Answer :
Answer:
direction = [tex]\frac{4}{\sqrt{97} }i+\frac{9}{\sqrt{97} }j[/tex]
velocity = [tex]\frac{30\times \sqrt{97} }{5}[/tex]
distance skid = 145.2 meters
Explanation:
Let the mass of the car 1 m1 = 1500 kg, m2 = 1000 kg, v1 = 90 km/h and v2 = 60 km/h.
v1 = 90 i and v2 = 60 j.
applying conservation of linear momentum in i and j direction since there is no external force involved.
along i direction:
[tex]m1\times v1 = (m1+m2)\times v\times \cos \Phi[/tex] 1
along j direction:
[tex]m2\times v2 = (m1+m2)\times v\times \sin \Phi[/tex] 2
where v is the new velocity of combined car and Ф is the angle made with x axis.
equation 2÷1
[tex]\tan \Phi =\frac{m2\times v2}{m1\times v1}[/tex]
therefore \tan \Phi = \frac{4}{9}
[tex]\sin \Phi =\frac{4}{\sqrt{97} }[/tex]
[tex]\cos \Phi =\frac{9}{\sqrt{97} }[/tex]
therefore the direction is[tex]\frac{4}{\sqrt{97} }i+\frac{9}{\sqrt{97} }j[/tex]
and velocity will be
[tex]\frac{30\times \sqrt{97} }{5}[/tex]
the car will skid until the total energy of the car is dissipiated by friction
therefore [tex]u\times (m1+m2)\times g\times x=\frac{1}{2}\times (m1+m2)\times v^{2}[/tex]
substituting the values we get x = 145.2 meters