Question

A mine car (mass mcar = 450 kg) rolls at a speed of vcar = 0.55 m/s on a horizontal track, as the drawing shows. An mcoal = 140 kg chunk of coal has a speed of vcoal = 0.73 m/s when it leaves the chute. Determine the speed of the car-coal system after the coal has come to rest in the car. (Assume the angle α = 26°.)

Answer 1

The speed of the car-coal system after the coal has come to rest in the car is 0.37 m/s.

Step-by-step explanation:

Given that,

Mass of car = 450 kg

Velocity of car = 0.55 m/s

Mass of coal = 140 kg

Velocity of coal = 0.73 m/s

Suppose the angle is horizontal angle .

`\theta=26^(\circ)`

We need to calculate the speed of the car-coal system after the coal has come to rest in the car

Using conservation of momentum

`m_(1)v_(1)\cos\theta=(m_(1)+m_(2))v_(x)`

Put the value into the formula

`450*0.55\cos26=(450+140)v_(x)`

`v_(x)=(450*0.55\cos26)/(450+140)`

`v_(x)=0.37\ m/s`

Hence, The speed of the car-coal system after the coal has come to rest in the car is 0.37 m/s.

Answer 2

0.575 m/s

Step-by-step explanation:

Data provided in the question:

Mass of the mine car, m₁ = 450 kg

Initial Velocity of the mine car, v₁ = 0.55 m/s

Mass of the coal chunk, m₂ = 140 kg

Initial velocity of the coal chunk, v₂ = 0.73 m/s

angle α = 26°

Now,

the horizontal component of the velocity of the coal chunk = v₂cos(α)

⇒ 0.73 × cos(26°)

⇒ 0.656

Now applying the concept of conservation of linear momentum in horizontal direction

we have

m₁v₁ + m₂v₂cos(α) = (m₁ + m₂)v

here v is the combined speed of the car-coal system

thus,

( 450 × 0.55 ) + ( 140 × 0.656) = (450 + 140) × v

or

247.5 + 91.84 = 590v

or

339.34 = 590v

or

⇒ v = 0.575 m/s