Physics, asked by nehadhurat91267, 1 month ago

Two bodies A and B have their masses in the ratio 3:2 and their potential energies are in the ratio 2:3. What is the ratio of the height to which they are raised?​

Answers

Answered by bahukhandirenu39
0

Explanation:

m

2

m

1

=

1

3

and

KE

2

KE

1

=

3

1

2

1

m

2

v

2

2

2

1

m

1

v

2

1

=

3

1

m

2

m

1

λ(

v

2

v

1

)

2

=

3

1

3(

v

2

v

1

)

2

=

3

1

v

2

v

1

=

9

1

=1:3

Answered by NewGeneEinstein
2

Let

  • masses are 3m and 2m
  • Potential energies=2p,3p

Acceleration due to gravity(g) is constant

We know

\boxed{\sf P.E=mgh}

From first case

\\ \sf\longmapsto 2p=3mgh_1

\\ \sf\longmapsto h_1=\dfrac{2p}{3mg}

From second case

\\ \sf\longmapsto 3p=2mgh_2

\\ \sf\longmapsto h_2=\dfrac{3p}{2mg}

Ratio

\\ \sf\longmapsto \dfrac{h_1}{h_2}=\dfrac{\dfrac{2p}{3mg}}{\dfrac{3p}{2mg}}

  • cancel p,m,g

\\ \sf\longmapsto \dfrac{h_1}{h_2}=\dfrac{2/3}{3/2}

\\ \sf\longmapsto \dfrac{h_1}{h_2}=\dfrac{2\times 2}{3\times 3}

\\ \sf\longmapsto \dfrac{h_1}{h_2}=\dfrac{4}{9}

\\ \bf\longmapsto h_1:h_2=4:9

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