Question 146: Centre of Mass and Rotational Motion

Question

A particle of mass m with an initial velocity $u\widehat i$
collides perfectly elastically with a mass 3 m at
rest. It moves with a velocity $v\widehat j$ after collision,
then, v is given by :

Accepted Answer

From momentum conservation
m(u)$\widehat i$ + 3m(0) = mv$\widehat j$ + 3m$\overrightarrow {v'} $
$ \Rightarrow $ 3m$\overrightarrow {v'} $ = m(u)$\widehat i$ - mv$\widehat j$
$ \Rightarrow $ $\overrightarrow {v'}  = {{u\widehat i - v\widehat j} \over 3}$
$ 	herefore $ $\left| {\overrightarrow {v'} } ight| = {{\sqrt {{u^2} + {v^2}} } \over 3}$
$ \Rightarrow $ ${\left| {\overrightarrow {v'} } ight|^2} = {{{u^2} + {v^2}} \over 9}$ ......(1)
As collision is perfectely elastic hence
KE
i
= KE
j
$ \Rightarrow $ ${1 \over 2}m{u^2} + {1 \over 2}\left( {3m} ight) 	imes {0^2} = {1 \over 2}m{v^2} + {1 \over 2}3m{\left( {v'} ight)^2}$
$ \Rightarrow $ u
2
= v
2
+ 3${\left( {v'} ight)^2}$
$ \Rightarrow $ u
2
= v
2
+ 3$\left( {{{{u^2} + {v^2}} \over 9}} ight)$
$ \Rightarrow $ 3u
2
= 3v
2
+ u
2
+ v
2
$ \Rightarrow $ 2u
2
= 4v
2
$ \Rightarrow $ $v = {u \over {\sqrt 2 }}$

Answer

$v = \sqrt {{2 \over 3}} u$

Answer

$v = {u \over {\sqrt 3 }}$

Answer

$v = {u \over {\sqrt 2 }}$

Answer

$v = {1 \over {\sqrt 6 }}u$

Centre of Mass and Rotational Motion

A particle of mass m with an initial velocity u\widehat i collides perfectly elastically with a mass 3 m at rest. It moves with a velocity v\widehat j after collision, then, v is given by :

Choose the correct answer: