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Question: In CID serial (on TV), Abhijeet fires a bullet of mass \[100g\] with a speed of \(100m{s^{ - 1}}\) o...

In CID serial (on TV), Abhijeet fires a bullet of mass 100g100g with a speed of 100ms1100m{s^{ - 1}} on a soft plywood of thickness 4cm4cm . The bullet emerges with 10%10\% of its initial KE. Find the emergent speed of the bullet.

Explanation

Solution

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=12mv2KE = \dfrac{1}{2}m{v^2}
Here, mm is the mass of the object.
vv is the velocity of the object.

Complete step by step answer:
Here we are given that,
Mass of the bullet fired by Abhijeet = 100g100g
Initial speed or velocity of the bullet when it is fired = 100ms1100m{s^{ - 1}}
Now, it is given in the question that the bullet emerges with only the 10%10\% of its initial Kinetic Energy. So we have,
Final Kinetic Energy = 10%10\% of Initial Kinetic Energy
This can also be written as,
Final Kinetic Energy = 0.10.1 ×\times Initial Kinetic Energy
As we know that, Kinetic Energy is represented by the expression,
KE=12mv2KE = \dfrac{1}{2}m{v^2}
Here, mm is the mass of the object which in this case is bullet.
vv is the velocity of the object.
Let the final velocity of the bullet when it emerges from the soft plywood = vf{v_f}
Now putting the expression for kinetic energy in the equation, Final Kinetic Energy = 0.10.1 ×\times Initial Kinetic Energy
We have,
12mvf2=0.1×12×m×(100ms1)2\dfrac{1}{2}mv_f^2 = 0.1 \times \dfrac{1}{2} \times m \times {(100m{s^{ - 1}})^2}
As mass mm is same on both the sides we can cancel it, also 12\dfrac{1}{2} is same on both sides it will also be cancelled, so we have,
vf2=0.1×10000v_f^2 = 0.1 \times 10000
On solving the RHS we get,
vf2=1000v_f^2 = 1000
After taking square root we have,
vf=31.62ms1{v_f} = 31.62m{s^{ - 1}}

Hence the emergent speed is 31.62ms131.62m{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.