According to question, the situation can be shown

AIIMS 2017 Physics - Laws of Motion Question 5 English Explanation

Case 1: For body projected up the plane

$$\begin{aligned} & v=0 \\ & v^2=u^2-2 a_1 s \\ & \text { i.e. } \quad 0=u^2-2 a_1 s \\ & \therefore \quad u=\sqrt{2 a_1 s} \\ & \text { Also, } \quad v=u-a_1 t \\ & \text { i.e. } \quad 0=u-a_1 t_1 \\ & t_1=u / a_1 \\ & \therefore \quad t_1=\frac{\sqrt{2 a_1 s}}{a_1}=\sqrt{\frac{2 s}{a_1}} \quad \text{...... (i)}\\ \end{aligned}$$

Retardation

$$a_1=\frac{\mu R+m g \sin 30^{\circ}}{m}$$

$$\begin{aligned} \text{As} \quad R & =m g \cos 30^{\circ} \\ a_1 & =\frac{\mu m g \cos 30^{\circ}+m g \sin 30^{\circ}}{m} \\ & =\mu g \frac{\sqrt{3}}{2}+\frac{g}{2} \quad \text{..... (ii)} \end{aligned}$$

Case 2: For body coming down the plane

$$\begin{aligned} & s=u t+\frac{1}{2} a_2 t_2^2 \\ & u=0 \\ & t_2=\sqrt{\frac{2 s}{a_2}} \quad \text{..... (iii)} \end{aligned}$$

Downward acceleration

$$a_2=\frac{m g \sin 30^{\circ}-\mu R}{m}$$

$$\begin{aligned} & =\frac{m g \sin 30^{\circ}-\mu m g \cos 30^{\circ}}{m} \\ & =\frac{g}{2}-\frac{\mu g \sqrt{3}}{2} \end{aligned}$$

Given, $$t_1=\frac{t_2}{2}$$

Substituting values of t$$_1$$ and t$$_2$$ from Eqs. (i) and (iii), we get

$$\sqrt{\frac{2 s}{a_1}}=\frac{1}{2} \sqrt{\frac{2 s}{a_2}}$$

Now, squaring, both sides we get

$$a_1=4 a_2$$ ..... (iv)

Substituting values of $$a_1$$ and $$a_2$$ from Eqs. (ii) and (iv), we get

$$\mu g \frac{\sqrt{3}}{2}+\frac{g}{2}=4\left(\frac{g}{2}-\frac{\mu g \sqrt{3}}{2}\right)$$

Solving for $$\mu$$, we get

$$\begin{aligned} \mu & =\frac{\sqrt{3}}{5}=\frac{1.732}{5} \\ & =0.346 \end{aligned}$$

Laws Of Motion

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