A 1050 kg car is traveling north at 18 m/s. It approaches an intersection, where a second 1900 kg truck is traveling east with a velocity of 13 m/s. All occupants of the vehicles are wearing seat belts and luckily there were no injuries, but they collide in the intersection. The car and truck stick together in the collision and move away from the collision point as a single object. Find the angle that the velocity of the combined system makes with the initial directions of the vehicles. (Do not enter units for this answer.)

Explanation:

Given a truck of mass

M1=1050kg

At velocity v1= 18m/s north

Second truck of mass

M2=1900kg

Moving at a velocity of v2= 13m/s eastward

This shows that the first truck is traveling northward from south

And the second truck is traveling eastward from the west.

Then they collide at the origin

Then,using law of conservation of momentum,

Since it is inelastic collision, the two body stick together and move together, let their common velocity be V

Momentum before collision= momentum after collision

M1v1+M2v2=(M1+M2)V

1050×18+1900×13=(1050+1900)V

43600=2950V

V=43600/2950

V=14.78m/s

Their common velocity is 14.78m/s, in the direction between north and east

Then,

The momentum along x axis is M1v1

1050×18=18900 kgm/s

The momentum along y axis is M2v2,

1900×13= 24700 kgm/s

Then the angle after collision is

θ=arctan(y/x)

θ=arctan(24700/18900)

θ=arctan(1.3069)

θ=52.58°

lol

explanation:

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