Question

A particle is projected at an angle 60 with speed $10\sqrt{3}{m}/{s}\;$from point A as shown in the figure. At the same time, the wedge is made to move with speed $10\sqrt{3}{m}/{s}\;$towards the right as shown in the figure. Then the time after which the particle will strike with the wedge is $(g=10{m}/{{{s}^{2}}}\;)$

Answer 1

Firstly we will compute the velocities of the wedge and the particle separately, then, we will compute the velocity of the particle concerning the wedge. Using these velocity values we will compute the angles. Finally, we will use the standard formula to compute the time taken.
Formula used:
$T=\dfrac{2u\sin (\alpha -\beta )}{g\cos \beta }$

Complete step-by-step solution:
Firstly we will compute the velocity of the wedge concerning the axis.
The velocity of the wedge along the x-axis is,
${{V}_{wx}}=10\sqrt{3}{m}/{s}\;$
The velocity of the wedge along the y-axis is,
${{V}_{wy}}=0{m}/{s}\;$
Therefore, the velocity of the wedge is, ${{V}_{w}}=10\sqrt{3}\,i$.
Now, we will compute the velocity of the particle concerning the axis.
The velocity of the particle along the x-axis is,

& {{V}_{px}}=10\sqrt{3}\cos 60{}^\circ \\\ & {{V}_{px}}=10\sqrt{3}\times \dfrac{1}{2} \\\ & {{V}_{px}}=5\sqrt{3}\,{m}/{s}\; \\\ \end{aligned}$$ The velocity of the particle along the y-axis is, $$\begin{aligned} & {{V}_{py}}=10\sqrt{3}\sin 60{}^\circ \\\ & {{V}_{py}}=10\sqrt{3}\times \dfrac{\sqrt{3}}{2} \\\ & {{V}_{py}}=15\,{m}/{s}\; \\\ \end{aligned}$$ Therefore, the velocity of the particle is, $${{V}_{p}}=5\sqrt{3}\,i+15j$$. The resultant velocity, that is, the velocity of the particle concerning the wedge is computed as follows. The resultant velocity along the x-axis is, $$\begin{aligned} & {{V}_{p/w\\_x}}=5\sqrt{3}-10\sqrt{3} \\\ & {{V}_{p/w\\_x}}=-5\sqrt{3}\,{m}/{s}\; \\\ \end{aligned}$$ The resultant velocity along the y-axis is, $$\begin{aligned} & {{V}_{p/w\\_y}}=15-0 \\\ & {{V}_{p/w\\_y}}=15\,{m}/{s}\; \\\ \end{aligned}$$ Therefore, the resultant velocity is, $${{V}_{p}}=5\sqrt{3}\,i+15j$$. $$\begin{aligned} & {{V}_{p/w}}={{V}_{p}}-{{V}_{w}} \\\ & {{V}_{p/w}}=5\sqrt{3}i+15j-10\sqrt{3}i \\\ & {{V}_{p/w}}=-5\sqrt{3}i+15j \\\ \end{aligned}$$ Now, we will compute the angle values. $$\begin{aligned} & \alpha =\theta \\\ & \beta =30{}^\circ \\\ \end{aligned}$$ We know that, $$\begin{aligned} & \tan \alpha =\dfrac{{{V}_{y}}}{{{V}_{x}}} \\\ & \Rightarrow \tan \alpha =\dfrac{15}{5\sqrt{3}} \\\ & \Rightarrow \tan \alpha =\sqrt{3} \\\ & \therefore \alpha =60{}^\circ \\\ \end{aligned}$$ The formula to be used to find the time after which particle will strike with the wedge is given as follows. $$T=\dfrac{2u\sin (\alpha -\beta )}{g\cos \beta }$$ Where $$T$$ is the time, $$u$$ is the initial velocity, $$\alpha ,\beta $$ are the angles and $$g$$ is the acceleration due to gravity. The diagram representing the parameters:  Substitute the values of the acceleration due to gravity, the initial velocity and the angle values in the above equation. $$T=\dfrac{2\times 10\sqrt{3}\times \sin (60{}^\circ -30{}^\circ )}{10\times \cos 30{}^\circ }$$ Continue the computation to find the value of the time taken. $$\begin{aligned} & T=\dfrac{2\times \sqrt{3}\times \sin 30{}^\circ }{\cos 30{}^\circ } \\\ & \Rightarrow T=\dfrac{2\times \sqrt{3}\times {}^{1}/{}_{2}}{{}^{\sqrt{3}}/{}_{2}} \\\ & \therefore T=2\,s \\\ \end{aligned}$$ **$$\therefore $$The value of the time after which the particle will strike with the wedge is $$2\,s$$.** **Note:** The computation of the angle values concerning the velocities of the particle and the wedge is very important. As, the parameters have both x and y components, so, the calculation gets a bit difficult at this step. By direct substitution also we can find the time taken.