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

Answer 1

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

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