To prove that there is conservation of momentum in both elastic and inelastic collisions.
To think about:
- With negligible friction and leveled tract, net force exerted on the cart right before collision is zero.
- Magnitude of force is at maximum when at one point during the collision, there's no velocity. As for FΔt=mv, when velocity reaches zero, force is at its maximum.
- Net
force right after collision is zero, since velocity reaches its maximum. And
since Δt=
contact time between the 2 objects, when they lose contact with one another,
net force goes to zero.
- The
collision takes approximately less than half a second. So ~ 0.2 second
- Given either a constant or non-constant force applied on the system, we can find the net impulse through calculation of area under the curve. As shown below, in the first case where the graph are composed of rectangles (force constant for every one second), we can use geometry to find the impulse from area of each rectangle combined. In the second case where force varies at any given time, we can use integration.
- Given either a constant or non-constant force applied on the system, we can find the net impulse through calculation of area under the curve. As shown below, in the first case where the graph are composed of rectangles (force constant for every one second), we can use geometry to find the impulse from area of each rectangle combined. In the second case where force varies at any given time, we can use integration.
Expt 1:
Set-up:
-Clamp a rod to the edge of table, and then clamp the cart
to the rod. The side with extended spring should face away from the rod.
- On another cart, attach a metal plate (for motion sensor)
and in front of it, force sensor.
- Hold force probe vertically, with hook pointing down. Then
use 1 kg mass to calibrate force (for 9.8N).
- Mount force probe in front of the metal plate, then attach
rubber stopper to the hook end, make sure that stopper collides on-center with
the extended spring.
- Make sure that force probe’s wire is out of the way of
motion sensor’s detection range.
- Place a motion sensor at the other end of track and make sure
it can see the metal plate without any interference.
- The sensor will detect velocity, position, and time. Force
probe will detect force and time.
- Open file Impulse and Momentum so the data could be recorded on graphs force vs. time and velocity vs. time. This file is also made to record motion at 50 data points per second.
- Hold the middle part of the cart and give it a sufficient push toward the extended spring, start collecting data, and
stop when the cart has travelled back toward motion sensor for a few more seconds. The force data should not be more than 10N as a higher force is likely to be misread by the force probe.
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| Motion sensor detect changes from the metal white plate as shown. |
Expt 2: A larger momentum change
Set-up: Everything remains the same. Increase the mass of the metal cart on the track by adding 500g to it. Hold the middle of cart and give the cart a push then record data. Force data shouldn't exceed 10N.
Expt 3: Impulse-momentum theorem in inelastic collision
Set-Up:
-Remove the blue cart with extended spring and the rod. Attach a vertical piece of wood (with clay sticking to it) using table clamp.
- Keep the 500g mass on the cart just like experiment 2. Replace the rubber stopper with a nail by taping it the hook of the force probe.
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| Nail should embed horizontally in the clay after collision takes place. |
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| Nail is taped on the other end of force probe pointing toward the pink clay. |
- The start position of the metal cart should be ~0.75m away from the clay. Hold the middle of the cart and give it a push then collect data. Again, force probe should not read more than 10N.
Data Analysis:
mass (cart+ force probe)= 0.673kg
Expt 1:
For this experiment and the second one, we could only produce a nearly perfectly elastic collision. Therefore we will see that there's some lost in final kinetic energy of the extended spring-cart system. In an effort to keep systematic error as minimum as possible, we decide to strike out some insufficient data. Note that there are two peaks in force vs. time graph below. We strike out data for the first peak and keep the second peak. It only makes sense that as an object is colliding, it's net force (with another object) increase then decrease as the collision is almost over. Therefore, for one collision, we can't expect two points of maximum net force.
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| Original graphs before striking out some data points |
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| After some alteration |
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| Note that the force jumps from -4 to -5N after deleting some points, which is more reasonable. |
-Now we take several integrations of 2 different points and compare this to the change in momentum (calculated by hand), and graph the two to see their linear relationship.
- From the graph, we see that there's a linear relationship between impulse and change of momentum given the same time interval. Hence with no systematic error, J= Δp= ∫ F*dt
- Time of contact during collision is ~ .2 second.
Expt 2:
- We see that right before and after collision, the cart's position and velocity are approximately the same. In this case, maximum compression of extended spring is ~0.65m. At max compression, cart's velocity is zero, net force is greatest here.
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| Force reaches its maximum and then go back down |
Integration: ∫ F*dt = -1.324 N.s
Δp= m(v2-v1)
= (0.673+0.5)(-0.490-0.550)= -1.220 kg.m/sec^2
Hence, our error is (1.220-1.324)*100/1.324 ~ 8%
- V2 and V1 are taken from t1 and t2, which are the limits of integration for impulse calculation.
Expt 3:
mass(nail)= 2g; mass (hook from force probe)= 2g; compensate one another. so, M-system= (0.673+0.5) kg
- For inelastic collision, we expect the final velocity of cart to be zero and the position doesn't change anymore at that point. Impulse-momentum theorem should still hold true.
- Limits of integration here is the time interval from when the collision starts to when the final velocity of the cart is zero or the time when position begins to look like a straight line.
- Again, to prove impulse-momemtum theorem in inelastic collision,
Integration: ∫ F*dt = -0.5459 N.s
Δp= m(v2-v1)
= (0.673+0.5)(-0.075-0.390)= -0.5454 kg.m/sec^2
Hence, our error is (0.5454-0.5459)*100/0.5459 ~ 0.08%
- V2 and V1 are taken from t1 and t2, which are the limits of integration for impulse calculation. Note that V2= -0.075m/sec which is very close to zero.
The difference between graphs of elastic and inelastic collisions:
elastic: In force vs. time graph, the force reaches it max peak and go to zero. In velocity vs. time, velocity right before and after collision are very similar. In position vs. time graph, position right before and after contact with extended spring are very similar.
Inelastic: In force vs. time, the force reaches it max peak (negative), go back to zero, then go once again to a peak (positive), then climb down to zero. In velocity vs. time, velocity right before is highest value (among other velocities during collision) and after collision, it goes to zero. In position vs. time graph, initially it climbs a slope. At the same time velocity goes to zero, position starts to stay at one point (a straight line).
Conclusion:
Within the error we calculated from each experiment, including measurement uncertainty, we conclude that impulse-momentum theorem that J= Δp= ∫ F*dt. Impulse-momentum theorem holds true for elastic and inelastic collision, and regardless of larger or smaller momentum change.
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