Question 143: 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: