A solid sphere of mass 2 kg is rolling on a frictionless horizontal surface with velocity 6m/ s. It collides on the free end of an ideal spring whose other end is fixed. The maximum compression produced in the spring will be (Force constant of the spring = 36N/m)
A. \[\sqrt {14} m\]
B. \[\sqrt {2.8} m\]
C. \[\sqrt {1.4} m\]
D. \[\sqrt {0.7} m\]
Answer
669.9k+ views
Hint: This question demands us to find the maximum compression produced in the spring, as mass of the solid sphere is 2 kg and velocity of the rolling sphere is 6m/s. As this rolling sphere provides us two different kinds of motion rotation and translation motion. So, we will find the required energy to derive our required answer.
Complete step-by-step answer:
Mass of the solid sphere= 2 kg
Velocity of the rolling sphere= 6m/s
So, total energy of the rolling sphere = \[{E_T} = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}I{\omega ^2}\] (cause of translation motion + cause of rotation motion)
As energy is produced so,
\[ \Rightarrow {E_T} = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}I{\omega ^2}\]
In rotation motion formula I = moment of inertia of the solid sphere I = \[\dfrac{2}{5}M{R^2}\]
So,
\[ \Rightarrow {E_T} = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}M{R^2}{\omega ^2}\]
As sphere is rolling so,
The velocity v = \[V = R\omega \]
\[ \Rightarrow {E_T} = \dfrac{1}{2}m{v^2} + \dfrac{1}{5}M{v^2}\]
\[ \Rightarrow {E_t} = m{v^2}\left\{ {\dfrac{1}{2} + \dfrac{1}{5}} \right\}\], as we took \[m{v^2}\] common from both formulas.
\[ \Rightarrow {E_t} = \dfrac{{7M{v^2}}}{{10}}\]
As we know that sphere is rolling and collides with spring the potential energy produced during this collision.
The potential energy of the spring on maximum compression would be x
Potential energy = \[\dfrac{1}{2}k{x^2} = \dfrac{{7M{v^2}}}{{10}}\]
\[ \Rightarrow {x^2} = \dfrac{{7 \times 2M{v^2}}}{{10k}}\] (where k is the spring constant)
Now, we need to keep the required value of mass velocity and spring constant. We can find the value
\[
\Rightarrow {x^2} = \dfrac{{7 \times 2 \times 2 \times {{(6)}^2}}}{{10 \times 36}} \\
\\
\Rightarrow x = \sqrt {2.8} m \\ \]
The maximum compression produced in the spring is \[\sqrt {2.8} m\].
Hence, B is the correct option.
Note- Compression, pressure loss of some item or material induced by tension. Solids, liquids, and gasses and breathing organisms may be exposed to stress. Early on, strain is calculated in the volume of the body at the normal stress under which an individual is subjected — e.g., the air pressure at the sea level is the norm, or equivalent, of the majority of the earth's species.
Complete step-by-step answer:
Mass of the solid sphere= 2 kg
Velocity of the rolling sphere= 6m/s
So, total energy of the rolling sphere = \[{E_T} = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}I{\omega ^2}\] (cause of translation motion + cause of rotation motion)
As energy is produced so,
\[ \Rightarrow {E_T} = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}I{\omega ^2}\]
In rotation motion formula I = moment of inertia of the solid sphere I = \[\dfrac{2}{5}M{R^2}\]
So,
\[ \Rightarrow {E_T} = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}M{R^2}{\omega ^2}\]
As sphere is rolling so,
The velocity v = \[V = R\omega \]
\[ \Rightarrow {E_T} = \dfrac{1}{2}m{v^2} + \dfrac{1}{5}M{v^2}\]
\[ \Rightarrow {E_t} = m{v^2}\left\{ {\dfrac{1}{2} + \dfrac{1}{5}} \right\}\], as we took \[m{v^2}\] common from both formulas.
\[ \Rightarrow {E_t} = \dfrac{{7M{v^2}}}{{10}}\]
As we know that sphere is rolling and collides with spring the potential energy produced during this collision.
The potential energy of the spring on maximum compression would be x
Potential energy = \[\dfrac{1}{2}k{x^2} = \dfrac{{7M{v^2}}}{{10}}\]
\[ \Rightarrow {x^2} = \dfrac{{7 \times 2M{v^2}}}{{10k}}\] (where k is the spring constant)
Now, we need to keep the required value of mass velocity and spring constant. We can find the value
\[
\Rightarrow {x^2} = \dfrac{{7 \times 2 \times 2 \times {{(6)}^2}}}{{10 \times 36}} \\
\\
\Rightarrow x = \sqrt {2.8} m \\ \]
The maximum compression produced in the spring is \[\sqrt {2.8} m\].
Hence, B is the correct option.
Note- Compression, pressure loss of some item or material induced by tension. Solids, liquids, and gasses and breathing organisms may be exposed to stress. Early on, strain is calculated in the volume of the body at the normal stress under which an individual is subjected — e.g., the air pressure at the sea level is the norm, or equivalent, of the majority of the earth's species.
Recently Updated Pages
Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

A Paragraph on Pollution in about 100-150 Words

XIX+XXX A 49 B 51 C 55 D 44 class 5 maths CBSE

If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

