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When a push becomes a collision...
When a push becomes a collision...
Perfectly Elastic Collisions - 4 Common Scenarios
Perfectly Elastic Collisions - 4 Common Scenarios
Bounce with no loss of energy
Scenario #1: Equal Masses - One at rest (p=0kgm/s)
Scenario #1: Equal Masses - One at rest (p=0kgm/s)
A 2kg red-cart mass is traveling to the right (+) at 2m/s. It collides with a stationary 2kg blue-cart of the same mass. The collision is perfectly elastic. The momentum of red-cart is completely transferred to the blue cart. The red cart comes to a stop and the blue cart moves to the right at +2.0 m/s.
The incoming object comest to a stop and the target gains the exact speed as the incoming object. This is often seen in the game of pool when the cue ball strikes another ball.
Scenario #2: Equal Masses - Opposite Velocities
Scenario #2: Equal Masses - Opposite Velocities
A 2-kg red cart is traveling to the right at +2.0m/s. It collides an identical blue cart traveling in the opposite direction. Both carts bounce elastically off of each other. This results in a 'swapping' of momentum, causing the two carts to bounce back with the same speed but opposite direction as before the collision.
When the two interacting objects meet, they have equal and opposite momenta (pl of momentum). In the scenario shown the two objects are identical, with equal masses and opposite velocities, so they also have equal and opposite momenta. However, this scenario also holds when the objects only have opposite momenta, they simply swap velocities.
Scenario #3: Large Mass into Light Mass at Rest
Scenario #3: Large Mass into Light Mass at Rest
This type of collision is often seen in videos on YouTube. When a large moving object collides with a small object at rest the smaller object flies forward at a high velocity.
This is also the basis for many sports that we play. Hitting a golf ball is an excellent example. Notice the club does not slow, however the ball leaves at a high rate of speed. This video also demonstrates the importance of following through on a shot as it will increase the time the force (club) is in contact with ball. (F t = m ∆v.) This also the basis for kicking any sort of ball.
A 3kg red cart is traveling to the right with a velocity of +2m/s, giving it a total of 6kgm/s of momentum. It collides with a much smaller mass of 0.2kg that is at rest. Following the collision, the velocity of the 3kg mass is reduced to +1.75m/s. This causes the smaller 0.2kg mass to move with a velocity of 3.75m/s. The momentum is conserved as the sum of the momentum of the two carts remains 6kgm/s.
Scenario #4: Light Mass into Large Mass at rest
Scenario #4: Light Mass into Large Mass at rest
Often the large object remains at rest and the smaller object simply bounces off with an opposite momentum. In the scenario shown the difference in mass is 15x, however the large mass does move. We can think of the Earth as Mass_2 when we are bouncing on it.
In the case of an object bouncing off a larger mass, a relatively small mass of 0.2kg bounces off a larger mass of 3kg. In a perfectly elastic collision, the larger mass does move, however in reality this motion may be undetectable. If a 0.2kg red cart traveling at 2m/s collides with a stationary 3kg mass. The red cart will rebound with a velocity of -1.75 m/s causing the blue cart to move with a velocity of +0.25m/s.
Sticky Collisions
Sticky Collisions
Scenario #1: Opposite direction, similar velocities
Scenario #1: Opposite direction, similar velocities
Suppose two objects are approaching each other and collide. After colliding they stick together. Mass_1 is 2 kg and has a positive velocity of 2 m/s. Mass_2 is also 2 kg but has a velocity of -3m/s (moving to the left). In our diagram below, Mass_1 is red and Mass_2 is blue.
The horizontal axis represents the mass of the object while the y-axis represents the velocity of the object. Objects moving to the right or upwards are positive, while objects moving leftwards or downwards are negative. If an object has a velocity of 0 m/s, the velocity js shown as a bar. The AREA of each square or rectangle represents the momentum of each object. The sum of the two rectangles is equal to the total momentum. This total value does not change.
We want to determine v_f
substituting in the values of the two objects:
this equals the p_after (p_2)
substitute the after values
Solve for v_f
velocity of final two masses:
Scenario #2: Equal masses, one stationary
Scenario #2: Equal masses, one stationary
In the second scenario, cart with a mass of 2kg is traveling at 4 m/s. It collides and sticks to a STATIONARY (v = 0 m/s) cart of mass 2kg. The momentum (p) before the collision is 8 kgm/s.
Momentum is conserved (always), the mass of the 2 cart system is now 4 kg. Combined these carts will travel to the right with a velocity of +2m/s.
This can be summarized in the equation above and solved in the same way as above.
Can you show that the v_f of scenario #2 is 2m/s?
Conservation of Momentum Problems
Conservation of Momentum Problems
LCM Data Analysis Whiteboards
LCM Data Analysis Whiteboards
Using the data from your experiment (from your spreadsheet), create a CER regarding the conservation of momentum in your scenario (Was momentum conserved in your scenario? YES, IT WAS!!! ) Use the models from the LCM Recap as support for your evidence. (Written description, Graph, Free-Body Diagrams, Momentum Bar Chart, Equation).
Table / Scenario
Table 1 - 1:2 (Bounce)
Table 2 - 2:1 (Bounce)
Table 3 - 1:1 (Stick)
Table 4 - 1:2 (Stick)
Table 5 - 2:1 (Stick)
Table 6 - 3:1 (Bounce)
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