JEE MAIN - Physics (2019 - 9th January Morning Slot - No. 10)
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 ?
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Explanation
As in elastic or in elastic clollision, momentum is conserved.
$$ \therefore $$ Pi = Pf
Pi = Initial momentum
Pf = Final Momentum
mv = (2m + m) Vf
$$ \Rightarrow $$ Vf = $${{mv} \over {2m + M}}$$
Here due to collision $${5 \over 6}$$th of kinetic energy is lost.
$$ \therefore $$ Remaining kinetic energy,
Kf = $${1 \over 6}$$ Ki
$$ \Rightarrow $$ $${1 \over 2}$$(2m + M) $$ \times $$ $${{{{\left( {mv} \right)}^2}} \over {{{\left( {2m + M} \right)}^2}}}$$ = $${1 \over 6} \times {1 \over 2}m{v^2}$$
$$ \Rightarrow $$ $${{{m^2}{v^2}} \over {2m + M}}$$ = $${1 \over 6}m{v^2}$$
$$ \Rightarrow $$ $${m \over {2m + M}}$$ = $${1 \over 6}$$
$$ \Rightarrow $$ 6m = 2m + M
$$ \Rightarrow $$ M = 4m
$$ \Rightarrow $$ $${M \over m}$$ = 4
$$ \therefore $$ Pi = Pf
Pi = Initial momentum
Pf = Final Momentum
mv = (2m + m) Vf
$$ \Rightarrow $$ Vf = $${{mv} \over {2m + M}}$$
Here due to collision $${5 \over 6}$$th of kinetic energy is lost.
$$ \therefore $$ Remaining kinetic energy,
Kf = $${1 \over 6}$$ Ki
$$ \Rightarrow $$ $${1 \over 2}$$(2m + M) $$ \times $$ $${{{{\left( {mv} \right)}^2}} \over {{{\left( {2m + M} \right)}^2}}}$$ = $${1 \over 6} \times {1 \over 2}m{v^2}$$
$$ \Rightarrow $$ $${{{m^2}{v^2}} \over {2m + M}}$$ = $${1 \over 6}m{v^2}$$
$$ \Rightarrow $$ $${m \over {2m + M}}$$ = $${1 \over 6}$$
$$ \Rightarrow $$ 6m = 2m + M
$$ \Rightarrow $$ M = 4m
$$ \Rightarrow $$ $${M \over m}$$ = 4
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