• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Smashing Gliders

Extracts from this document...


IB Physics 11 HL                            Joy Fan (Raymond Tang, Candice Lin)

January 5, 2009  

Blk. D

Smashing Gliders


To determine the amount of momentum and kinetic energy conserved in elastic and inelastic collisions.


        Air track, two gliders (one heavier than the other), force spring, 500g-weight, timer, ruler, paper, two carts, masking tape.

Background Theory:

Momentum ρ= mv

- Momentum is conserved in a closed isolated system (no external forces)

- Kinetic energy is conserved in an elastic collision

- Inelastic collision- stick together

Inelastic collision:



Elastic collision:



Cart Explosion:




Part I: Inelastic and Elastic Collisions

  1. Set up an air track with two gilders of different masses on it.
  2. Tape cardboard to the gilders so that the timers can read them, measure the lengths of the cardboard and record in Table 1.
  3. Weigh the gliders and record in Table 1.
  4. In the inelastic collision, the smaller glider is at rest, gently push the larger glider so that it sticks onto the smaller glider across the timer with minimum space in between.
...read more.


Cart A

(with weight)

2.09m ± 0.005m

5.1 ± 0.1s

1.67m ± 0.005m

5.3 ± 0.1s

2.01m ± 0.005m

5.7 ± 0.1s

Cart B

2.61m ± 0.005m

6.0 ± 0.1s

2.93m ± 0.005m

5.5 ± 0.1s

3.08m ± 0.005m

6.3 ± 0.1s

Sample Calculation for momentum and kinetic energy:

- 1st Trial in Inelastic Collision

Σρbefore = Σρafter

Σρbefore = mAvA = (mAdA)/ tA

= (82.52g ± 0.005g × 6.50cm ± 0.05cm) / 0.1825s ± 0.00005s

= (536.38 g×cm ± 4.126 g×cm) / 0.1825s ± 0.00005s

= 2939.06 g×cm/s ± 22.62 g×cm/s

= 2.94 kg×cm/s ± 0.02 kg×cm/s  

Σρafter = (mA + mB) vAB = [(mA + mB) (dA + dB)]/ tB

=[(82.52g ± 0.005g + 43.00g ± 0.005g) (6.50cm ± 0.05cm+ 7.70cm ± 0.05cm)]/ 0.638s ± 0.0005s

= (125.52g ± 0.007g)(14.2cm ± 0.070cm)/ 0.638s ± 0.0005s

= 1782.38gcm ± 8.876 gcm/ 0.638s ± 0.0005s

= 2.79 kg×cm/s ± 0.01 kg×cm/s

% loss= [(2.94 kg×m/s ± 0.02 kg×m/s - 2.79 kg×m/s ± 0.01 kg×m/s)/ 2.94 kg×m/s ± 0.02     jksfdkg×m/s]×100%

= 6.80 ± 0.8 %

Ek before = (mAvA2)/2 = 0.174 ± 0.002 J

Ek after = (mAvA2)/2 + (mBvB2)/2

= (mAdA2)/ 2×tA2+ (mBdB2)/ 2×tB2

= [(82.52g ± 0.005g × (6.50cm± 0.05cm)2]/ 2(0.638s ± 0.0005s)2+ [(43.00g ± 0.005g× (7.70cm ± 0.05cm)2]/ 2(0.638s ± 0.0005s)2

=1740 g×cm ± 20 gcm + 1270 g×cm ± 10 g×cm

= 0.622 ± 0.04 J

Table 5 Trials 2 and 3 in the Inelastic Collision

2nd Trial

3rd Trial


3.23 kgcm ± 20 kgcm

5.93 kgcm ± 50 kgcm


3.34 kgcm ± 20 kgcm


...read more.


In conclusion, this lab had only limited success in showing the conservation of momentum and kinetic energy. One aspect that still puzzles me is the large gain in momentum in trial 1 of our elastic collision lab, may it be a faulty operation and should have been seen as an anomoly I have yet to find out.

Sources of Uncertainty:

When doing the cart explosion, we found that the cart rarely travelled on a straight line. This could be due to factors such as friction, position of the cart on the floor and position of the weight on the cart. The existence of this uncertainty resulted in our data being somewhat off. In future experiments, we shall attempt to limit this by aligning the cart with a meter stick or such.

...read more.

This student written piece of work is one of many that can be found in our International Baccalaureate Physics section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Physics essays

  1. Oblique Collisions in Two Dimensions

    In this experiment the main source of error will not come from the idea that the experiment is not perfectly elastic but due to errors in calculations and measurement Variables: In this experiment, the independent variable would definitely be the angle of the target ball to the incident ball.

  2. Suspension Bridges. this extended essay is an investigation to study the variation in tension ...

    hanging bob which is in fact its own weight acting vertically downwards. Even though the bob is not supported by a hand or any other rigid support beneath it, it does not fall but rather stays in equilibrium. This can only be possible if there is an equal and opposite

  1. Investigating the Breaking Distance of a Cart

    Obviously there are multiple factors affecting the braking distance of a moving object. Very important factor is the frictional force between the track and wheels of the crate. It has become clear that even slight increase in speed greatly increases braking distances.

  2. Aim: ...

    angle of the collision was, conservation of momentum and energy should still occur. With regards to the dependant variable, the thing that changed throughout the experiment and that relied on the independent variable was the distances and the angles at which the metal spheres landed on the carbon paper.

  1. Finding the Spring Constant

    The theoretical value I obtained from my previous lab for the value of k (Spring constant)was 28.701 Nm-1 Thus, %Difference = [(Theoretical value - Experimental value) � (Theoretical value) (x 100)] = > % Difference = [(28.701 - 25.980) � (28.701)

  2. HL Physics Revision Notes

    Greenhouse Gas Natural Sources Anthropogenic/Human Sources Water Vapor Evaporation of water from oceans, rivers and lakes Carbon Dioxide Forest fires, volcanoes, evaporation of water from oceans Burning fossil fuels in power plants and cars, burning forests Methane Wetlands, oceans, lakes and rivers Flooded rice fields, farm animals, termites, processing of

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work