Question 211: Centre of Mass and Rotational Motion

Question

A particle of mass $$m$$ moving in the $$x$$ direction with speed $$2v$$ is hit by another particle of mass $$2m$$  moving in the $$y$$ direction with speed $$v.$$  If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to:

Accepted Answer

Applying conservation of linear momentum in x direction
$$2mv = \left( {2m + m} ight){V_x}$$
$$ \Rightarrow {V_x} = {2 \over 3}v$$
Applying conservation of linear momentum in y direction
$$2mv = \left( {2m + m} ight){V_y}$$
$$ \Rightarrow {V_y} = {2 \over 3}v$$
Final speed of 3m mass, $${V_f}$$  = $$\sqrt {V_x^2 + V_y^2} $$
= $$\sqrt {{{4{v^2}} \over 9} + {{4{v^2}} \over 9}} $$
= $$\sqrt {{{8{v^2}} \over 9}} $$
Initial kinetic energy
$${E_i} = {1 \over 2}m{\left( {2v} ight)^2} + {1 \over 2}\left( {2m} ight){\left( v ight)^2}$$
$$ = 2m{v^2} + m{v^2}$$
= $$3m{v^2}$$
Final kinetic energy,
$${E_f} = {1 \over 2}\left( {3m} ight)$$$${V_f^2}$$
$$ = {{3m} \over 2}\left[ {{{8{v^2}} \over 9}} ight]$$
$$ = {{4m{v^2}} \over 3}$$
Energy loss = $${E_i} - {E_f}$$
= $$3m{v^2} - {{4m{v^2}} \over 3}$$
= $${{5m{v^2}} \over 3}$$
Percentage loss in the energy during the collision
= $${{{E_i} - {E_f}} \over {{E_i}}}$$$$ 	imes 100$$
= $${{{{5m{v^2}} \over 3}} \over {3m{v^2}}}$$$$ 	imes 100$$
= $${5 \over 9} 	imes 100$$
$$ \simeq 56\% $$

Answer

$$56\% $$

Answer

$$62\% $$

Answer

$$44\% $$

Answer

$$50\% $$

Centre of Mass and Rotational Motion

A particle of mass $m moving in the x direction with speed 2v is hit by another particle of mass 2m moving in the y direction with speed v.$ If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to:

Choose the correct answer: