Respuesta :
Answer:
velocity
a). [tex]v_{f}=235.2 (m/s)[/tex]
direction
b).[tex]\beta =71.3[/tex]
kinetic energy
c).[tex]E_{K}=225.97N[/tex]
Explanation:
Those kind of question have two typical questions inside to response the final velocity and the direction finding the angle and finally the kinetic energy so each the information you need can find there so
a).
[tex]m_{b}*v_{b}+m_{t}*v_{t}=(m_{b}+m_{t})*v_{f}[/tex]
solve to vf
[tex]v_{fx}=\frac{m_{b}*v_{b}*cos(21.3)+m_{t}*v_{t}*cos(15.4)}{m_{b}+m_{t}}=\frac{5.12x10^{-3}kg*208(m/s)0.931+3.05x10^{-3}kg*228(m/s)*0.964}{(5.12x10^{-3}+3.05x10^{-3}kg)}[/tex]
[tex]v_{fx}=222.94(m/s)[/tex]
[tex]v_{fy}=\frac{m_{b}*v_{b}*sin(21.3)+m_{t}*v_{t}*sin(15.4)}{m_{b}+m_{t}}=\frac{5.12x10^{-3}kg*208(m/s)0.363+3.05x10^{-3}kg*228(m/s)*0.265}{(5.12x10^{-3}+3.05x10^{-3}kg)}[/tex]
[tex]v_{fy}=75.21(m/s)[/tex]
[tex]V_{f}=\sqrt{V_{fx}^2+V_{fy}}=\sqrt{222.94^2+75.21^2}[/tex]
[tex]V_{f}=235.28\frac{m}{s}[/tex]
b).
[tex]Tan(\beta)=\frac{v_{fx}}{v_{fy}}[/tex]
solve to β
[tex]\beta =tan^-1*(\frac{222.94}{75.4})[/tex]
[tex]\beta=tan^{-1}*2.956[/tex]
[tex]\beta =71.3[/tex]
c).
Final kinetic energy
[tex]E_{k}=\frac{1}{2}m_{t}*V_{f}^2[/tex]
[tex]E_{k}=\frac{1}{2}(5.12x10^{-3}+3.05x10^{-3})kg*(235.2m/s)^2[/tex]
[tex]E_{K}=225.97N[/tex]
The velocity after the collision of the two bullets In an intense battle, gunfire is 235.28 m/s.
What is conservation of momentum?
Momentum of a object is the force of speed of it in motion. Momentum of a moving body is the product of mass times velocity.
When the two objects collides, then the initial collision of the two body is equal to the final collision of two bodies by the law of conservation of momentum.
[tex]m_1v_1+m_2v_2=(m_1+m_2)v'[/tex]
In an intense battle, gunfire can so concentrated that bullets from opposite sides collide in midair.
The bullet, one with mass 5.12 g moving to the right at a speed 208 m/s directed 21.3∘ above the horizontal.
This bullet collides and fuses with another with mass 3.05 g moving to the left at a speed 282 m/s directed 15.4∘ above the horizontal.
Thus by the above formula for x- direction,
[tex](5.12\times10^{-3})(208\cos 21.3)+(3.05\times10^{-3})(282\cos 15.4)=(5.12\times10^{-3}+3.05\times10^{-3})v_x\\v_x=222.94\rm m/s[/tex]
For y-direction,
[tex](5.12\times10^{-3})(208\sin21.3)+(3.05\times10^{-3})(282\sin15.4)=(5.12\times10^{-3}+3.05\times10^{-3})v_y\\v_y=75.21\rm m/s[/tex]
The value of velocity after the collision,
[tex]V=\sqrt{v_x^2+v_y^2}\\V=\sqrt{22.94^2+75.21^2}\\V=235.28\rm \; m/s[/tex]
Hence, the velocity after the collision of the two bullets In an intense battle, gunfire is 235.28 m/s.
Learn more about the conservation of momentum here;
https://brainly.com/question/7538238
