A ball of mass 2 kg and another of mass 6 kg are dropped together from a 80 feet tall building. After a fall of 50 feet each towards the earth, their respective kinetic energies will be in the ratio of

Question

A ball of mass 2 kg and another of mass 6 kg are dropped together from a 80 feet tall building. After a fall of 50 feet each towards the earth, their respective kinetic energies will be in the ratio of (A)  1: 3 (B) 1: √2 (C) 2: 1 (D) √3: 1

Tardigrade Admin · Accepted Answer

Decrease in P.E = Gain in K.E energies. m g(h1-h2)=kf-ki (ki=0) (m1 g(h1-h2)=kf1/m2 g(h1-h2)=kf2) ⇒ (m1/m2)=(kf1/kf2) ⇒ (1/3)=(kf1/kf2)

Q. A ball of mass $2 k g$ and another of mass $6 k g$ are dropped together from a $80$ feet tall building. After a fall of $50$ feet each towards the earth, their respective kinetic energies will be in the ratio of

$1 : 3$

$1 : 2$

$2 : 1$

$3 : 1$

Q. A ball of mass 2kg and another of mass 6kg are dropped together from a 80 feet tall building. After a fall of 50 feet each towards the earth, their respective kinetic energies will be in the ratio of

1:3

1:2​

2:1

3​:1

Decrease in P.E = Gain in K.E energies. mg(h1​−h2​)=kf​−ki​(ki​=0) m2​g(h1​−h2​)=kf2​​m1​g(h1​−h2​)=kf1​​​⇒m2​m1​​=kf2​​kf1​​​⇒31​=kf2​​kf1​​​

Q. A ball of mass $2 k g$ and another of mass $6 k g$ are dropped together from a $80$ feet tall building. After a fall of $50$ feet each towards the earth, their respective kinetic energies will be in the ratio of

$1 : 3$

$1 : 2$

$2 : 1$

$3 : 1$