Explanation
After applying horizontal impulse $$J$$ horizontal velocity of sphere becomes $$v$$ as initial linear momentum was $$0$$
$$J=Mv$$
Angular impulse about C.M
$$=hMv=Iw$$
$$hMv=\cfrac{2}5M{R}^{2}w$$
$$w=\cfrac { 5vh }{ 2{ R }^{ 2 } } $$
While rolling $$w=\cfrac { 5vh }{ 2{ R }^{ 2 } } $$
$$\cfrac { h }{ R } =\cfrac { 2 }{ 5 } $$
From work- energy theorem, for upward motion against resistance
$$\frac{1}{2}m{{v}^{2}}\,=mgh\,+\,W.................(1)$$
For downward motion,
$$\frac{1}{2}m{{v}^{2}}\,=\,mgh-W............(2)$$
From equation (1) and (2)
$$ \frac{1}{2}m({{16}^{2}}-{{8}^{2}})\,=2mgh $$
$$ \frac{1}{2}(256-64)\,=\,2\times 10\times h $$
$$ h=\,4.8m $$
Given,
$$ x=3t-4{{t}^{2}}+{{t}^{3}} $$
$$ dx=d\left( 3t-4{{t}^{2}}+{{t}^{3}} \right)=\left( 3-8t+3{{t}^{2}} \right)dt $$
Acceleration
$$a(x)=\dfrac{{{d}^{2}}x}{dt}=\dfrac{{{d}^{2}}x}{dt}=\dfrac{d\left( 3-8t+3{{t}^{2}} \right)}{dt}=-8+6t$$
Work done =$$dw=ma.dx=m\left( 6t-8 \right)\left( 3-8t+3{{t}^{2}} \right)dt$$
$$ \int_{0}^{W}{dw}=m\int_{0}^{4}{\left( 6t-8 \right)\left( 3-8t+3{{t}^{2}} \right)dt} $$
$$ W=m\times 176=0.003\times 176=528\times {{10}^{-3}}\,J $$
$$ W=528\times {{10}^{-3}} J $$
A block of mass m is connected to the lower end of a massless vertically suspended spring. The ball is displaced slightly downward and released; it oscillates up and down
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