In a shipping company distribution center, an open cart of mass 45.0kg is rollin

Question

In a shipping company distribution center, an open cart of mass 45.0kg is rolling to the left at a speed of 5.50m/s (see the figure). You can ignore friction between the cart and the floor. A 18.0kg package slides down a chute that is inclined at 37 degree from the horizontal and leaves the end of the chute with a speed of 3.20. m/sThe package lands in the cart and they roll off together. A)If the lower end of the chute is a vertical distance of 4.00above the bottom of the cart, what is the speed of the package just before it lands in the cart? B)What is the final speed of the cart?

Explanation / Answer

a) velocity of the package= 3.2 m/sec

horizontal component of velocity= Vcos= 3.2*co37=2.556 m/sec

Vx=2.556 m/sec

vertical component of velocity Vy=3.2*sin 37=1.9258 m/sec

h=4

final velocity be Vy'

Vy'=sqrt(Vy^2+2*g*h)

Vy'=sqrt(3.7+78.4)=sqrt(82.1)=9.06 m/sec

final velocity of package V'=sqrt(Vx^2+Vy'^2)

V'=sqrt(2.556^2+9.06^2)=9.4136 m/sec

b) conservation of momentum along X direction

(m1*v1-m2*v2=m1+m1)V

45*5.5-18*2.556=63*V

V=247.5-46.008/63

V= 3.198 m/sec

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