# Kinematics bar problem

1. Apr 23, 2007

### drivet

Hi guys, I was flicking through some past paper IB questions to prepare for the upcoming exam and I came across this:

1. The problem statement, all variables and given/known data

(diagram: a large metal bar poised to drop above a smaller bar which is on the ground, and the resulting drop would drive the smaller bar into the ground. )

In the situation shown, the object has a mass 20103.×kg and the metal bar has a has a mass of 400 kg.

The object strikes the bar at a speed of 6ms−. It comes to rest on the bar without bouncing.
As a result of the collision, the bar is driven into the ground to a depth of 0.75 meters.

2. Relevant questions

(a) Determine the speed of the bar immediately after the object strikes it.
(b) Determine the average frictional force exerted by the ground on the bar.

3. The attempt at a solution

Unsure of which concepts to use in this case. I tried to solve it with potential energy/kinetic energy as well as conservation of momentum.

Last edited: Apr 23, 2007
2. Apr 23, 2007

### mezarashi

Let's start by trying to solve part (a). You have a conservation of momentum and energy problem. Let us say that the impact occurs at t=0.

Just at t = -0.0000001 seconds, there was only the 20 ton object moving with a certain momentum and certain kinetic energy. At t = 0.000001 seconds, both the object and the metal bar are moving together with a conserved momentum and kinetic energy. Hint: the energy and momentum situation before and after the collision.

Can you write out some equations based on that?

3. Apr 23, 2007

### andrevdh

http://www.physicsclassroom.com/Class/energy/U5L2a.html" [Broken]

Last edited by a moderator: May 2, 2017
4. Apr 23, 2007

### drivet

I think I see what you mean,

m1v1=(m1+m2)*v2
?

Still kind of confused as to where kinetic energy comes into play

5. Apr 23, 2007

### andrevdh

After the collision the objects are moving as a unit. Therefore their speed will be the same.

6. Apr 23, 2007

### drivet

So is that the answer then?

v2=(m1*v1)/(m1+m2)
?

7. Apr 23, 2007

### drivet

......anyone?

8. Apr 23, 2007

### mezarashi

I over generalized my hints earlier. For this particular part of the problem, only the conservation of momentum equation is needed. The equation you posted seems correct. The KE equation will come in handy for part (b) however.

9. Apr 23, 2007

### andrevdh

To solvethis problem you need to apply the Work - Kinetic energy theorem - see the link I posted previously.

10. Apr 23, 2007

### drivet

ok, but could someone please clarify the term "Average frictional force" . Do they mean friction as in the earth rubbing against the sides of the smaller bar whilst it is being pushed into the ground?

Last edited: Apr 23, 2007
11. Apr 24, 2007

### andrevdh

When calculating the work done by a force one needs to take into account the direction and magnitude of the force along the path that the object is moving. This means that we need to integrate along the path to obtain the work done by the force. It the direction and magnitude of the force do not change things are much simpler and the work done can be calculated by a simple multiplication of the constant force and the total displacement of the object - which is what you are to assume about this situation.