Question

A ball of mass is fired from a spring loaded gun that is mounted on a table.If ball strikes to point B on ground. then find out initial compression of the spring (Assume table is smooth and spring constant of spring is k):

• A. $\frac{d\sqrt{2h}}{2mg}$
• B. $\frac{dmg}{2kh}$
• C. $\frac{d\sqrt{2h}}{mg}$
• D. $d\sqrt{\frac{mg}{2kh}}$

Answer 1

The correct option is D $d\sqrt{\frac{mg}{2kh}}$

R.E.F image

Let the initial compression of spring

be x and horizontal vel of the ball

at point -c be v. Then;

by horizontal projection, we get

$v.t=d[t\Rightarrow$ time of Height]

By second $e{q}^n$ of motion

$h=ocps+\frac{g}{2}{t}^2$

$\Rightarrow \sqrt{\frac{2h}{g}}=t$

So, $v=\frac{d}{t}=(\sqrt{\frac{2h}{g}}{)}^{-1}d=d\sqrt{\frac{g}{2h}}$

$\Delta P{E}_{Total}=\Delta PEspring=0-\frac{k{x}^2}{2}=\frac{-k{x}^2}{2}$

$\Delta K{E}_{Total}=\Delta KEBall=\frac{m{v}^2}{2}=\frac{m}{2}(d\sqrt{\frac{g}{2h}}{)}^2$

These above two $e{q}^n$ has been

written considering motion of

ball from initial point to Point.c such

that graving has not done any work

so, the work is done only by spring

force which is conservative,so.

$\Delta P{E}_{Total}+\Delta K{E}_{Total}=0\Rightarrow K{E}_{Total}=-\Delta P{E}_{Total}$

$\Rightarrow \frac{m}{2}(d\sqrt{\frac{g}{2n}}{)}^2=\frac{k}{2}\times x^2\Rightarrow x=d\sqrt{\frac{mg}{2kh}}$

Option - D is correct.