2023 A/L Physics Past Paper · Question 41 (English Medium)

Answer: (5) $2g$

What the question asks

We want: the downward acceleration $a_A$ of the upper block, just after the string breaks.

What each letter means:

$m$ ($\text{kg}$): the mass of each block. The two blocks are identical.

$g$ ($\text{m s}^{-2}$): the acceleration of gravity.

$F_s$ ($\text{N}$): the force of the spring.

$a_A$ ($\text{m s}^{-2}$): the downward acceleration of the block $A$, just after the string breaks.

Relevant theory

Newton's first law states that a resultant force can change the state of motion of an object. Newton's second law states about this resultant force.

where k is the constant of proportionality and is called the force constant of the wire. The

The first quote gives Newton's second law, $F = ma$: the resultant force on a body gives it an acceleration. The second quote says that the force of a stretched wire or spring grows with its extension, $F = ke$.

For example, hold a stretched elastic band with a bag hanging at its lower end. The band pulls the bag up and pulls your hand down with the same force.

Let go of your hand. The band cannot shrink in an instant, so for that instant it still pulls the same.

Required equations

$F_s = mg$

$mg + F_s = m a_A$

$mg + mg = m a_A$

$a_A = 2g$

Reasoning and calculation

Before the string breaks:

The block $B$ hangs at rest. Two forces act on it: its weight $mg$ down, and the pull of the spring $F_s$ up. They balance, so $F_s = mg$.

The spring pulls $B$ up. It pulls $A$ down with the same force, $mg$.

$mg$

$F_s = mg$

Just after the string breaks:

The string pull on $A$ is gone. The spring has not had time to change its length, so it still pulls with $F_s = mg$.

Now the block $A$ has two forces, both downward: its weight $mg$ and the spring pull $mg$. Newton's second law gives $mg + mg = m a_A$.

$2mg = m a_A$

$mg + mg = m a_A$

$2g$

$a_A = \dfrac{2mg}{m}$

$10$

$F_s = 1 \times 10$

$20$

$mg + F_s = 10 + 10$

$a_A = \dfrac{20}{1}$

$(m * g + m * g) / m$

Reading the options

Now compare $2g$ with the five options.

Option (5) is $2g$. It matches.

Option (1) is zero, option (2) is $\dfrac{g}{2}$, option (3) is $g$ and option (4) is $\sqrt{2}\,g$. None of these is $2g$.

Checks

Check the units. In $mg + F_s = m a_A$, each term is a force, in $\text{N}$. Divide by $m$ and each term is an acceleration, in $\text{m s}^{-2}$.

So $a_A = 2g$ is in $\text{m s}^{-2}$, as an acceleration must be.

Does it make sense?

Think of the block $B$. Just after the break the spring pulls it up with $mg$ and its weight pulls it down with $mg$. They balance, so $B$ has no acceleration for that instant.

If the spring were not there, the block $A$ would fall with $g$. The spring pull adds one more $g$, so $2g$ is a fair size.

Conclusion

$a_A = 2g$

Just after the string breaks, the downward acceleration of the block $A$ is $2g$.

The answer is (5).

The quotes give Newton's second law and the force of a stretched spring. The steps that the spring keeps its pull at the instant the string breaks, and that the block $B$ is at rest before it, are our reasoning from them.

Teacher's note

Draw the forces on the block you want, just after the change. The spring keeps its pull for that instant. Add the forces on the one line, then use $F = ma$.

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