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physics.fisikastudycenter.com- An example of collision in two dimensions problem. Some literatures of our high school physics start to use this topic as an enrichment. The previous ones just discuss about one dimension collision.

This one a sample for understanding this topic especially for totally inelastic collision when the objects stick together just after collision and have the same final velocity.

Question
In an intersection, a truck of 7500 kg travelling east with velocity of 32 m/s collides with another truck of 7500 kg travelling north at 24 m/s.
After this collision the two trucks stick together with velocity v'.
a) What is the velocity of the trucks just after collision?
b) What is the direction of the trucks just after collision?

Discussion
Consider this case as inelastic collisions in two dimensions. The initial velocities and masses of the trucks are:
m₁ = 7500 kg
m₂ = 7500 kg
v₁ = 32 m/s
v₂ = 24 m/s

These calculations including both two axis, x and y with v' is the final velocity just after collision and θ is an angle formed by the final velocity.

By applying the conservation of momentum along x axis and considering an inelastic collision we have the first equation :
m₁v₁ + m₂v₂ = (m₁ + m₂)v'
7500(32) + 7500(0) = (7500 + 7500)v'cosθ
7500(32) = (15000)v'cosθ
16 = v'cosθ ....eq(i)

The conservation of momentum along Y axis:
m₁v₁ + m₂v₂ = (m₁ + m₂)v'
7500(0) + 7500(24) = (7500 + 7500)v'sinθ
7500(24) = (15000)v'sinθ
12 = v'sinθ ....eq(ii)

By combining (dividing) eq (ii) and (i) we have the direction of final velocity:
12 = v'sinθ
------------------
16 = v'cosθ

sinθ/cosθ = 12/16
tanθ = 12/16 = 3/4
θ = 37°

Finding the solution of v'
v'sinθ = 12
v' = 12/sinθ
v' = 12/sin37°
v' ]= 12/(0,6) = 20 m/s

So then the answers are:
a) What is the velocity of the trucks just after collision?
20 m/s
b) What is the direction of the trucks just after collision?
37°