Question

A small ball is projected along surface of a smooth include plane with speed $10ms$ along the direction shown at $t=0$. The point of projected is origin, $z-$axis is along the vertical. The acceleration due to gravity is $10m/s^2$. Column $I$ lists the value of certain parameter related to motion of ball and column $II$ lists different time instants. Math appropriately.

Answer 1

R.E.F image

Normal balances $mgcos\theta$ $N=mgcos\theta$

So only component along the plane is $mgsin\theta$

$\Rightarrow$ acc. along plane $gsin\theta$

Along the plane it will be projectile motion with $a=gsin\theta$

In Ball + Earth system no external force is present

$\Rightarrow ME$ is conserved.

$\Rightarrow$ when ball reaches same height as initial height

total velocity with be same, but $(V_y{)}_f=-{\left(V_y\right)}_i$ & ${\left(V_x\right)}_f=(V_x{)}_i$

To find total flight time : $(V_y{)}_F={\left({\mu }_y\right)}_i+a_{y}t$

$\Rightarrow -10sin{37}^{\circ }=10sin{37}^{\circ }-gsin{37}^{\circ }\times t$

$\Rightarrow t=2s$

$Ans:C\to 4$

So max height will be at t = 15

$Ans:D\to 2$

t = 15 M = const = k+u

$\Rightarrow$ k is min $V_y=0$

$\Rightarrow$ Speed is min $Ans:B\to 2$

max height $:S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$

$\Rightarrow S_y=10sin{37}^{\circ }\times 1+\frac{1}{2}\times -10sin{37}^{\circ }\times 1$

$=\frac{1}{2}\times 10sin{37}^{\circ }$

$=\frac{1}{2}\times 10\times \frac{3}{5}=3m$

So the distance will be less than 3 m just after and

before $t=10,$ symmetrically,

$Ans:A\to 1,3$