2022 A/L Physics Past Paper · Question 18 (English Medium)

Answer: (2) $25\ \text{m s}^{-1}$

Given information

What we know:

The mass of the body is $m = 2\ \text{kg}$.

It starts from rest, so $u = 0$.

The surface has no friction, so the force $F$ in the graph is the only horizontal force.

From the graph: $F = 10\ \text{N}$ from $0$ to $4\ \text{s}$. Then $F$ falls in a straight line to $0$ at $6\ \text{s}$.

What the question asks

We want: the final velocity $v$ of the body at $t = 6\ \text{s}$.

What each letter means:

$m$ ($\text{kg}$): mass of the body.

$u$ ($\text{m s}^{-1}$): the velocity at the start.

$v$ ($\text{m s}^{-1}$): the velocity at the end.

$F$ ($\text{N}$): the horizontal force.

$I$ ($\text{N s}$): the impulse of the force.

Relevant theory

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

For example, push a shopping trolley that is standing still.

Push it harder and it gets faster. Push for longer and it gets faster. What counts is the push multiplied by the time of the push.

$$I = F \times t$$

In words: the impulse is the force times the time it acts.

The book also tells us what the impulse does. It changes the momentum.

$$I = mv - mu$$

In words: the impulse equals the final momentum minus the starting momentum.

When the force changes, we cut the time into very short pieces. In each piece $F \times \Delta t$ is the area of a thin strip under the graph. Adding all the strips gives the whole area.

In words: the area under the $F$–$t$ graph is the impulse.

Required equations

$I = \text{area under the } F\text{–}t \text{ graph}$

$I = mv - mu$

$v = \dfrac{I}{m}\ (u = 0)$

Applying it to this question

The force is in newtons and the time in seconds, so the area comes out in newton seconds. Then $v = \dfrac{I}{m}$ comes out in metres per second.

$40$

$I_1 = 10 \times 4$

$10$

$I_2 = \dfrac{1}{2} \times (6 - 4) \times 10$

$50$

$I = I_1 + I_2 = 40 + 10$

$25$

$v = \dfrac{I}{m} = \dfrac{50}{2}$

Reasoning and calculation

The area under the graph is the impulse, $50\ \text{N s}$.

Dividing by the mass $2\ \text{kg}$ gives the final velocity $25\ \text{m s}^{-1}$. That is option (2).

Checks

$(F * t1 + 0.5 * F * (t2 - t1)) / m$

Does it make sense?

The average force over the $6\ \text{s}$ is $\dfrac{50}{6} \approx 8.3\ \text{N}$. That is a little less than the biggest force $10\ \text{N}$, because the force dies away at the end.

Then the average acceleration is about $4.2\ \text{m s}^{-2}$, and in $6\ \text{s}$ that gives about $25\ \text{m s}^{-1}$.

Conclusion

$v = 25\ \text{m s}^{-1}$

The area under the graph is $I = 50\ \text{N s}$. So $v = \dfrac{I}{m} = \dfrac{50}{2} = 25\ \text{m s}^{-1}$.

The answer is (2).

The area under a force–time graph is the impulse, and the impulse equals the change of momentum.

Do not forget the last step: divide the impulse by the mass.

Exam tip