JEE MAIN - Physics (2017 - 8th April Morning Slot - No. 19)
An object is dropped from a height h from the ground. Every time it hits the ground it looses 50% of its kinetic energy. The total distance covered as t $$ \to $$ $$\infty $$ is :
3h
$$\infty $$
$${5 \over 3}$$h
$${8 \over 3}$$h
Explanation
Let,
Kinetic energy (k) = $${1 \over 2}$$ m $$\upsilon $$2 before it hit the ground.
After hitting the ground kinetic energy
(k') = $${1 \over 2}$$ m $$\upsilon $$$$_1^2$$
$$\therefore\,\,\,$$According to the question,
$${1 \over 2}$$ m$$\upsilon $$$$_1^2$$ = $${1 \over 2}$$ $$ \times $$ $${1 \over 2}$$ m$$\upsilon $$2
$$ \Rightarrow $$$$\,\,\,$$ $$\upsilon $$1 = $${v \over {\sqrt 2 }}$$
After hitting the ground the object will bounce
h' = $${{v_1^2} \over {2g}}$$ = $${{{v^2}} \over {4g}}$$ = $${h \over 2}$$ [ as h = $${{{v^2}} \over {2g}}$$ ]
Total distance travelled from the time it first hits the ground to the next time it hits the ground is = $${h \over 2}$$ + $${h \over 2}$$ = h
So, this will create a infinite geometric progression with the common ration $${1 \over 2}$$.
$$\therefore\,\,\,$$ Total distance covered
= h (distance travelled by the obhect when first dropped, before it hits the ground)
+ (h + $${h \over 2}$$ + $${h \over 4}$$ + . . . . . . . .$$ \propto $$)
= h + $${h \over {1 - {1 \over 2}}}$$
= h + 2h
= 3h
Kinetic energy (k) = $${1 \over 2}$$ m $$\upsilon $$2 before it hit the ground.
After hitting the ground kinetic energy
(k') = $${1 \over 2}$$ m $$\upsilon $$$$_1^2$$
$$\therefore\,\,\,$$According to the question,
$${1 \over 2}$$ m$$\upsilon $$$$_1^2$$ = $${1 \over 2}$$ $$ \times $$ $${1 \over 2}$$ m$$\upsilon $$2
$$ \Rightarrow $$$$\,\,\,$$ $$\upsilon $$1 = $${v \over {\sqrt 2 }}$$
After hitting the ground the object will bounce
h' = $${{v_1^2} \over {2g}}$$ = $${{{v^2}} \over {4g}}$$ = $${h \over 2}$$ [ as h = $${{{v^2}} \over {2g}}$$ ]
Total distance travelled from the time it first hits the ground to the next time it hits the ground is = $${h \over 2}$$ + $${h \over 2}$$ = h
So, this will create a infinite geometric progression with the common ration $${1 \over 2}$$.
$$\therefore\,\,\,$$ Total distance covered
= h (distance travelled by the obhect when first dropped, before it hits the ground)
+ (h + $${h \over 2}$$ + $${h \over 4}$$ + . . . . . . . .$$ \propto $$)
= h + $${h \over {1 - {1 \over 2}}}$$
= h + 2h
= 3h
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