2023 A/L Physics Past Paper · Question 17 (English Medium)

Answer: (4) $\dfrac{K}{4}$

Given information

What we know:

The ball is thrown up at an angle of $\theta = 60^\circ$ with the horizontal.

Its kinetic energy at the start is $K$.

We neglect air resistance. So the force on the ball is its weight, which pulls straight down.

What the question asks

We want: the kinetic energy $K_{\text{top}}$ of the ball at its maximum height, in terms of $K$.

What each letter means:

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

$u$ ($\text{m s}^{-1}$): the speed of the ball when it is thrown.

$\theta$ ($^\circ$): the angle of the throw with the horizontal.

$v_{\text{top}}$ ($\text{m s}^{-1}$): the speed of the ball at its maximum height.

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

$K_{\text{top}}$ ($\text{J}$): the kinetic energy at the maximum height.

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 nothing pushes the ball sideways, so its speed along the ground stays the same. The second quote says the upward speed gets smaller until it is zero. That moment is the maximum height. The third quote gives the kinetic energy of a moving body.

So at the top, the ball is not still. It has lost its upward speed, but it still has its sideways speed.

For example, think of a cricket ball hit high into the air. At the very top it stops going up. But it keeps drifting across the field towards the fielder.

That sideways drift is the motion it has at the top. Its energy there is the energy of that sideways speed alone.

Required equations

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

$v_{\text{top}} = u\cos\theta$

$K_{\text{top}} = \dfrac{1}{2}mv_{\text{top}}^2$

$K_{\text{top}} = \dfrac{1}{2}mu^2\cos^2\theta = K\cos^2\theta$

Applying it to this question

Both energies are in $\text{J}$ and the answer is a fraction of $K$, so no change of units is needed.

$0.5u$

$v_{\text{top}} = u\cos 60^\circ = u \times 0.5$

$0.25u^2$

$v_{\text{top}}^2 = (0.5u)^2$

$0.25 \times \dfrac{1}{2}mu^2$

$K_{\text{top}} = \dfrac{1}{2}m \times 0.25u^2$

$0.25K$

$K_{\text{top}} = 0.25K$

$K/4$

$K_{\text{top}} = \dfrac{K}{4}$

Reasoning and calculation

At the top, half of the launch speed is left. Half of the speed means a quarter of the energy, because the energy goes with the speed squared.

So the ball has $\dfrac{K}{4}$ at the top. That is option (4).

Checks

$c * c$

The unit: $K_{\text{top}}$ is a fraction of $K$, so it is in $\text{J}$, as an energy should be.

Does it make sense?

The ball is still moving at the top, so $K_{\text{top}}$ cannot be $0$. It has lost some energy to height, so it cannot be $K$ either.

Check the two ends. A throw straight up ($\theta = 90^\circ$) gives $\cos 90^\circ = 0$ and no energy at the top. A throw along the ground ($\theta = 0^\circ$) gives $\cos 0^\circ = 1$ and the full $K$. Our angle $60^\circ$ is between them, and $\dfrac{K}{4}$ is between $0$ and $K$.

Conclusion

$K_{\text{top}} = \dfrac{K}{4}$

The kinetic energy at the maximum height is $K_{\text{top}} = K\cos^2 60^\circ = \dfrac{K}{4}$.

The answer is (4).

At the top of the flight, the sideways speed $u\cos\theta$ is what is left.

The energy goes with the speed squared, so $K_{\text{top}} = K\cos^2\theta$. Put the angle in, and then check the two ends, $\theta = 0^\circ$ and $\theta = 90^\circ$.

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