Dylan Humenik Ben Daily Srikrishnan Varadarajan Double Cart Collisions

Transcription

1 Double Cart Collisions Objective: -Apply knowledge of collisions in analysis of collision -Find momentum and kinetic energy of two different collisions (elastic and inelastic) Data: Mass (kg) Cart 1 (moving) Velocity of Cart 1 (before impact) Mass (kg) Cart 2 Velocity of Cart 2 (before impact) Velocity Together (after impact) Velocity (cart 1) (m/s) time Velocity (system togather after impact) time Sticky Collision Bouncy Collision Mass (kg) Cart 1 (moving) Velocity of Cart 1 (before impact) Mass (kg) Cart 2 Velocity of Cart 2 (before impact) Velocity of Cart 1 (after impact) Velocity of Cart 2 (after impact) Velocity (cart 1) time Velocity (cart 2) time kg.23 m/s 1 kg m/s.7 m/s.5 kg.41 m/s 1 kg m/s m/s.27 m/s

2 Calculations: -Sticky Collision: Mass of System (after impact) Velocity Together (after impact) Energy (before impact) Momentum (initial) Momentum (final) Energy (after impact) energy of cart 1 (N) energy of cart 2 (N) energy of system (N) Momentu m cart 1 (N*,m) Momentu m of cart 2 (N*m) 1.5 kg.87 m/s N.115 N*m.15 N*m.3675 N Momentu m of system (N*m) Time Time (s) (s) Sticky Collision: Mass of System after Impact: Mass of Cart 1 + Mass of Cart 2 Velocity Together (after impact): Avg velocity based off of data.5 kg + 1 kg = 1.5 kg =.87 m/s Energy before Impact: KE = (1/2)mv 2 KE = (1/2)(.5 kg)(.23 m/s) 2 KE = N Momentum (initial): Momentum (final): Momentum = (.5 kg)(.23 m/s) Momentum =.115 N*m Momentum = (1.5 kg)(.7 m/s) Momentum =.15 N*m

3 Energy after Impact: KE = (1/2)mv 2 KE = (1/2)(1.5 kg)(.7 m/s) 2 KE =.3675 N Energy of Cart 1: KE = (1/2)mv 2 KE = (1/2)(.5 kg)(.141 m/s) 2 KE =.497 N Energy of Cart 2: KE = (1/2)mv 2 KE = (1/2)(1 kg)(. m/s) 2 KE =. N Energy of System: KE = KE Cart 1 + KE Cart 2 KE = (.497 N) + (. N) KE =.497 N Momentum of Cart 1: Momentum = (.5 kg)(.141 m/s) Momentum =.75 N*m Momentum of Cart 2: Momentum = (1 kg)(. m/s) Momentum =. N*m Momentum of System: Momentum = Momentum Cart 1 + Momentum Cart 2 Momentum = (.75 N*m) + (. N*m) Momentum =.75 N*m

4 -Bouncy Collision: Velocity of Cart 1 (after impact) Velocity of Cart 2 (after impact) Energy (before impact) Momentum (initial) Momentum (final-based of raw data) Energy (after impact- based on raw) energy of cart 1 (N) energy of cart 2 (N) energy of system (N) Momentum cart 1 (N*m) Momentum of cart 2 (N*s) m/s m/s.4225 N.25 N*m.1945 N*m N Momentum of system (N*m) Time (s) Bouncy Collision: Velocity of Cart 1 (after impact): Avg. velocity based off of data Velocity of Cart 2 (after impact): Avg. velocity based off of data = m/s = m/s Energy of System before Impact: KE = (1/2)mv 2 KE = (1/2)(.5 kg)(.41 m/s) 2 KE =.4225 N Momentum (initial): Momentum = (.5 kg)(.41 m/s) Momentum =.25 N*m Momentum (final): Momentum final = (Momentum Cart1 +Momentum Cart 2 ) Momentum = ((.5 kg)(-.155 m/s)+(1 kg)(.27 m/s)) Momentum =.1945 N*m

