A mine car, whose mass is 440kg, rolls at a speed of 0.50m/s on ahorizontal track, as the drawing shows. A 150kg chunk of coalhas a speed of 0.80m/s when it leaves the chute. Determinethe velocity of the car/coal system after the coal has come to restin the car.

so there are X and Y components for the momentum in this problem. The Y component is not conserved as when the coal gets in the cart, the normal force exerted by the surface reduces it to 0.

Now, the X component is definitely conserved here.

so you have the momentum of the cart which is 440*0.5 added to the momentum of the chunk which is 150*0.8*cos(25°), that is the momentum before the coupling between the objects. Afterwards both objects will have the same velocity, so we write the equation like this:

[tex]440*0.5 + 150*0.8*cos(25) = 440*v_{final} + 150*v_{final} \\ => 220+120*cos(25) = (440+150)v_{final} => v_{final} = \frac{220+120*cos(25)}{590} = 0.56 m/s[/tex]

onyebuchinnaji onyebuchinnaji · Answer 2 · 2021-10-15T16:21:13+02:00

The velocity of the car/coal system after the coal has come to rest in the car is 0.58 m/s.

The given parameters;

mass of the mine car, m₁ = 440 kg
initial speed of the mine car, u₁ = 0.5 m/s
mass of the coal, m₂ = 150 kg
initial speed of the coal, u₂ = 0.8 m/s

The final velocity of the car/coal system after the coal has come to rest in the car is calculated by applying the principle of conservation of linear momentum for inelastic collision.

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the final velocity of the car/coal system

(440 x 0.5) + (150 x 0.8) = v(440 + 150)

340 = v(590)

[tex]v = \frac{340}{590} \\\\v = 0.58 \ m/s[/tex]

Thus, the velocity of the car/coal system after the coal has come to rest in the car is 0.58 m/s.

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