# Homework Help: Question about momentum and collision

1. Feb 10, 2004

### Arfrce8729

I have the following problem...

A truck of mass 1000 kg moving at a speed of 13 m/s hits a second truck of 1600 kg at rest at a stop sign. The bumpers of the trucks act like springs, each with a spring constant of 3*10^4 N/m. Neglecting friction, find the following:

(1)The speed of the two trucks when the bumpers compress the most ( the trucks have the same speed at the point ).

(2)Energy stored in the bumpers at the point.

(3)Maximum bumper compression.

(4)The speed and direction of each car after they bumpers push them apart again.

I'm not really sure where to start this problem. But here is what I was thinking of doing...

(1) Use the conservation of linear momentum...

m1 * v1<initial> + m2 * v2<initial> = m1 * v1<final> + m2 * v2<final>

(2) Use the equation for kinetic eergy

Energy = .5 * m1 * v1<final ^ 2

(3) Set the elastic potential energy equation to the opposite of kinetic energy equation

.5 * m1 * v1<final ^ 2 = -0.5 * k<spring constant> * x^2

and solve for x

(4) Split it up into x and y components and take ithe inverse tangent of

ycomponent / xcomponent

Will this properly answer the question? Thanks in advance

2. Feb 11, 2004

### HallsofIvy

Yes, except that you are told that v1<final> and v2<final> are the same: call it v<final>. You also know that v1<initial>= 13 m/s and v2<initial>= 0.
m1= 1000 kg and m2= 1600 kg. That is enough to find v<final>.

This is the final kinetic energy of only one truck. Since both trucks are moving at the same velocity, use (m1+m2) to find the total kinetic energy. The energy stored in the bumper (which is what was asked) is the initial kinetic energy minus the final kinetic energy.

Except that the left side of that is the final kinetic energy of one truck. You need to use the energy found in (2), the initial kinetic energy (of the one truck that was moving) minus the final kinetic energy of both trucks.

x and y components? Everything is moving in a straight line, isn't it? By "direction" the problem only means "left" or "right"- negative or positive velocity. Now forget about the bumper. Use conservation of momentum and conservation of energy to solve for v1<final> and v2<final> (which will now be different).