Question

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.

Answer 1

See the explanation below

Step-by-step 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