5 Energy after Impact: KE = ((1/2)mv 2 Cart 1)+((1/2)mv 2 Cart 2) KE = (1/2)(.5 kg)(-.155 m/s) 2 + (1/2)(1 kg)(.27 m/s) 2 KE = N Energy of Cart 1: KE = (1/2)mv 2 KE = (1/2)(.5 kg)(.418 m/s) 2 KE = N Energy of Cart 2: KE = (1/2)mv 2 KE = (1/2)(1 kg)(. m/s) 2 KE =. N Energy of System: KE = KE Cart 1 + KE Cart 2 KE = ( N) + (. N) KE = N Momentum of Cart 1: Momentum = (.5 kg)(.418 m/s) Momentum =.29 N*m Momentum of Cart 2: Momentum = (1 kg)(. m/s) Momentum =. N*m Momentum of System: Momentum = Momentum Cart 1 + Momentum Cart 2 Momentum = (.29 N*m) + (. N*m) Momentum =.29 N*m

6 Energy (N) Energy (N) Momentum (N*s) Momentum (N*s) Dylan Humenik Graphs: -Sticky Collision: Momentum of Cart 1 and Cart 2 as a function of Time (Sticky) Momentum of System as a function of Time (Sticky) Cart 1 Cart System Energy of Cart 1 and Cart 2 as a function of Time (Sticky) Energy of System as a function of Time (Sticky) Cart 1 Cart System

7 Energy (N) Energy (N) Momentum (N*s) Momentum (N*s) Dylan Humenik -Bouncy Collision: Momentum of Cart 1 and Cart 2 as a function of Time (Bouncy) Momentum of System as a function of Time (Bouncy) Cart 1 Cart System Energy of Cart 1 and Cart 2 as a function of Time (Bouncy) Energy of System as a function of Time (Bouncy) Cart 1 Cart System

8 Results: - Knowledge of collisions was applied in analysis of collision. - The momentum for the inelastic (sticky) collision was.115 N*m before collision and.15 N*m after collision. The kinetic energy for the inelastic (sticky) collision was N before collision and.3675 N after collision. The momentum for the elastic (bouncy) collision was.25 N*m before collision and.1945 N*m after collision. The kinetic energy for the elastic (bouncy) collision was.4225 N before collision and N after collision. Qualitative Questions: 1. Explain the appearance of the Energy Graph for each impact (focus on the graph of the system). Both graphs (elastic and inelastic) appear to show the kinetic energy as staying relatively constant throughout the duration of the collision. 2. Is the graph of momentum of the system as a function of time consistent with expectations? Yes, in both graphs (elastic and inelastic) it appears as though momentum is not lost throughout the duration of the collision (conservation of momentum). 3. How does the force cart 1 applied to cart 2 during impact compare to the force cart 1 applied to cart 2 during impact? The force applied by Cart 1 to Cart 2 to during impact of the elastic (bouncy) collision is greater that the force applied by Cart 1 to Cart 2 during impact of the inelastic (sticky) collision. This is because during the elastic (bouncy) collision, the cart was pushed with a greater force, resulting in a greater initial velocity, which resulted in more force being applied to Cart 2. In both collisions, the force applied to Cart 1 was equal to the force applied to Cart 2, but because Cart 2 had more mass, this is why Cart 1 is pushed backwards during the elastic collision. 4. How does the impulse of cart 1 compare to the impulse of cart 2 during impact? For the inelastic (sticky) collision, the impulse (change in momentum) of Cart 1 is equal to the impulse of Cart 2. The same can be said for the elastic (bouncy collision.

9 5. Draw a qualitative graph of velocity of the center of mass of the system as a function of time for each of the two lab collisions Draw a qualitative graph of Force on cart 1 and Force on cart 2 as a function to time. (clearly show a period of time before, during, and after impact) Explain in detail how you could construct a Force as a function of time graph using the momentum vs time graph. (the process should be the same for either collision) To find the force as a function of time graph, the integral would have to be taken of the momentum vs. time graph

10 1. If the approaching cart is moving at the same speed with the same mass in each of the two collision types, a. in which collision will the second cart experience a greater change in momentum. The second cart will experience a greater change in momentum during the inelastic collision. b. IF the collisions occurred over an identical time for each collision, which collision occurs with greater force. Relate your answers to crumple zones in cars. The inelastic collision occurs with greater force because the crumple zone is less than (only the length of the clay- about 1cm) the crumple zone for the elastic collision (the length of the magnetic field- a few inches).