Answer:
Percentage loss is 99.5%
Solution:
As per the question:
Mass of the bullet, m = 20 g = 0.020 kg
Mass of plate, [tex]M_{1} = 1\ kg[/tex]
Mass of the second plate, [tex]M_{2} = 2.98\ kg[/tex]
Now,
To calculate the percentage loss:
By using the principle of conservation of energy:
Since, the velocity with which the bullet is fired is [tex]v_{b}[/tex] and then after piercing through the plates the bullet gets embed in it and hence moves with the same velocity, v as that of the plates:
[tex]mv_{b} = (M_{1} + M_{2} + m)v[/tex]
[tex]v_{b}[/tex] = initial velocity of the bullet
v = final velocity of the combination
Thus by substituting the appropriate values in the eqn to find 'v':
[tex]v = \frac{0.020}{1 + 2.98 + 0.020}v_{b} = 0.005v_{b}[/tex]
Now, the loss in the velocity of the bullet:
[tex]v_{b} - 0.005v_{b} = 0.995v_{b}[/tex]
Percentage loss = [tex]\frac{0.995v_{b}}{v_{b}}\times 100[/tex] = 99.5%
