Examination of the affect of an explosion on motion of carts

See attached file for graphs, tables and formulas.

Conservation of Momentum in an Explosion

Impulses and Momentum

When a resultant force acts on an object, the object is accelerated. The acceleration of the object is proportional to the force and inversely proportional to the mass of the object. Algebraically this is expressed as a= F/m, or with the correct selection of units, F= ma. Since we know that acceleration is the rate of change of velocity, or a= dv/dt we can write

F = ma = m*dv/dt

or

If the mass remains constant, m*dv = d(mv), and we have Fdt = d(mv)

The product of an object's mass and its velocity (mv) is known as the momentum, a vector quantity. The product of the resultant force and the time interval during which it acts (Fdt) is known as the inpulse. The relationship above thus states that the change in an object's momentum is equal to the applied impulse.

The Problem

In this experiment we shall allow an "explosion" caused by a compressed spring to push two loaded carts apart. We can measure their masses and the comparative resulting velocities. This enables us to compare their momenta. Knowing their comarative momenta allows us to compare the impulses acting on the two objects and the forces the objects exerted on each other during the explosion.

Comparing the Velocities of the Carts

Our apparatus of two carts, one of which has a spring- actuated plunger. The spring is compressed and the carts are placed together.

The plunger can be released by a trigger, and the explosion causes the carts to fly apart. We shall use a trick to determine their comparative velocities. The carts are attached by a length of string that is slack before the explosion. The carts fly apart and move until the string is pulled tight; they are then abruptly stopped. Since the two carts (represented in the following by subscripts 1 and 2) move for the same length of time, the distances they travel ( s1 and s2 ) away from their starting points must be in the same ratio as their speeds (v1 and v2 ), or v1/v2 = s1/s2

Knowing the masses of the carts, we can calculate the ratio of their momenta after the explosion has occurred.

Making the Measurements

Mark the starting position of the cart with a piece of masking tape on the table top. With another piece of masking tape, mark the position the cart reaches after the explosion. You will want to repeat this several times to make sure that you have marked the correct distance. Place the two carts together in the starting position and mark the original position of the second cart. By placing the first in its final position and pulling the string tight, you can locate the final position of the second cart. You can now measure the distance each cart travels. Repeat this experiment for a variety of combinations of masses of the carts.

Obtaining results from the Data

Compute the ratio of the momenta of the carts. What do the momentum ratios suggest about the comparative magnitudes of the momenta of the carts after the explosion? Recalling that momentum is a vector quality, make a general statement about the total momentum of the combination before the explosion and after the explosion.

Compare the times during which the spring pushed on the carts. Compare the forces on the two carts during the explosion.

Cart 1

Mass (g) position
released (cm) position
caught (cm) Displacement (centimeters) Mass*Displacement
700 100 65.5
1,200 100 41.1
1,700 100 20.7
2,300 100 11.1
2,700 100 4.5
1,700 100 43.1

Cart 2

Mass (g) position
released (cm) position
caught (cm) Displacement (centimeters) Mass*Displacement
620 100 61.5
620 100 84.7
620 100 106.8
620 100 114.4
620 100 121
1, 620 100 32.6

Questions

1. If both carts had exactly the same mass in trial 1, how would you expect their velocities to compare with each other? Why?
2. In the two- fragment explosion that results when a gun is fired, explain why the bullet acquires a high velocity while the gun does not.
3. When the spring is released, the rod pushes against one cart with a given force. This cart pushes back with an equal force. Explain why this means that the total force on the system of the two carts is zero.
4. If Ea and Eb are the kinetic energies of the two carts after the spring is released, prove that the ratio Ea /Eb is equal to the ratio mb /ma