Respuesta :
Answer:
0.575 m/s
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
Answer:
The speed of the car-coal system after the coal has come to rest in the car is 0.37 m/s.
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 .
[tex]\theta=26^{\circ}[/tex]
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
[tex]m_{1}v_{1}\cos\theta=(m_{1}+m_{2})v_{x}[/tex]
Put the value into the formula
[tex]450\times0.55\cos26=(450+140)v_{x}[/tex]
[tex]v_{x}=\dfrac{450\times0.55\cos26}{450+140}[/tex]
[tex]v_{x}=0.37\ m/s[/tex]
Hence, The speed of the car-coal system after the coal has come to rest in the car is 0.37 m/s.
