Answer:
A. The initial velocity of the bullet is [tex]= 300.6m/s[/tex]
B. Mechanical energy of the system before and after collision: 451.80 J, 0.9018 J
C. Percentage of K.E lost to heat is = 99.8 %
Explanation:
From conservation of linear momentum,
[tex](m_{1}v_{1} +m_{2}v_{2})= (m_{1}+m_{2})v[/tex]
let the mass of the block be m1 and velocity = v1
let the mass of the bullet be m2 and velocity = v2
Let the final velocity of the system be v.
A. Plugging our parameters into the equation, we have:
[tex][(5 \times 0) +(0.01\times v_{2})]= 5.01 \times 0.6[/tex]
[tex]v_{2}=\frac{3.006}{0.01}= 300.6m/s[/tex]
Hence, the initial velocity of the bullet is [tex]= 300.6m/s[/tex]
B. The mechanical energies of the system exist in form of kinetic energy.
I. Kinetic energy of the system before collision:
[tex]0.5 \times 5\times 0^{2} + 0.5 \times 0.01 \times 300.6^{2}= 451.80 J[/tex]
II. Kinetic energy after collision:
[tex]0.5\times 5.01 \times 0.6^{2}= 0.9018 J[/tex]
C. Change in Mechanical Energy = [tex]451.8 - 0.9018 J= 450.9J[/tex]
[tex]\frac{450.9}{451.8} \times 100 =99.8%[/tex]
Percentage of K.E lost to heat is = 99.8 %
