Question 195: Centre of Mass and Rotational Motion

Question

Three blocks A, B and C are lying on a smooth horizontal surface, as shown in the figure. A and B have equal masses, m while C has mass M, Block A is given an initial speed $$\upsilon $$ towards B due to which it collides with B perfectly inelastically. The combined mass collides with C, also perfectly inelastically $${5 \over 6}$$th of the initial kinetic energy is lost in whole process. What is value of M/m ?

Accepted Answer

As in elastic or in elastic clollision, momentum is conserved.
$$ 	herefore $$   P
i
= P
f
P
i
= Initial momentum
P
f
= Final Momentum
mv = (2m + m) V
f
$$ \Rightarrow $$   V
f
= $${{mv} \over {2m + M}}$$
Here due to collision $${5 \over 6}$$th of kinetic energy is lost.
$$ 	herefore $$    Remaining kinetic energy,
K
f
= $${1 \over 6}$$ Ki
$$ \Rightarrow $$    $${1 \over 2}$$(2m + M) $$ 	imes $$ $${{{{\left( {mv} ight)}^2}} \over {{{\left( {2m + M} ight)}^2}}}$$ = $${1 \over 6} 	imes {1 \over 2}m{v^2}$$
$$ \Rightarrow $$   $${{{m^2}{v^2}} \over {2m + M}}$$ = $${1 \over 6}m{v^2}$$
$$ \Rightarrow $$   $${m \over {2m + M}}$$ = $${1 \over 6}$$
$$ \Rightarrow $$   6m = 2m + M
$$ \Rightarrow $$   M = 4m
$$ \Rightarrow $$    $${M \over m}$$ = 4

Answer

5

Answer

2

Answer

4

Answer

3

Centre of Mass and Rotational Motion

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