Challenges in mechanics by Ronak Agarwal (Part 2)

Initial state Image 1

Final state Image 2

Since the collision was elastic we write ${ v }_{ seperation }={ v }_{ approach }$

hence we have ${ v }_{ 1 }+{ v }_{ 2 }sin(\theta )=vcos(\theta )\quad (i)$

Also applying conservation of momentum in x-direction we get $0=m({ v }_{ 1 }sin(\theta )+vsin(\theta )cos(\theta ))-m({ v }_{ 2 })$

$\Rightarrow \quad { v }_{ 2 }={ v }_{ 1 }sin(\theta )+vsin(\theta )cos(\theta )\quad (ii)$

From (i) and (ii) we get ${ v }_{ 1 }=\frac { v{ cos }^{ 3 }(\theta ) }{ 1+{ sin }^{ 2 }(\theta ) } \quad \quad (iii)$

Now we will find ${ v }_{ x }\quad and\quad { v }_{ y }.\\ { v }_{ x }={ v }_{ 1 }sin(\theta )+vsin(\theta )cos(\theta )\quad =\quad \frac { 2vsin(\theta )cos(\theta ) }{ 1+{ sin }^{ 2 }(\theta ) } \\ { v }_{ y }={ v }_{ 1 }cos(\theta )-v{ sin }^{ 2 }(\theta )\quad =\quad \frac { v({ cos }^{ 2 }(\theta )-2{ sin }^{ 2 }(\theta )) }{ 1+{ sin }^{ 2 }(\theta ) }$

Since $tan(\theta )=\frac { 1 }{ \sqrt { 2 } }$ hence ${ v }_{ x }=\frac { v }{ \sqrt { 2 } } ,{ v }_{ y }=0$

Also note that $v=\sqrt { 2g(h-xtan\theta ) } \Rightarrow { v }_{ x }=\sqrt { g(h-xtan\theta ) }$

Range = ${ v }_{ x }\sqrt { \frac { 2xtan\theta }{ g } } =\sqrt { 2xtan\theta (h-xtan\theta ) }$

Now putting $x=1,h=2\sqrt { 2 } m$ we get Range = $\sqrt { 3 }$

Now $AB=Range-x\quad \Rightarrow AB=\sqrt { 3 } -1 \\ \Rightarrow \boxed { \sqrt { 3 } AB=3-\sqrt { 3 } }$