2) and (3) have the 2 in the wrong place.

Options (4) and (5) have no $\mu$ at all.

Checks

Check the units.

$$\frac{[v_0^2]}{[g]} = \frac{L^2 T^{-2}}{L T^{-2}} = L$$

$\mu$ has no unit.

So $s$ is a length. Good.

Check the extremes.

No friction ($\mu \to 0$): the object never stops, so $s \to \infty$. Our answer does this. Good.

No starting speed ($v_0 \to 0$): it does not move, so $s \to 0$. Good.

Options (4) and (5) still give a finite distance with no friction. So they are wrong.

Conclusion

$s = \frac{v_0^2}{2\mu g}$

Friction slows it: $a = -\mu g$.

Then $v^2 = u^2 + 2as$ gives

$$s = \frac{v_0^2}{2\mu g}$$

The mass $m$ cancels out. The answer is option (1).

The mass $m$ is not in the answer.

Why? Friction grows with $m$, and so does the weight. They cancel.

Exam tip

Explore

A kick sets the object off across the frozen pond at speed $v_0$.

Only the friction between the ice and the object holds it back.

So its speed drops by the same amount every second.

When the speed reaches zero it stops, after a distance $s = \dfrac{v_0^2}{2\mu g}$.