Answer:
a).[tex]d=848.3m[/tex]
b).K=1600J
Explanation:
Those kind of question have the typical questions like the final distance and the total energy of the motion so those question are solve and any other information you can get from here
a).How far from the point of firing does the other fragment strike if the terrain is level?
Using Newton's laws for a accelerated motion
The motion have two steps as we see in the image so:
[tex]v_{iy}=V*sin(60)=80*sin(60)=69.28\frac{m}{s}[/tex]
[tex]v_{fy}=v_{iy}+a_{y}*t[/tex]
[tex]a_{t}=g=9.8\frac{m}{s^2}[/tex]
Solve to t
[tex]t=\frac{69.28 m/s}{9.8 m/s^2}=7.069s[/tex]
Now determinate the height of the motion
[tex]h_{f}=v_{if}*t+\frac{1}{2}*a_{y}*t^2[/tex]
[tex]h_{f}=69.2m/s*7.07s+\frac{1}{2}*-9.8m/s^2*(7.07)^2[/tex]
[tex]h_{f}=244.88m[/tex]
[tex]v_{ix}=V*cos(60)=80*cos(60)=40\frac{m}{s}[/tex]
[tex]x_{1}=v_{ix}*t=40m/s*7.07s=282.8m[/tex]
Now the momentum of each particle using conversation of momentum can find the speed in the second step of the motion
[tex]P=m*v=20kg*40m/s=800 \frac{kg*m}{s}[/tex]
[tex]P_{a}=P_{b}=\frac{m}{2}*v_{x2}[/tex]
[tex]V_{x2}=\frac{800}{\frac{20}{2}}=80m/s[/tex]
[tex]y_{f}=y_{i}+v_{iy2}*t_{2}+\frac{1}{2}*a_{y}*t_{2}^2=0-245=-4.9m/s^2*t_{2}^2[/tex]
[tex]t_{2}=\sqrt{\frac{245m}{4.9m/s^2}}=7.07s[/tex]
[tex]x_{2}=t_{2}*v_{x2}=7.07*80m/s=565.4m[/tex]
Total distance:
[tex]d=565.4+282.8m=848.3m[/tex]
b).
How much energy is released during the explosion?
Kinetic energy
ΔK=K2-K1
[tex]K=\frac{1}{2}*m*v_{x2}^2+\frac{1}{2}*m*v_{x1}^2-\frac{1}{2}*m*v_{x}^2[/tex]
[tex]K=\frac{1}{2}*20*(80m/s)^2-\frac{1}{2}*20*(40m/s)^2[/tex]
K=1600J
