Amine 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.

0.56 m/s

Explanation:

hello, there is 25° inclination angle for the chute in the drawing. Thankfully, I know this problem. The conservation of momentum.

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:

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;

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)

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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what is the question because that is not a question.

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