2022 A/L Physics Past Paper · Question 22 (English Medium)

Answer: (3) $C$

Given information

What we know:

The sheet $PQRS$ is a uniform square of side $a$. Its weight acts at its centre, which is the point $B$ on the dotted line.

The triangle $QBR$ has its base $QR = a$ and its tip at $B$, a distance $\dfrac{a}{2}$ from $QR$.

It is cut off and joined on the other side of $QR$, with its tip now at $B^{\prime}$.

Nothing is lost, so the composite sheet has the same area and the same weight as the square.

What the question asks

We want: the point on the dotted line where the centre of gravity of the composite sheet lies.

What each letter means:

$a$ (the unit of length): the side of the square.

$W$ ($\text{N}$): the weight of the whole sheet.

$w$ ($\text{N}$): the weight of the triangle.

$x_1$, $x_2$ ($a$): the distances of the centre of gravity of the triangle from $PS$, before and after it is moved.

$x_G$ ($a$): the distance of the centre of gravity of the composite sheet from $PS$.

Relevant theory

The centre of gravity of a body is the point at which the resultant force of gravity or weight of the body acts.

For example, think of a seesaw with a child at each end. If one child moves a step outwards, the balance point moves a little towards that child. A small weight that moves only a short way shifts the balance point only a little.

The same happens here. A part of the weight moves to the right, so the centre of gravity moves to the right.

The weight of the whole sheet acts at the centre of gravity $G$. So its moment about $PS$ equals the sum of the moments of the parts. The square keeps its own centre $B$. The triangle leaves $x_1$ and arrives at $x_2$.

$$W\,x_G = W \times \dfrac{a}{2} - w\,x_1 + w\,x_2$$

In words: the moment of the whole equals the moment of the square, take away the triangle where it was, add the triangle where it is now.

The sheet is uniform, so the weight is in proportion to the area. A triangular sheet has its centre of gravity where its medians meet, two thirds of the way from the tip to the base.

Required equations

$W\,x_G = W \times \dfrac{a}{2} - w\,x_1 + w\,x_2$

$x_G = \dfrac{a}{2} + \dfrac{w}{W}\,(x_2 - x_1)$

Applying it to this question

All distances are measured in units of the side $a$. The weight ratio $\dfrac{w}{W}$ has no unit.

$0.25$

$\dfrac{w}{W} = \dfrac{\frac{1}{2} \times a \times \frac{a}{2}}{a^2} = \dfrac{1}{4}$

$0.8333333$

$x_1 = \dfrac{1}{2} + \dfrac{2}{3} \times \dfrac{1}{2} = \dfrac{5}{6}$

$1.1666667$

$x_2 = \dfrac{3}{2} - \dfrac{2}{3} \times \dfrac{1}{2} = \dfrac{7}{6}$

$0.3333333$

$x_2 - x_1 = \dfrac{7}{6} - \dfrac{5}{6} = \dfrac{1}{3}$

$0.0833333$

$\dfrac{w}{W}\,(x_2 - x_1) = \dfrac{1}{4} \times \dfrac{1}{3} = \dfrac{1}{12}$

$0.5833333$

$x_G = \dfrac{1}{2} + \dfrac{1}{12} = \dfrac{7}{12}$

Reasoning and calculation

$B$ is at $\dfrac{1}{2}a$. The centre of gravity is at $\dfrac{7}{12}a$, only $\dfrac{1}{12}a$ to the right of $B$.

$D$ is at $a$ and $E$ is at $\dfrac{7}{6}a$. Both are far beyond. The point just to the right of $B$ is $C$. That is option (3).

Checks

$0.5 + f * (x2 - x1)$

Does it make sense?

Only $\dfrac{1}{4}$ of the weight moved, and it moved only $\dfrac{a}{3}$. So the sheet is nearly the same as the square and its centre of gravity stays close to $B$.

The weight moved to the right, so the shift is to the right. That agrees with our result.

Conclusion

$x_G = \dfrac{7a}{12}$

The centre of gravity is at $x_G = \dfrac{7}{12}a$ from $PS$. That is $\dfrac{a}{12}$ to the right of $B$, at the point $C$.

The answer is (3).

When part of a sheet is moved, the centre of gravity moves towards where the weight went. The shift is the fraction of the weight moved times the distance it moved.

Here only a quarter of the weight moved a short way, so the shift is small.

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