Answer
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Hint: The elastic energy stored in the spring due to compression of the spring by the ball is transferred as kinetic energy of the ball. Use the formula for elastic potential energy to calculate the elastic potential energy stored in the spring. Equate this energy with the kinetic energy of the ball and calculate the velocity.
Formula used:
The elastic potential energy of the spring is,
\[U = \dfrac{1}{2}k{x^2}\]
Here, x is the compression in the spring.
The kinetic energy of the ball is,
\[K = \dfrac{1}{2}m{v^2}\]
Here, m is the mass and v is the velocity of the ball.
Complete step by step answer:
We know that according to conservation of energy, the loss in potential energy is equal to gain in kinetic energy of the body. As the spring is compressed by the ball, the elastic potential energy gets stored in the spring.
We know that elastic energy stored in the spring of spring constant k due to compression in the spring is given by,
\[U = \dfrac{1}{2}k{x^2}\]
Here, x is the compression in the spring.
We have given the spring constant \[k = 90\,N/cm = 9000\,N/m\] and compression in the spring \[x = 12\,cm = 0.12\,m\].
Therefore, we substitute \[k = 9000\,N/m\] and \[x = 0.12\,m\] in the above equation.
\[U = \dfrac{1}{2}\left( {9000} \right){\left( {0.12} \right)^2}\]
\[ \Rightarrow U = 64.8\,J\]
Now, this elastic potential energy is converted into a kinetic energy of the ball. We can write,
\[U = K = \dfrac{1}{2}m{v^2}\]
\[ \Rightarrow 64.8\,J = \dfrac{1}{2}m{v^2}\]
Here, m is the mass of the ball.
Substitute \[m = 16\,g = 16 \times {10^{ - 3}}\,kg\] in the above equation.
\[ \Rightarrow 64.8\,J = \dfrac{1}{2}\left( {16 \times {{10}^{ - 3}}} \right){v^2}\]
\[ \Rightarrow 8100 = {v^2}\]
\[ \Rightarrow v = \sqrt {8100} \]
\[ \Rightarrow v = 90\,m/s\]
So, the correct answer is “Option B”.
Note:
Students might think how come the velocity of the ball would be \[90\,m/s\]. This is almost 324 km/hr. The reason is the huge spring constant of \[9000\,N/m\]. So, if you think the velocity of the ball is huge, then you should cross check the spring constant of the spring.
Formula used:
The elastic potential energy of the spring is,
\[U = \dfrac{1}{2}k{x^2}\]
Here, x is the compression in the spring.
The kinetic energy of the ball is,
\[K = \dfrac{1}{2}m{v^2}\]
Here, m is the mass and v is the velocity of the ball.
Complete step by step answer:
We know that according to conservation of energy, the loss in potential energy is equal to gain in kinetic energy of the body. As the spring is compressed by the ball, the elastic potential energy gets stored in the spring.
We know that elastic energy stored in the spring of spring constant k due to compression in the spring is given by,
\[U = \dfrac{1}{2}k{x^2}\]
Here, x is the compression in the spring.
We have given the spring constant \[k = 90\,N/cm = 9000\,N/m\] and compression in the spring \[x = 12\,cm = 0.12\,m\].
Therefore, we substitute \[k = 9000\,N/m\] and \[x = 0.12\,m\] in the above equation.
\[U = \dfrac{1}{2}\left( {9000} \right){\left( {0.12} \right)^2}\]
\[ \Rightarrow U = 64.8\,J\]
Now, this elastic potential energy is converted into a kinetic energy of the ball. We can write,
\[U = K = \dfrac{1}{2}m{v^2}\]
\[ \Rightarrow 64.8\,J = \dfrac{1}{2}m{v^2}\]
Here, m is the mass of the ball.
Substitute \[m = 16\,g = 16 \times {10^{ - 3}}\,kg\] in the above equation.
\[ \Rightarrow 64.8\,J = \dfrac{1}{2}\left( {16 \times {{10}^{ - 3}}} \right){v^2}\]
\[ \Rightarrow 8100 = {v^2}\]
\[ \Rightarrow v = \sqrt {8100} \]
\[ \Rightarrow v = 90\,m/s\]
So, the correct answer is “Option B”.
Note:
Students might think how come the velocity of the ball would be \[90\,m/s\]. This is almost 324 km/hr. The reason is the huge spring constant of \[9000\,N/m\]. So, if you think the velocity of the ball is huge, then you should cross check the spring constant of the spring.
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