Answer:
(a)0.531m/s
(b)0.00169
Explanation:
We are given that
Mass of bullet, m=4.67 g=[tex]4.67\times 10^{-3} kg[/tex]
1 kg =1000 g
Speed of bullet, v=357m/s
Mass of block 1,[tex]m_1=1177g=1.177kg[/tex]
Mass of block 2,[tex]m_2=1626 g=1.626 kg[/tex]
Velocity of block 1,[tex]v_1=0.681m/s[/tex]
(a)
Let velocity of the second block after the bullet imbeds itself=v2
Using conservation of momentum
Initial momentum=Final momentum
[tex]mv=m_1v_1+(m+m_2)v_2[/tex]
[tex]4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2[/tex]
[tex]1.66719=0.801537+1.63067v_2[/tex]
[tex]1.66719-0.801537=1.63067v_2[/tex]
[tex]0.865653=1.63067v_2[/tex]
[tex]v_2=\frac{0.865653}{1.63067}[/tex]
[tex]v_2=0.531m/s[/tex]
Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s
(b)Initial kinetic energy before collision
[tex]K_i=\frac{1}{2}mv^2[/tex]
[tex]k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)[/tex]
[tex]k_i=297.59 J[/tex]
Final kinetic energy after collision
[tex]K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2[/tex]
[tex]K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2[/tex]
[tex]K_f=0.5028 J[/tex]
Now, he ratio of the total kinetic energy after the collision to that before the collision
=[tex]\frac{k_f}{k_i}=\frac{0.5028}{297.59}[/tex]
=0.00169
