Question

A coal car of mass m₁ = 3608 kg is rolling down the tracks at a velocity of v_i = 2.5 m/s when a loader drops a load of coal m₂ = 6976 kg into the car.

m₁ = 3608 kg
m₂ = 6976 kg
v_i = 2.5 m/s

  33% Part (a) What is the momentum of the system in (kg⋅m)/s before the coal is added?

  33% Part (b) What is the velocity after the coal is added in meters per second?
  33% Part (c) If you measured the final velocity of the car to be 1 m/s and the initial mass of the car remains the same, how much coal was added to the car in kg?

Answer 1

here,

m1 = 3608 kg

m2 = 6976 kg

vi = 2.5 m/s

a)

the momentum of the system in before the coal is added , Pi = m1 * vi

Pi = 3608 * 2.5 kg.m/s

Pi = 9020 kg.m/s

b)

let the velocity after the coal is added be v

using conservation of momentum

Pi = (m1 + m2) * v

9020 = (3608 + 6976) * v

v = 0.85 m/s

the velocity after the coal is added is 0.85 m/s

c)

when the final velocity , v' = 1 m/s

let the mass of coal added be m2'

using conservation of momentum

Pi = (m1 + m2') * v'

9020 = ( 3608 + m2') * 1

m2' = 5412 kg

the mass of coal added is 5412 kg