2024 A/L Physics Past Paper · Question 03 (English Medium)

Answer: (5) $60^{\circ}$

Given information

What we know:

At the maximum height the kinetic energy is $\dfrac{1}{4}$ of the initial kinetic energy.

The air resistance is neglected.

What the question asks

We want: the angle of projection $\theta$.

What each letter means:

$u$ ($\text{m s}^{-1}$): the speed of projection.

$\theta$: the angle of projection with the horizontal.

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

$K_0$ and $K$ ($\text{J}$): the kinetic energy at the start and at the top.

Relevant theory

Considering the effect on the horizontal component, $\rightarrow u\cos\theta$, Since horizontally, Acceleration = 0 $v = u + at$ $v_1 = u\cos\theta$

Horizontal component remains constant.

Vertical component of velocity decreases.

This decreases until it reaches zero.

Translational kinetic energy of the body, $$W = \frac{1}{2}mv^2$$

The first quote says that the horizontal component of the velocity stays at $u\cos\theta$. The second says that the vertical component falls to zero at the top. So at the maximum height the speed is $u\cos\theta$. The third quote gives the kinetic energy, $\dfrac{1}{2}mv^2$.

For example, a stone thrown over a wall rises, slows down and reaches the top. At the top it is not going up or down any more. It is moving sideways, and this sideways speed is the same as it had at the start.

Required equations

$K_0 = \dfrac{1}{2}mu^2$

$v = u\cos\theta$

$K = \dfrac{1}{2}m(u\cos\theta)^2$

$\dfrac{K}{K_0} = \cos^2\theta$

Applying it to this question

Divide $K$ by $K_0$. The factor $\dfrac{1}{2}mu^2$ is in both, so it cancels.

$$\dfrac{K}{K_0} = \cos^2\theta$$

$0.25$

$\cos^2\theta = \dfrac{1}{4}$

$0.5$

$\cos\theta = \sqrt{\dfrac{1}{4}}$

$60$

$\theta$

Reasoning and calculation

$cos(t * pi / 180) ^ 2$

Now match it with the five options.

Option (5) is $60^{\circ}$. It matches.

The others give a ratio $\cos^2\theta$ of about $0.97$, $0.88$, $\dfrac{3}{4}$ and $\dfrac{1}{2}$. None of them is $\dfrac{1}{4}$.

Checks

Check the units. $K$ and $K_0$ are both in joules, so their ratio is a plain number, and $\cos^2\theta$ is a plain number too.

Does it make sense?

If the ball is thrown straight up, $\theta = 90^{\circ}$, it stops at the top and its kinetic energy there is $0$. If it is thrown flat, $\theta = 0^{\circ}$, it keeps the whole of its kinetic energy.

$60^{\circ}$ lies between these two, and it keeps a quarter. That fits.

Conclusion

$\theta = 60^{\circ}$

The angle of projection is $\theta = 60^{\circ}$.

The answer is (5).

At the top of the path just the horizontal speed $u\cos\theta$ is left. So the kinetic energy there is $\cos^2\theta$ times the first.

Set this equal to the given fraction and solve for $\theta$.

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