Question
Question: In a ballistic demonstration a police officer fires a bullet of mass \[50g\] with speed \[200m{{s}^{...
In a ballistic demonstration a police officer fires a bullet of mass 50g with speed 200ms−1 on soft plywood of thickness 2cm. The bullet emerges with only 10% of its initial kinetic energy. The emergent speed of the bullet is:
(A)210ms−1
(B)2010ms−1
(C)102ms−1
(D)1020ms−1
Solution
Here initial speed and final speed are given; also the mass of the bullet is given. In this case, initial kinetic energy and final kinetic energy can be calculated from the given data. After those two equations are generated and solving them leads to the solution of this sum.
Formula used:
K=21mv2
Complete answer:
In this question given that,
Mass of the bullet mb=50g=50×10−3kg
Initial speed of the bullet vi=200ms−1
Final speed of the bullet vf ms−1
Now we have to find the initial and final energy of the bullet.
For initial kinetic energy of bullet;
Ki=21mbvi2
∴Ki=21×(50×10−3)×(200)2
∴Ki=1000 kg ms−1
For final kinetic energy of bullet;
Kf=21mbvf2
∴Kf=21×(50×10−3)×vf2 .....(1)
But in question given that;
⇒Kf=10% of Ki
⇒Kf=10010×1000 .....(2)
Equating both the equations (1)and(2),
⇒21×(50×10−3)×vf2=10010×1000
⇒vf2=50×10−3100×2
⇒vf2=4000
⇒vf=2010 ms−1
So, the correct answer is “Option B”.
Additional Information:
The kinetic energy of the particle is dependent on how much force is given to the particle and because of that how much speed the particle will gain. Kinetic energy also depends on the mass of the particle. In all cases to conserve the momentum of the system, if the body has less mass comparatively another part of the system it has high speed and high kinetic energy. In this particular question, kinetic energy plays a very important role. In mechanics total energy of the system is conserved but in the case of a particular energy, it may not be conserved.
Note:
The energy conservation law is not applicable here because final energy is 10% of initial energy so 90 of energy lost during bullet and plywood collision. So it is better to not use energy conservation law if only one form of energy is given and another form is not present.