Question

A crane is used to swing a 450 kg wrecking ball into a 500 kg wall. The crane takes 5 s to push the ball into the building. The crane provides 2000 N of force. The ball is left moving 10 m/s after the collision. How fast is the wall moving afterwards? *

How does the kinetic energy change in the crane problem?
1. KE increases
2. KE decreases
3. KE stays the same


What type of collision is this?
elasti
inelastic

Answer 1

This is a momentum problem, so let's find the velocity of the ball initially. It can be given by:

`\vec{F} = m\vec{a}`
Let's plug in values and solve for a:

`\vec{a} = (2000)/(450) = 4.44 m/s^2`
Now, we can multiply by time to get velocity:

`\vec{v} = \vec{a}t = 4.44(5) = 22.2m/s`

We have all the information we need, so let's set up an equation to solve for the velocity of the wall after the collision:

`mv_(ib)= mv_(fb) + mv_(w)`

Solve for velocity of wall finally:

`v_w = (mv_(ib)-mv_(fb))/(m)`

Plug in all values:

`v_w = (450(22.2)-450(10))/(500) = 10.98 (m)/(s)`

So, the final velocity of the wall will be 10.98 m/s.
The KE in this problem initially was 110889 J (1/2mv^2). After the collision it was 52640.1 so, the KE decreased.
Since there was no mashing of objects together, this was an elastic collision.