Answer:
6. 22 m/s
Explanation:
Assuming no external forces present during the collision, total momentum must be conserved. As momentum is a vector, if we decompose it in two components mutually perpendicular, both components must be conserved too.
We must have into account that after the collision, both masses keep moving together, in a direction 20º North of East.
In our case, it is advisable to project the momentum vector along directions N-S (our y-axis) and W-E (our x-axis), so we can write the following algebraic equations:
Δpx = 0⇒ px₀ = pxf
⇒ mcar*vcar₀ = (mc+mT)* vf* cos 20º
⇒ 1000 kg*vcar₀ = 3000 kg*vf*cos 20º (1)
Δpy = 0 ⇒ py₀ = pyf
⇒ mT*vT₀ = (mc+mT)*vf*sin 20º (2)
⇒2000 kg*4m/s = 3000 kg*vf* sin 20º
Dividing both sides of (2) by (1), we have:
tg 20º = 8000 kg*m/s / 1000 kg*vcar₀
Solving for vcar₀, we get:
vcar₀ = 8 / tg 20º m/s = 22 m/s