A classical mechanics problem by jatin yadav

This problem can be solved conservation of energy.

Let $V_{BW}=V_{1}$ be the velocity of ball wrt wedge and $V_{B}$ and $V_{W}=V_{2}$ be the velocities of ball and wedge respectively.

Let $\theta=x$

${ V }_{ BW }=V_{ 1 }sin(x)\hat { i } -\quad { V }_{ 1 }cos(x)\hat { J\quad }$

Now by concept of relative velocity

${ V }_{ B }=\quad { V }_{ BW }+{ V }_{ W }$

So ${ V }_{ B }=V_{ 1 }sin(x)-V_{ 2 }\hat { i } -\quad { V }_{ 1 }cos(x)\hat { J\quad }$

On putting values we get

${ V }_{ B }^{ 2 }=({ V }_{ 1 }sin(x)-{ V }_{ 2 })^{ 2 }\quad +\quad (-{ V }_{ 1 }cos(x))^{ 2 }$

As no external force acts on the system in horizontal direction so horizontal velocity of the CoM of the system is 0.

So

${ V }_{ 1 }sin(x)-{ V }_{ 2 }-{ V }_{ 2 }$ =0

${ V }_{ 1 }sin(x)=2V_{ 2 }$

Now by energy conservation

Loss in potential energy is equal to gain in kinetic energy.

So $2mgsin(x)=\frac { 1m }{ 2 } \left\{ { V }_{ B }^{ 2 }+{ V }_{ W }^{ 2 } \right\}$

On putting values

$4mgsin(x)=\quad ({ V }_{ 1 }sin(x)-{ V }_{ 2 })^{ 2\quad }+\quad ({ V }_{ 1 }cos(x))^{ 2 }\quad +\quad (-{ { V }_{ 2 } })^{ 2 })$

${ V }_{ 2 }=\quad \sqrt { \frac { 2y{ sin }^{ 3 }\quad (x) }{ 2-{ sin }^{ 2 }(x) } }$

On differentiating this equation with respect to $x$ and then putting the value of $x=45$ we get

$100f^{ ' }\left( x \right) =394.049$