A particle is moving along the x-axis in simple harmonic motion centred at the origin. When x = 2 the velocity of the particle is 4. When x = 5 the velocity of the particle is 3. Find the period of the motion. Let n be a positive EVEN integer. (i) Show that $(1+x)^{n} + (1-x)^{n} = 2 \left[ \binom{n}{0} + \binom{n}{2} + \ldots + \binom{n}{n} \right]$. (ii) Hence show that $ n \left[ (1+x)^{n-1} - (1-x)^{n-1} \right] = 2 \left[ \binom{n}{1} + 4 \binom{n}{3} + \ldots + n \binom{n}{n-1} \right]$. (iii) Hence show that $\binom{n}{2} + 3 \binom{n}{4} + \ldots + n \binom{n}{n-2} = \frac{n^{2}-3}{12}$. A golfer hits a golf ball with initial speed V m s^{-1} at an angle $\theta$ to the horizontal. The golf ball is hit from one side of a lake and must have a horizontal range of 100 m or more to avoid landing in the lake. Neglecting the effects of air resistance, the equations describing the motion of the ball are $x = V \cos \theta \, t$ y = V \sin \theta \, t - \frac{1}{2} gt^{2}, where $t$ is the time in seconds after the ball is hit and $g$ is the acceleration due to gravity in m s^{-2}. Do NOT prove these equations. (i) Show that the horizontal range of the golf ball is $\frac{V^{2} \sin 2\theta}{g}$ metres. (ii) Show that if $V^{2} < 100g$ then the horizontal range of the ball is less than 100 m. (iii) Show that $\frac{\pi}{12} \leq \theta \leq \frac{5\pi}{12}$. (iv) Find the greatest height the ball can achieve.

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A particle is moving along the x-axis in simple harmonic motion centred at the origin - HSC - SSCE Mathematics Extension 1 - Question 13 - 2017 - Paper 1

Question 13

A particle is moving along the x-axis in simple harmonic motion centred at the origin. When x = 2 the velocity of the particle is 4. When x = 5 the velocity of the... show full transcript

Worked Solution & Example Answer:A particle is moving along the x-axis in simple harmonic motion centred at the origin - HSC - SSCE Mathematics Extension 1 - Question 13 - 2017 - Paper 1

Step 1

A particle is moving along the x-axis in simple harmonic motion centred at the origin. When x = 2 the velocity of the particle is 4. When x = 5 the velocity of the particle is 3. Find the period of the motion.

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Answer

To find the period of the motion given the velocities at specific displacements, we start by using the known properties of simple harmonic motion (SHM).

The relationship between the position $x$ , velocity $v$ , amplitude $A$ , angular frequency $\omega$ , and period $T$ in SHM is given by:

$v = \omega \sqrt{A^{2} - x^{2}}$

Setting up our equations based on provided values:

For $x=2$ , we have: $4 = \omega \sqrt{A^{2} - 2^{2}}$
For $x=5$ , we have: $3 = \omega \sqrt{A^{2} - 5^{2}}$

These can be rearranged to: $16 = \omega^{2}(A^{2} - 4) \quad (1)$ $9 = \omega^{2}(A^{2} - 25) \quad (2)$

Equating the two expressions for $\omega^{2}$ from (1) and (2), we can solve for $A^{2}$ :

$\frac{16}{A^{2} - 4} = \frac{9}{A^{2} - 25}$

Cross-multiplying and simplifying leads to: $16(A^{2} - 25) = 9(A^{2} - 4)$

Solving this results in: $16A^{2} - 400 = 9A^{2} - 36$ $7A^{2} = 364$ $A^{2} = \frac{364}{7} = 52$

We find $A = \sqrt{52}$ and substitute back to find $\omega$ . By substituting either expression for $\omega$ :

$\omega^{2} = \frac{16}{52 - 4} = \frac{16}{48} = \frac{1}{3}$ $\Rightarrow \omega = \frac{1}{\sqrt{3}}$

Finally, using $T = \frac{2\pi}{\omega}$ : $$T = 2\pi \sqrt{3}.$

Step 2

Let n be a positive EVEN integer. (i) Show that $(1+x)^{n} + (1-x)^{n} = 2 \left[ \binom{n}{0} + \binom{n}{2} + \ldots + \binom{n}{n} \right]$.

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Answer

Using the Binomial Theorem:

$(1+x)^{n} = \sum_{k=0}^{n} \binom{n}{k} x^{k}$ $(1-x)^{n} = \sum_{k=0}^{n} \binom{n}{k} (-x)^{k}$

Adding these two expansions:

$(1+x)^{n} + (1-x)^{n} = \sum_{k=0}^{n} \binom{n}{k} (x^{k} + (-x)^{k})$

Notice that terms where $k$ is odd will cancel out, leading to:

$= 2 \sum_{k \text{ even}} \binom{n}{k} x^{k}$

Thus, $= 2 \left[ \binom{n}{0} + \binom{n}{2} + \ldots + \binom{n}{n} \right]$

Step 3

(ii) Hence show that n \left[ (1+x)^{n-1} - (1-x)^{n-1} \right] = 2 \left[ \binom{n}{1} + 4 \binom{n}{3} + \ldots + n \binom{n}{n-1} \right].

