Coming back home after a few strikes

Classical Mechanics Level 5

From one corner A of a fixed regular billiards table ABCD placed on a horizontal surface. A ball of mass m=2kg and negligible dimensions is projected in the horizontal plane. The coefficient of restitution is 0.5 for every collision. It strikes the sides in order BC , AD , DC ,BC , and then returns to the same point A then the angle of initial projection it makes with side AB is θ \theta

If the ratio of sides AB:BC is 5:1, find ⌊ 100 t a n θ ⌋ \lfloor 100tan\theta \rfloor .


The answer is 4.

Milun Moghe by Milun Moghe , 25, India

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1 solution

Milun Moghe
Mar 20, 2014

Image Image t 1 = a u c o s θ = B P 1 u s i n θ t_{1}=\frac{a}{ucos\theta}=\frac{BP_{1}}{usin\theta} ... A → P 1 A\rightarrow P_{1}

t 2 = a e u c o s θ = K P 2 u s i n θ t_{2}=\frac{a}{eucos\theta}=\frac{KP_{2}}{usin\theta} .... P 1 → P 2 P_{1}\rightarrow P_{2} .....K is the mid point of A P 2 AP_{2}

t 3 = D P 3 e 2 u c o s θ = P 2 D u s i n θ t_{3}=\frac{DP_{3}}{e^{2}ucos\theta}=\frac{P_{2}D}{usin\theta} ..... P 2 → P 3 P_{2}\rightarrow P_{3}

t 4 = P 3 C e 2 u c o s θ = C P 4 e u s i n θ t_{4}=\frac{P_{3}C}{e^{2}ucos\theta}=\frac{CP_{4}}{eusin\theta} ..... P 3 → P 4 P_{3}\rightarrow P_{4}

t 5 = a e 3 u c o s θ = P 4 B e u s i n θ t_{5}=\frac{a}{e^{3}ucos\theta}=\frac{P_{4}B}{eusin\theta} .... P 4 → A P_{4}\rightarrow A

a u c o s θ + a e u c o s θ + D P 3 e 2 u c o s θ + P 3 C e 2 u c o s θ + a e 3 u c o s θ = B P 1 u s i n θ + K P 2 u s i n θ + P 2 D u s i n θ + C P 4 e u s i n θ + P 4 B e u s i n θ \frac{a}{ucos\theta}+\frac{a}{eucos\theta}+\frac{DP_{3}}{e^{2}ucos\theta}+\frac{P_{3}C}{e^{2}ucos\theta}+\frac{a}{e^{3}ucos\theta}=\frac{BP_{1}}{usin\theta}+\frac{KP_{2}}{usin\theta}+\frac{P_{2}D}{usin\theta}+\frac{CP_{4}}{eusin\theta}+\frac{P_{4}B}{eusin\theta}

a c o s θ ( 1 + 1 e + 1 e 2 + 1 e 3 ) = b s i n θ ( 1 + 1 e ) \frac{a}{cos\theta}(1+\frac{1}{e}+\frac{1}{e^{2}}+\frac{1}{e^{3}})=\frac{b}{sin\theta}(1+\frac{1}{e})

e = a s i n θ b c o s θ − a s i n θ e=\sqrt{\frac{asin\theta}{bcos\theta-asin\theta}}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice work Milun,

Although I think that you have an extra = sign in the penultimate line also K isn't the midpoint of A P 2 AP_2 .

The last lines should read a c o s θ ( 1 + 1 e + 1 e 2 + 1 e 3 ) = b s i n θ ( 1 + 1 e ) \frac{a}{cos \theta} (1+\frac{1}{e} + \frac{1}{e^2} + \frac{1}{e^3} )= \frac{b}{sin \theta}(1+ \frac{1}{e})

We are given e = 0.5 e=0.5 and a / b = 5 a/b =5 in the question. Substituting in these values gives 5 t a n θ × ( 1 + 2 + 4 + 8 ) = ( 1 + 2 ) 5tan \theta \times( 1+ 2+ 4+8)=(1+2) 25 t a n θ = 1 25tan\theta = 1 Or 100 t a n θ = 4 100tan \theta = 4

sirio quintavalle - 7 years, 2 months ago

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Well k is the midpoint of AP2 I've just not shown it in the diagram. BTW thanks for pointing out the typos

Milun Moghe - 7 years, 2 months ago

What a lengthy question! Well, It felt awesome, when answer matched!

Md Zuhair - 3 years, 2 months ago

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