Sliding coins

Since the angle is $30$ degrees and $g = -9.8 m/s^2$ then the acceleration of the coin is equal to $-9.8 cos30 m/s^2 = -4.9 m/s^2$ . Using the equation change in $x = \frac {1}{2} at^2$ and substituting $a = -4.9m/s^2, x = 1m$ we get $t = 0.639s$ .

Since the object starts at rest, $v = at \rightarrow v = (-4.9)(0.639) = -3.13 m/s$ . In addition, $Speed = |v| = 3.13m/s$ .

$a=g~\sin\theta=9.8~\textrm{m/s}^2 \times \sin 30°=9.8~\textrm{m/s}^2 \times 0.5=4.9~\textrm{m/s}^2$

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$v^2=u^2+2as\Longrightarrow v^2 =0^2+2\times 4.9~\textrm{m/s}^2 \times 1~\textrm{m}=9.8~\textrm{m}^2/\textrm{s}^2$

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$\therefore v = \sqrt{9.8~\textrm{m}^2/\textrm{s}^2} \approx 3.13~\textrm{m/s}$

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Therefore, the required answer is $~\Large{\fbox{3.13}}$ !

Given that - ramp is angled at $30 ^ \circ$ with respect to (w.r.t) the horizontal .

Now , since angle of acceleration due to gravity is $90 ^ \circ$ w.r.t horizontal ,

Thus we obtain a triangle between the slope of ramp (motion of coin) , the horizontal and the acceleration due to gravity .

Thus acceleration (in $m /s^2)$ of the coin along its motion (i.e along the ramp) will be :

$g \cdot \cos 60^\circ = \dfrac{g}{2} = 4.9$

Thus by using formula (where initial velocity $= u = 0$ ) :

$\begin{aligned} v^2 - u^2 &= 2 \cdot a \cdot s \\ \Rightarrow v^2 &= 2 \cdot 4.9 \ cdot 1 \\ \Rightarrow v^2 &= 9.8 \\ \Rightarrow v &= \boxed{3.13} m/s \end{aligned}$

The acceleration of the coin is 4.9~\text{m/s\$^2\$} . That means that the speed of the coin at time $t$ will be $4.9t$ . To find $t$ , we can use the formula $d=v_0t+\frac{1}{2}at^2$ to get $1=\frac{4.9}{2}t^2\Rightarrow t=\sqrt{\frac{2}{4.9}}$ , so the speed of the coin after 1 metre will be $4.9\times \sqrt{\frac{2}{4.9}}=\boxed{3.13}$

Let us look at the event more closely. A coin sliding down the incline, the angle being θ. Since it starts from rest and gains some speed after a time period ‘t’, the coin is moving with some acceleration ‘a’. The weight is acting vertically downwards. Its cosine component balances the normal reaction. Its sine component acts in the direction of motion (perhaps the secret behind the coin’s motion!). Taking into consideration frictional force ‘f’ we deduce the following relation – mgsinθ – f = ma …………….(i) mgsinθ - µmgcosθ = ma Here, ‘µ’ is the coefficient of friction and frictional force being equal to the product of normal reaction and µ. Since, the problem assumes the absence of frictional force (an ideal case indeed!) we deduce the following – mgsinθ = ma ………………(ii) gsinθ = a ………………(iii) Now, let the coin move with a velocity ‘v’. We know, v2 = v02 – 2aS ………………(iv) Here, S being the displacement and v0 being the initial speed. The coin being initially at rest we come to the following conclusion – v2 = – 2aS ………………..(v) v2 = – 2(gsinθ)S ………………..(vi) There you are. Take the square root of (vi) to get the velocity with which the coin is sliding down the incline.

accleration(a)=g sin30 initial velosity(u)=0 displacement down the plane(s)=1m final velosity(v)=? v^2-u^2=2 a s v^2-0=2 9.8*1=(3.1304951684997055749728431362238)^2 therefore v=3.1304951684997055749728431362238