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Smashing Gliders

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Introduction

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

January 5, 2009  

Blk. D

Smashing Gliders

Purpose:

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

Materials:

        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:

Before:

After:

Elastic collision:

Before:

After:

Cart Explosion:

Before:

After:

Procedure:

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.

Middle

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

Σρbefore

3.23 kgcm ± 20 kgcm

5.93 kgcm ± 50 kgcm

Σρafter

3.34 kgcm ± 20 kgcm

5.

...read more.

Conclusion

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.

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