Question

the ratio of the heights from which two bodies are dropped is 3:5, respectively.what is the ratio of their final velocity​

Answer 1

SoluTion :

✴ Given :-

▪ The ratio of the heights from which two bodies are dropped is 3:5 respectively.

✴ To Find :-

▪ Ratio of their final velocity

✴ Concept :-

✈ This question is completely base on concept of 'Energy conservation'.

✴ Calculation :-

$\implies\sf\:\red{initial\:potential\:energy=final\:kinetic\:energy}\\ \\ \implies\sf\:\dfrac{m_1gH_1}{m_2gH_2}=\dfrac{\frac{1}{2}m_1{v_1}^2}{\frac{1}{2}m_2{v_2}^2}\\ \\ \implies\sf\:\dfrac{H_1}{H_2}=\dfrac{{v_1}^2}{{v_2}^2}\\ \\ \implies\sf\:\dfrac{3}{5}=\dfrac{{v_1}^2}{{v_2}^2}\\ \\ \implies\sf\:\dfrac{v_1}{v_2}= \sqrt{\dfrac{3}{5}}\\ \\ \implies\:\boxed{\sf{\pink{\large{v_1:v_2=\sqrt{3}:\sqrt{5}}}}}$

Answer 2

GiveN :

• Ratio of heights of the two bodies are dropped is 3:5

To FinD :

• Ratio of their final velocity

SolutioN :

As we know from the law of conservation of energy :

Kinetic Energy = Potential Energy

$\dashrightarrow \tt{\dfrac{\dfrac{1}{2} m_1 v_1 ^2}{\dfrac{1}{2} m_2 v_2 ^2} \: = \: \dfrac{m_1 g h_1}{m_2 g h_2}} \\ \\ \dashrightarrow \tt{\dfrac{v_1^2}{v_2 ^2} \: = \: \dfrac{3}{5}} \\ \\ \dashrightarrow \tt{\bigg( \dfrac{v_1}{v_2} \bigg) ^2 \: = \: \dfrac{3}{5}} \\ \\ \dashrightarrow \tt{\dfrac{v_1}{v_2} \: = \: \sqrt{\dfrac{3}{5}}} \\ \\ \dashrightarrow \tt{v_1 : v_2 \: = \: \sqrt{3} : \sqrt{5}}$

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Additional information :

• Kinetic energy is the energy that comes into play when the object is in state of motion.

• Potential energy comes to play when object is in state of rest.

• SI unit of energy is Joule (J)