Answer:21.97 m/s
Step-by-step explanation:
Mass of truck
![m_T=2000 kg](https://img.qammunity.org/2020/formulas/physics/high-school/mzciodmx7ic6icq6t415auj10xnyhc6azg.png)
Velocity of truck
![v_c=4 m/s](https://img.qammunity.org/2020/formulas/physics/high-school/c1377bzp4jst7cfl77fl0di2jh2kra3qz5.png)
Mass of car
![m_c=1000 kg](https://img.qammunity.org/2020/formulas/physics/high-school/bzq9ryv64mwgpzkkozf8ugxgjbo9ni63sa.png)
let
be the car velocity and u be the velocity of combined system after collision at angle of
![20^(\circ)](https://img.qammunity.org/2020/formulas/physics/college/z7iop5e386i9vi03xgh1x42yf5ysadeqjv.png)
Conserving momentum in east and North direction Respectively
In east Direction
-------1
In North direction
---------2
Divide 1 and 2 we get
![(m_T\cdot v_T)/(m_c\cdot v_c)=((m_c+m_T)u\sin 20)/((m_c+m_T)u\cos 20)](https://img.qammunity.org/2020/formulas/physics/high-school/xs5ae7dqt479r4mftqlg49zaku8febksx2.png)
![v_c=(m_T\cdot v_T)/(m_c\cdot \tan 20)](https://img.qammunity.org/2020/formulas/physics/high-school/sj6e0wg46y9cxbnq49dhfwdtko6pxpiuyr.png)
![v_c=(2000)/(1000)* (4)/(\tan 20)](https://img.qammunity.org/2020/formulas/physics/high-school/91uyct822zko08d3o3ccghp8to57nwdyzy.png)
![v_c=21.97 m/s](https://img.qammunity.org/2020/formulas/physics/high-school/o0xfseaqjki9cayt5sv9v15gwtl8j0ldo9.png)