Question

A toy car having mass m = 1.50 kg collides inelastically with a toy train of mass M = 3.60 kg. Before the collision, the toy train is moving in the positive x-direction with a velocity of Vi = 2.45 m/s and the toy car is also moving in the positive x-direction with a velocity of vi = 4.75 m/s. Immediately after the collision, the toy car is observed moving in the positive x-direction with a velocity of 2.15 m/s.

(a) Determine Vf, the final velocity of the toy train.

Answer 1

`v_(f)` = 3,126 m / s

Step-by-step explanation:

In a crash exercise the moment is conserved, for this a system formed by all the bodies before and after the crash is defined, so that the forces involved have been internalized.

the car has a mass of m = 1.50 kg a speed of v1 = 4.758 m / s and the mass of the train is M = 3.60 kg and its speed v2 = 2.45 m / s

Before the crash

p₀ = m v₁₀ + M v₂₀

After the inelastic shock

`p_(f)`= m
`v_(1f)` + M
`v_(2f)`

p₀ =
`p_(f)`

m v₀ + M v₂₀ = m
`v_(1f)` + M
`v_(2f)`

We cleared the end of the train

M
`v_(2f)` = m (v₁₀ - v1f) + M v₂₀

Let's calculate

3.60 v2f = 1.50 (2.15-4.75) + 3.60 2.45

`v_(2f)` = (-3.9 + 8.82) /3.60

`v_(2f)` = 1.36 m / s

As we can see, this speed is lower than the speed of the car, so the two bodies are joined

set speed must be

m v₁₀ + M v₂₀ = (m + M)
`v_(f)`

`v_(f)` = (m v₁₀ + M v₂₀) / (m + M)

`v_(f)` = (1.50 4.75 + 3.60 2.45) /(1.50 + 3.60)

`v_(f)` = 3,126 m / s