Question 139: Centre of Mass and Rotational Motion

Question

Blocks of masses m, 2m, 4m and 8m are arranged in a line on a frictionless floor. Another block of
mass m, moving with speed v along the same line (see figure) collides with mass m in perfectly
inelastic manner. All the subsequent collisions are also perfectly inelastic. By the time the last block
of mass 8m starts moving the total energy loss is p% of the original energy. Value of 'p' is close to :

Accepted Answer

All collisions are perfectly inelastic, so after the final collision, all blocks are moving together. So let the
final velocity be v', so on applying momentum conservation :
mv = 16mv'
$$ \Rightarrow $$ v' = $${v \over {16}}$$
Now initial energy E
i
= $${1 \over 2}m{v^2}$$
Final energy : E
f
= $${1 \over 2} 	imes 16m{\left( {{v \over {16}}} ight)^2}$$
Energy loss : E
i
Ã¢ÂÂ E
f
= $${1 \over 2}m{v^2}$$ - $${1 \over 2} 	imes m{\left( {{v \over {16}}} ight)^2}$$
= $${1 \over 2}m{v^2}\left[ {1 - {1 \over {16}}} ight]$$
= $${1 \over 2}m{v^2}\left[ {{{15} \over {16}}} ight]$$
Total energy loss is p% =
Energy loss
Original energy
$$ 	imes $$ 100
= $${{{1 \over 2}m{v^2}\left[ {{{15} \over {16}}} ight]} \over {{1 \over 2}m{v^2}}} 	imes 100$$
= 93.75% $$ \simeq $$ 94 %

Answer

37

Answer

77

Answer

87

Answer

94

Centre of Mass and Rotational Motion

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