A mine car (mass mcar = 444 kg) rolls at a speed of vcar = 0.52 m/s on a horizontal track, as the drawing shows. An mcoal = 141 kg chunk of coal has a speed of vcoal = 0.79 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 = 27°.)
m/s
To determine the speed of the car-coal system after the coal has come to rest in the car, we can use the principle of conservation of momentum.
1. First, let's calculate the initial momentum of the coal and the car. Momentum is given by the product of mass and velocity.
Initial momentum of coal (mc) = mass of coal (mcoal) * velocity of coal (vcoal)
Initial momentum of car (mccar) = mass of car (mcar) * velocity of car (vcar)
2. Next, we need to calculate the final momentum of the car-coal system. Since the coal comes to rest, its final velocity is zero.
Final momentum of car-coal system (mcfinal) = (mass of car + mass of coal) * velocity of car-coal system (vcfinal)
3. According to the principle of conservation of momentum, the initial momentum equals the final momentum.
Initial momentum of coal + Initial momentum of car = Final momentum of car-coal system
mc * vcoal + mccar * vcar = mcfinal * vcfinal
4. Now, let's substitute the known values into the equation and solve for the final velocity of the car-coal system.
mc = 141 kg (mass of coal)
vcoal = 0.79 m/s (velocity of coal)
mcar = 444 kg (mass of car)
vcar = 0.52 m/s (velocity of car)
α = 27° (angle)
(mcoal * vcoal) + (mcar * vcar) = (mcar + mcoal) * vcfinal
(141 kg * 0.79 m/s) + (444 kg * 0.52 m/s) = (444 kg + 141 kg) * vcfinal
111.39 kg∙m/s + 231.36 kg∙m/s = 585 kg * vcfinal
342.75 kg∙m/s = 585 kg * vcfinal
5. Finally, divide both sides of the equation by the mass of the car-coal system (mcar + mcoal) to solve for the final velocity of the car-coal system (vcfinal).
vcfinal = (342.75 kg∙m/s) / (585 kg)
vcfinal ≈ 0.585 m/s
Therefore, the speed of the car-coal system after the coal has come to rest in the car is approximately 0.585 m/s.