A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiting truck. Un

Question

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiting truck. Unfortunately, the truck driver went on a break without having removed the previous package, of mass 2m, from the bottom of the chute. Suppose the packages stick together. What is their common speed after the collision

in progress 0
Hưng Khoa 5 years 2021-07-20T18:56:53+00:00 1 Answers 24 views 0

Answers ( )

  1. Cherry
    0
    2021-07-20T18:58:02+00:00
    Reply

    Answer:

    The speed of combined package after the collision is \frac{\sqrt(2gh)}{4}\\.

    Explanation:

    mass of first package = m

    height , h = 3 m

    mass of second package = 3 m

    let the speed of first package is u as it strikes with the second package.

    Use the third equation of motion

    v^2 = u^2 + 2 gh \\\\u^2 = 0 + 2 g h \\\\u = \sqrt {2gh}

    Let the velocity of combined package after the collision is v.

    Use the conservation of momentum

    m\times u + 3m \times 0= (m + 3m)\times v\\\\m u = 4mv \\\\v=\frac {u}{4}\\\\v = \frac{\sqrt(2gh)}{4}\\\\

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )