
In CID serial (on TV), Abhijeet fires a bullet of mass \[100g\] with a speed of $100m{s^{ - 1}}$ on a soft plywood of thickness $4cm$ . The bullet emerges with $10\% $ of its initial KE. Find the emergent speed of the bullet.
Answer
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Hint: In order to solve this question, the concept of energy conservation is important. Here initial energy of the bullet is given and also the energy lost when the bullet hits the soft plywood is also given. The formula of Kinetic energy contains the velocity in it. Hence, for calculating the emergent speed of velocity use the concept of kinetic energy.
Formula Used:
Kinetic Energy is given by,
$KE = \dfrac{1}{2}m{v^2}$
Here, $m$ is the mass of the object.
$v$ is the velocity of the object.
Complete step by step answer:
Here we are given that,
Mass of the bullet fired by Abhijeet = \[100g\]
Initial speed or velocity of the bullet when it is fired = $100m{s^{ - 1}}$
Now, it is given in the question that the bullet emerges with only the $10\% $ of its initial Kinetic Energy. So we have,
Final Kinetic Energy = $10\% $ of Initial Kinetic Energy
This can also be written as,
Final Kinetic Energy = $0.1$ $ \times $ Initial Kinetic Energy
As we know that, Kinetic Energy is represented by the expression,
$KE = \dfrac{1}{2}m{v^2}$
Here, $m$ is the mass of the object which in this case is bullet.
$v$ is the velocity of the object.
Let the final velocity of the bullet when it emerges from the soft plywood = ${v_f}$
Now putting the expression for kinetic energy in the equation, Final Kinetic Energy = $0.1$ $ \times $ Initial Kinetic Energy
We have,
$\dfrac{1}{2}mv_f^2 = 0.1 \times \dfrac{1}{2} \times m \times {(100m{s^{ - 1}})^2}$
As mass $m$ is same on both the sides we can cancel it, also $\dfrac{1}{2}$ is same on both sides it will also be cancelled, so we have,
$v_f^2 = 0.1 \times 10000$
On solving the RHS we get,
$v_f^2 = 1000$
After taking square root we have,
${v_f} = 31.62m{s^{ - 1}}$
Hence the emergent speed is $31.62m{s^{ - 1}}$.
Note: Here in the above question we have used speed and velocity interchangeably yet these two the different physical quantities, the reason is that the magnitude of both speed and velocity is the same. The difference is that speed is a scalar quantity but velocity is a vector quantity. Velocity in simple terms is speed with direction.
Formula Used:
Kinetic Energy is given by,
$KE = \dfrac{1}{2}m{v^2}$
Here, $m$ is the mass of the object.
$v$ is the velocity of the object.
Complete step by step answer:
Here we are given that,
Mass of the bullet fired by Abhijeet = \[100g\]
Initial speed or velocity of the bullet when it is fired = $100m{s^{ - 1}}$
Now, it is given in the question that the bullet emerges with only the $10\% $ of its initial Kinetic Energy. So we have,
Final Kinetic Energy = $10\% $ of Initial Kinetic Energy
This can also be written as,
Final Kinetic Energy = $0.1$ $ \times $ Initial Kinetic Energy
As we know that, Kinetic Energy is represented by the expression,
$KE = \dfrac{1}{2}m{v^2}$
Here, $m$ is the mass of the object which in this case is bullet.
$v$ is the velocity of the object.
Let the final velocity of the bullet when it emerges from the soft plywood = ${v_f}$
Now putting the expression for kinetic energy in the equation, Final Kinetic Energy = $0.1$ $ \times $ Initial Kinetic Energy
We have,
$\dfrac{1}{2}mv_f^2 = 0.1 \times \dfrac{1}{2} \times m \times {(100m{s^{ - 1}})^2}$
As mass $m$ is same on both the sides we can cancel it, also $\dfrac{1}{2}$ is same on both sides it will also be cancelled, so we have,
$v_f^2 = 0.1 \times 10000$
On solving the RHS we get,
$v_f^2 = 1000$
After taking square root we have,
${v_f} = 31.62m{s^{ - 1}}$
Hence the emergent speed is $31.62m{s^{ - 1}}$.
Note: Here in the above question we have used speed and velocity interchangeably yet these two the different physical quantities, the reason is that the magnitude of both speed and velocity is the same. The difference is that speed is a scalar quantity but velocity is a vector quantity. Velocity in simple terms is speed with direction.
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