Answer:
a
The value is [tex]v = 0.8351 \ m/s [/tex]
b
The ratio is [tex]\frac{K_a}{K_b} = 5.94 11 *10^{-6}[/tex]
Explanation:
From the question we are told that
The mass of the bullet is [tex]m_b = 6.16 \ g = 0.00616 \ kg[/tex]
The velocity is [tex]u_b = 342 \ m/s[/tex]
mass of the first block is [tex]m__{{B_1}}} = 1155 \ g = 1.155 \ kg[/tex]
The velocity of the first block after bullet passes is [tex]v__{{B_1}}} = 0.728 m/s[/tex]
The mass of the second block is [tex]m__{{ B_2}}} = 1517 g = 1.517 \ kg[/tex]
Gnerally according to the law of momentum conservation
[tex]m_b * u_b + m__{{B_1}}} * u__{{B_1}}} = m__{{B_1}}} * v__{{B_1}}} + v_b * m_b[/tex]
Here [tex]v_b[/tex] is the velocity of the bullet emerging from the first block
and [tex]u__{{B_1}}}[/tex] is zero because initial the first block was at rest
So
[tex]0.00616 * 342 + 1.155* 0 = v_b * 0.00616 + 1.155 * 0.728 [/tex]
[tex]v_b = 206.5 \ m/s [/tex]
Considering the second block
Gnerally according to the law of momentum conservation
[tex]m_b * v_b + m__{{B_2}}} * u__{{B_2}}} = [m__{{B_2}}} +m_b] v[/tex]
Here [tex]u__{{B_2}}}[/tex] is zero because initial the second block was at rest
=> [tex]0.00616 * 206.5 + 1.517* 0= [1.517 +m_b] v [/tex]
=> [tex]0.00616 * 206.5 = [1.517 +0.00616] v [/tex]
=> [tex]v = 0.8351 \ m/s [/tex]
The kinetic energy of the bullet before collision is
[tex]K_b = \frac{1}{2} * m * u_b^2[/tex]
=> [tex]K_b = 0.5 * 0.00616 * 342^2[/tex]
=> [tex]K_b = 360.2 \ J [/tex]
The kinetic energy of the bullet after collision is
[tex]K_a = \frac{1}{2} * m * v^2[/tex]
=> [tex]K_a = 0.5 * 0.00616 * 0.8351^2[/tex]
=> [tex]K_a = 0.00214 \ J [/tex]
Generally the ratio of the kinetic energy is mathematically represented as
[tex]\frac{K_a}{K_b} = \frac{0.00214}{360.2 }[/tex]
=> [tex]\frac{K_a}{K_b} = \frac{0.00214}{360.2 }[/tex]
=> [tex]\frac{K_a}{K_b} = 5.94 11 *10^{-6}[/tex]
