Answer:
Before:
[tex]p_{truck}=16400\ kg.m/s[/tex]
[tex]p_{car}=10000\ kg.m/s[/tex]
After:
[tex]p_{truck}=8000\ kg.m/s[/tex]
[tex]p_{car}=8400\ kg.m/s[/tex]
[tex]v_{fcar}=8.4\ m/s[/tex]
[tex]F=9333.33 \ Nw[/tex]
Explanation:
Conservation of Momentum
Two objects of masses m1 and m2 moving at speeds v1o and v2o respectively have a total momentum of
[tex]p_1=m_1v_{1o}+m_2v_{2o}[/tex]
After the collision, they have speeds of v1f and v2f and the total momentum is
[tex]p_2=m_1v_{1f}+m_2v_{2f}[/tex]
Impulse J is defined as
[tex]J=F.t[/tex]
Where F is the average impact force and t is the time it lasted
Also, the impulse is equal to the change of momentum
[tex]J=\Delta p[/tex]
As the total momentum is conserved:
[tex]p_1=p_2[/tex]
[tex]m_1v_{1o}+m_2v_{2o}=m_1v_{1f}+m_2v_{2f}[/tex]
We can compute the speed of the second object by solving the above equation for v2f
[tex]\displaystyle v_{2f}=\frac{ m_1v_{1o}+m_2v_{2o} -m_1v_{1f} }{ m_2 }[/tex]
The given data is
[tex]m_1=2000\ kg[/tex]
[tex]m_2=1000\ kg\\v_{1o}=8.2\ m/s\\v_{2o}=0\ m/s\\v_{1f}=4\ m/s[/tex]
a) The impulse will be computed at the very end of the answer
b) Before the collision
[tex]p_{truck}=2000\cdot 8.2=16400\ kg.m/s[/tex]
[tex]p_{car}=1000\cdot 0=0\ kg.m/s[/tex]
c) After collision
[tex]p_{truck}=2000\cdot 4=8000\ kg.m/s[/tex]
Compute the car's speed:
[tex]\displaystyle v_{2f}=\frac{ 16400+0 -8000 }{ 1000 }[/tex]
[tex]v_{2f}=8.4\ m/s[/tex]
And the car's momentum is
[tex]p_{car}=1000\cdot 8.4=8400\ kg.m/s[/tex]
The Impulse J of the system is zero because the total momentum is conserved, i.e. \Delta p=0.
We can compute the impulse for each object
[tex]J_1=\Delta p_1=2000(4-8.2)=-8400 \N.s[/tex]
The force can be computed as
[tex]\displaystyle F=\frac{J}{t}=-\frac{8400}{0.9}=-9333.33 \ Nw[/tex]
The force on the car has the same magnitude and opposite sign
