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    Conservation of Momentum

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    See attached figure.

    In a shipping company distribution center, an open cart of mass 53.0 kg is rolling to the left at a speed of 5.40 m/s (see the figure ). You can ignore friction between the cart and the floor. A 14.0 kg package slides down a chute that is inclined at 37 degrees from the horizontal and leaves the end of the chute with a speed of 3.10 m/s . The package lands in the cart and they roll off together. See attached figure.

    A) If the lower end of the chute is a vertical distance of 4.00 m above 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?

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    ** Please see the attached file for the complete solution response **

    A) Velocity at the instant of leaving the chute (initial velocity) = u = 3.10 m/s

    Vertical component of the initial velocity = u_y = 3.10sin37^O = 1.87 m/s

    Horizontal component of the initial velocity = u_x = 3.10cos37^O = 2.48 m/s

    On leaving the chute, the package behaves like a projectile. The horizontal component of the initial velocity remains constant (neglecting air friction). However, vertical component is subject to acceleration due to gravity. ...

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