A car of mass 2200 kg collides with a truck of mass 4800 kg, and just after the collision the car and truck slide along, stuck together. The car's velocity just before the collision was < 31, 0, 0 > m/s, and the truck's velocity just before the collision was < -14, 0, 22 > m/s. (a) What is the velocity of the stuck-together car and truck just after the collision? explanation please :)

Explanation:

Given that

Mass of car

Mc = 2200kg

Mass of truck

Mt = 4800kg

After collision, the car and the truck stuck together, i.e. they are moving together at a constant velocity, this is called inelastic collision

Car velocity before collision

Vc = (31, 0, 0) m/s

Truck velocity before collision

Vt = (-14, 0, 22)m/s

Since they have a common velocity after collision, let called the velocity V.

So, applying conservation of linear momentum

Momentum before collision equal momentum after collision

P(before) = P(after)

Mc•Vc + Mt•Vt = (Mc+Mt)•V

2200(31,0,0) + 4800(-14,0,22) = (2200+4800)•V

(68,200, 0, 0)+(-67,200, 0, 105,600) = 7000•V

(68,200—67,200, 0+ 0, 0 +105,600)

= 7000V

(1000, 0, 105,600) = 7000V

Divide through by 7000

V = (0.143, 0, 15.1) m/s

Then,

Magnitude of the velocity is

V = √Vx²+Vy²+Vz²

V = √0.143²+0²+15.1²

V =15.1 m/s

its c i just checked it

not too sure thinking all accelerations are pointed down