Question

A body is allowed to fall from a height $h$ above the ground. Then match the following:

a) $PE=KE$	e) at height $h/2$
b) $PE=2KE$	f) constant at any point
c) $KE=2PE$	g) at height $2h/3$
d) $PE+KE$	h) at height $h/3$

• A. $a\to e,b\to g,c\to h,d\to f$
• B. $a\to g,b\to e,c\to f,d\to h$
• C. $a\to f,b\to g,c\to e,d\to h$
• D. $a\to e,b\to h,c\to f,d\to f$

Answer 1

The correct option is A $a\to e,b\to g,c\to h,d\to f$

From law of conservation of energy, the total energy of a system is conserved. Hence at all points, $TE+KE=constant$

Hence $d\to f$.

When the $PE=KE$, half the total energy is converted to kinetic energy. Hence the initial potential energy has dropped to half. Therefore the ball must have fallen by $h/2$.

Hence $a\to e$

When the $PE=2KE$, one third the total energy is converted to kinetic energy. Hence the initial potential energy has dropped to two-third of it Therefore the ball must have fallen by $h/3$.

Hence $b\to g$

When the $2PE=KE$, two third of the total energy is converted to kinetic energy. Hence the initial potential energy has dropped to one third. Therefore the ball must have fallen by $2h/3$.
Hence $c\to h$