A 2,922 kg truck moving with a velocity of 11 m/s hits a 1,318 kg parked car. The impact causes the 1,318 kg car to be set in motion at 18 m/s. What is the velocity of the truck immediately after the collision?

v₃ = 22.67 [m/s]

Explanation:

In order to solve this problem, we must use the principle of conservation of the quantity of linear momentum. Where momentum is conserved before and after the collision, i.e. remains the same.

The terms on the left of the equation represent the amount of linear momentum before the collision and the members on the right represent the momentum after the collision.

where:

m₁ = mass of the truck = 3000 [kg]

m₂ = mass of the car = 1000 [kg]

v₁ = velocity of the truck before the coliision = 25 [m/s]

v₂ = velocity of the car parked = 0 (without movement)

v₃ = velocity of the truck after the collision [m/s]

v₄ = velocity of the car after the collision = 7 [m/s]

Now replacing:

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .

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