conservation of linear momentum

N 23 (number on Roster) Student Name: _________________

Table: _________ Partner: __________________ _________________

Professor: Dr.BorisAsherov

Course: Physics 433.0 Lab

Section: _________

Date of Experiment: ____________________________

Date of Experiment handed in: ____________________

learning objectives:

In this laboratory activity we study elastic and inelastic collisions in one dimension and verify the principles of conservation of linear momentum and conservation of energy in an elastic collision in one dimension. We also verify the principle of conservation of linear momentum in explosions and in an inelastic collision in one dimension.

Theory:

The momentum p of an object is the product of its mass and its velocity: p = mv Momentum is a vector quantity, since it comes from velocity multiplied by mass

. The law of conservation of momentum states that the total momentum of all bodies within an isolated system, p total = p1 + p2 + ......., is constant. That is, if the total momentum has some initial value pi

, then, whatever happens later, the final value of the total momentum pf must equal the initial value. So we can write the law of conservation of momentum like this: pf = pi Conservation of momentum is usually studied in problems that involve collisions. In this experiment, you’ll look at collisions between two gliders on an air track. You will measure the final momentum of an initially stationary glider, struck by another glider which is initially moving. You’ll do this experiment for two different types of collisions, elastic and inelastic.

Elastic collisions are ones where kinetic energy is conserved. Inelastic collisions do not conserve kinetic energy. The kinetic energy of an object is defined as k=1/2mv^2 where m is the object’s mass and v is its velocity. Kinetic energy is not a vector: it’s a scalar, and its units are Joules (J).

procedure:

1. set up the dynamics track and photogates.

2. measure the flag's length and the masses of carts.

3. connect the science workshop interface box to the computer, turn on the interface and computer.

4. connect the photogates accessories plug into Digital channel.

5. set up the sensors in the software.

6. explosions.

7. massive projectile, target at rest.

8. Head-on collision.

9. setup the dynamics cart track and the photogates and repeat steps 7 and 8 for inelastic collisions.

sample calculation: v=x/t=0.049m/0.0442s=1.11m/s difference=(0.55-0.52)/0.52=5.94%

K=0.5mv^2=0.5*0.294kg*1.11m/s=0.182j

Total K=0.055j+0.108j=0.163j

Analysis of data and possible sources of error:

There may be a small error about calculation; flag position may not be very accurate; or copied the wrong date and used the wrong formula or Photogate placement position is not accurate.

Answer to the question:

1. The momentum before an inelastic bigger than the momentum after the collision.

2. The momentum before an elastic bigger than the momentum after the collision.

3. The Kinetic energy before an elastic bigger than the Kinetic energy after the collision.

4. The momentum before an elastic bigger than the Kinetic energy after the collision.

5. The velocities of the two bodies after the elastic collisions are given by V1=(M1-M2)U1/(M1+M2)+2M2U2/(M1+M2) V2=(M2-M1)U2/(M1+M2)+2U1M1/(M1+M2) Where, V1,V2 are the velocities of the two bodies after collision. U1,U2 are the velocities of the two bodies before colision.(U1>U2) M1,M2 are the masses of the two bodies. when the mass of two bodies are equal that is M1= M2 then V1=0+2MU2/2M=U2 V2=0+2MU1/2M=U1 Thus when two bodies…