Answer:
a) [tex]v_{3} =8.43 m/s[/tex]
b) [tex]v_{2}=2.15m/s[/tex]
c) ΔK=[tex]-28.18x10^4J[/tex]
d)ΔK=[tex]-10.33x10^4J[/tex]
Explanation:
From the exercise we know that there is a collision of a sports car and a truck.
So, the sport car is going to be our object number 1 and the truck object number 2.
[tex]m_{1}=1050kg\\v_{1}=-13m/s\\m_{2}=6320kg\\v_{2}=12m/s[/tex]
Since the two vehicles remain locked together after the collision the final mass is:
[tex]m_{3}=7370kg[/tex]
a) To find the velocity of the two vehicles just after the collision we must use linear's momentum principle
[tex]p_{1}=p_{2}[/tex]
[tex]m_{1}v_{1}+ m_{2}v_{2}=m_{3}v_{3}[/tex]
[tex]v_{3}=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{3}}=\frac{(1050kg)(-13m/s)+(6320kg)(12m/s)}{7370kg}[/tex]
[tex]v_{3}=8.43m/s[/tex]
b) To find the speed the truck should have had so both vehicles stopped in the collision we need to use the same principle used before
[tex]m_{1}v_{1}+ m_{2}v_{2}=0[/tex]
[tex]v_{2}=\frac{-m_{1}v_{1}}{m_{2} }=\frac{-(1050kg)(-13m/s)}{(6320kg)}=2.15m/s[/tex]
c) To find the change in kinetic energy we need to do the following steps:
ΔK=[tex]k_{2}-k_{1}=\frac{1}{2}m_{3}v_{3}^{2}-(\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2} )[/tex]
ΔK=[tex]\frac{1}{2}(7370)(8.43)^{2}-(\frac{1}{2}(1050)(-13)^{2}+\frac{1}{2}(6320)(12)^{2} )=-28.18x10^{4}J[/tex]
d) The change in kinetic energy where the two vehicles stopped in the collision is:
ΔK=[tex]k_{2}-k_{1}=0-(\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2} )[/tex]
ΔK=[tex]-(\frac{1}{2}(1050)(-13)^{2}+\frac{1}{2}(6320)(2.15)^{2} )=-10.33x10^4J[/tex]
