Answer:
0.07339 m
-42857.1428571 N
Explanation:
[tex]m_1[/tex] = Mass of bullet= 0.01 kg
[tex]m_2[/tex] = Mass of block = 5 kg
[tex]u_1[/tex] = Initial Velocity of bullet = 1000 m/s
[tex]u_2[/tex] = Initial Velocity of block = 0 m/s
[tex]v_1[/tex] = Initial Velocity of bullet = 1000 m/s
[tex]v_2[/tex] = Final Velocity of block
[tex]m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow v_2=\frac{m_{1}u_{1}+m_{2}u_{2}-m_1v_1}{m_2}\\\Rightarrow v_2=\frac{0.01\times 1000+0-0.01\times 400}{5}\\\Rightarrow v_2=1.2\ m/s[/tex]
[tex]KE=PE\\\Rightarrow \frac{1}{2}mv_2^2=mgh\\\Rightarrow h=\frac{1}{2}\frac{v^2}{g}\\\Rightarrow h=\frac{1}{2}\frac{1.2^2}{9.81}\\\Rightarrow h=0.07339\ m[/tex]
The maximum height the block rises above its initial position is 0.07339 m
[tex]v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{400-1000}{1.4\times 10^{-4}}\\\Rightarrow a=-4285714.28571\ m/s^2[/tex]
[tex]F=ma\\\Rightarrow F=0.01\times -4285714.28571\\\Rightarrow F=-42857.1428571\ N[/tex]
The force that the bullet exerts on the block is -42857.1428571 N
