Answer:21.97 m/s
Explanation:
Mass of truck [tex]m_T=2000 kg[/tex]
Velocity of truck [tex]v_c=4 m/s[/tex]
Mass of car [tex]m_c=1000 kg[/tex]
let [tex]v_c[/tex] be the car velocity and u be the velocity of combined system after collision at angle of [tex]20^{\circ}[/tex]
Conserving momentum in east and North direction Respectively
In east Direction
[tex]m_c\cdot v_c=(m_c+m_T)u\cos 20[/tex]-------1
In North direction
[tex]m_T\cdot v_T=(m_c+m_T)u\sin 20[/tex]---------2
Divide 1 and 2 we get
[tex]\frac{m_T\cdot v_T}{m_c\cdot v_c}=\frac{(m_c+m_T)u\sin 20}{(m_c+m_T)u\cos 20}[/tex]
[tex]v_c=\frac{m_T\cdot v_T}{m_c\cdot \tan 20}[/tex]
[tex]v_c=\frac{2000}{1000}\times \frac{4}{\tan 20}[/tex]
[tex]v_c=21.97 m/s[/tex]
