A bumper car with a mass of 240 kg is moving to the right with a velocity of 2.3 m/s. A second bumper car with a mass of 260 kg is moving to the left with a
velocity of 2.7 m/s. After the collision the first car's velocity is 2.9 m/s left, the
velocity of the second car is 2.1 m/s to the right. Prove conservation of momentum.
Assume the surface is frictionless.

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Explanation:

To solve this problem we will take the definition of linear momentum, which tells us that momentum is equal to the product of mass by the velocity vector. Since velocity is a vector, we will take the right-hand movement as positive and the left-hand movement as negative, the left-hand members are taken as before the collision and the right-hand members as after the collision

ΣM1 = ΣM2

(m1*v1) + (m2*v1) = (m1*v2) + (m2*v2)

V1 = velocity before the collision [m/s]

V2 = velocity after the collision [m/s]

m1 and m2 = mass of the vehicles [kg]

Therefore:

(240*2.3) - (260*2.7) = - (240*2.9) + (260*2.1)

Resolving this arithmetic operation we will have:

720 - 702 = - 696 + 546

- 150 = - 150

We can see that before the crash and after the crash the momentum is preserved

speed

ratio of distance traveled to time interval

net force

vector sum of forces on object

object

source of diverging light rays either luminous or illuminated

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explanation: