A 1350 kg car moving north at a speed of 35.0 m/s colliges with a truck that is moving at an agnale of north of west at a speed of 27.5 the car and teh truck stick togeter after the collision what is the magnituge of their common velocity after the collision?

Question:

A 1350 kg car moving north at a speed of 35.0 m/s collides with a 2300 kg truck that is moving at an angle of 32 degrees north of west at a speed of 23.0 m/s. The car and the truck stick together after the collision. What is the magnitude of their common velocity after the collision?

The magnitude of their common velocity after the collision is 24 m/s

Explanation:

Here we have

Sum of total initial momentum = Sum of total final momentum

Initial momentum = m₁·v₁ + m₂·v₂

Where:

m₁ = Mass of car = 1350 kg

v₁ = Initial velocity of car = 35.0 m/s north

m₂ = Mass of truck = 2300 kg

v₂ = Velocity of truck = 23 m/s 32° north of west

Therefore

v₂ north = 23 × sin 32 = 12.2 m/s and

v₂ west = 23 × cos 32 = 19.51 m/s

Sum of intitial momentum

= 1350 × 35 i  + 2300× 12.2 i + 2300×19.51 j = 47250 i + 28032.73 i + 44861.744 j

= 75282.73 i + 44861.744 j

Magnitude of the total momentum =

∴ Magnitude of the total momentum = 87635.98 kg·m/s

Total mass after collision = 1350 + 2300 = 3650 kg

Therefore, the magnitude of their common velocity after the collision is given by;

Magnitude of the total momentum ÷ Total mass after collision

= 87635.98 kg·m/s ÷ 3650 kg  = 24.009 m/s  ≈ 24 m/s.

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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