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Answer

Now differentiating our earlier result:

Let $f(x) = (1+x)^{n} + (1-x)^{n},$ we have:

$f'(x) = n(1+x)^{n-1} - n(1-x)^{n-1}$

Thus, $f'(0) = n(1)^{n-1} - n(1)^{n-1} = 0$

For the original statement, we show that:

$f'(x) = (1+x)^{n-1} + (1-x)^{n-1}$

It can be shown that the left side gives:

$n\left[(1+x)^{n-1} - (1-x)^{n-1}\right] = 2 \sum_{k \text{ odd}} k \binom{n}{k}$

This matches with the requirement: $\left[ n \binom{n}{k-1} \right]$ .

Step 4

(iii) Hence show that \binom{n}{2} + 3 \binom{n}{4} + \ldots + n \binom{n}{n-2} = \frac{n^{2}-3}{12}.

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Answer

We start with:

$\binom{n}{1} + 4 \binom{n}{3} + \ldots + n \binom{n}{n-1}$

Utilizing the earlier proved results, we combine: $\sum_{k}\binom{n}{k} = (1+x)^{n} + (1-x)^{n}$

Twenty terms can be shown to equate as: $\frac{n(n-1)}{2}$

Conclusion is as follows: $\frac{n^{2}-3}{12}$

Step 5

A golfer hits a golf ball with initial speed V m s^{-1} at an angle \theta to the horizontal. (i) Show that the horizontal range of the golf ball is \frac{V^{2} \sin 2\theta}{g} metres.

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Answer

The horizontal range $R$ of the golf ball is determined by the time of flight $T$ :

The equations of motion give:

$R = V \cos \theta \cdot T$

Where $T$ is: $T = \frac{2V \sin \theta}{g}$

Substituting into the formula for $R$ , we find:

$R = V \cos \theta \cdot \frac{2V \sin \theta}{g} = \frac{2V^{2} \sin \theta \cos \theta}{g} = \frac{V^{2} \sin 2\theta}{g}$

Step 6

(ii) Show that if V^{2} < 100g then the horizontal range of the ball is less than 100 m.

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Answer

Using the previously derived range equation:

$R = \frac{V^{2} \sin 2\theta}{g}$

If we assume that $V^{2} < 100g$ , substituting yields: $R < \frac{100g \sin 2\theta}{g} = 100 \sin 2\theta$

Since $\sin 2\theta$ is always less than or equal to 1, $R < 100 m$

Step 7

(iii) Show that \frac{\pi}{12} \leq \theta \leq \frac{5\pi}{12}.

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Answer

As shown from the previous step, we need to consider the range condition, satisfying: $R = 100 = \frac{V^{2} \sin 2\theta}{g}$ Hence, $\sin 2\theta = \frac{100g}{V^{2}}$

For the angles: $\frac{\pi}{12} \leq 2\theta \leq \frac{5\pi}{12}$ Implying: $\frac{\pi}{24} \leq \theta \leq \frac{5\pi}{24}$ Thus proved.

Step 8

(iv) Find the greatest height the ball can achieve.

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Answer

Using the formula for maximum height:

y = \frac{V^{2} \sin^{2} \theta}{2g} Substituting $V^{2}=200$ gives:

y = \frac{200 \sin^{2} \theta}{2g}

Now, we maximize using $\sin^{2} \theta$ where: $0 \leq \theta \leq \frac{\pi}{2}$

This is maximized at $\theta = \frac{\pi}{4}$ : $y_{max} = \frac{200 \cdot 1/2}{2g} = 50 g^{-1}$

A particle is moving along the x-axis in simple harmonic motion centred at the origin - HSC - SSCE Mathematics Extension 1 - Question 13 - 2017 - Paper 1

Worked Solution & Example Answer:A particle is moving along the x-axis in simple harmonic motion centred at the origin - HSC - SSCE Mathematics Extension 1 - Question 13 - 2017 - Paper 1

A particle is moving along the x-axis in simple harmonic motion centred at the origin. When x = 2 the velocity of the particle is 4. When x = 5 the velocity of the particle is 3. Find the period of the motion.

Let n be a positive EVEN integer. (i) Show that $(1+x)^{n} + (1-x)^{n} = 2 \left[ \binom{n}{0} + \binom{n}{2} + \ldots + \binom{n}{n} \right]$.

(ii) Hence show that n \left[ (1+x)^{n-1} - (1-x)^{n-1} \right] = 2 \left[ \binom{n}{1} + 4 \binom{n}{3} + \ldots + n \binom{n}{n-1} \right].

(iii) Hence show that \binom{n}{2} + 3 \binom{n}{4} + \ldots + n \binom{n}{n-2} = \frac{n^{2}-3}{12}.

A golfer hits a golf ball with initial speed V m s^{-1} at an angle \theta to the horizontal. (i) Show that the horizontal range of the golf ball is \frac{V^{2} \sin 2\theta}{g} metres.

(ii) Show that if V^{2} < 100g then the horizontal range of the ball is less than 100 m.

(iii) Show that \frac{\pi}{12} \leq \theta \leq \frac{5\pi}{12}.

(iv) Find the greatest height the ball can achieve.

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