![]() ![]() It would exert almost the same amount of force on the body as a solid surface. The airbag would not be nearly as effective. Would they be as effective in protecting a passenger in a collision? Suppose airbags were not vented to allow the gas inside to escape, but remained inflated (like a balloon). An airbag acts gradually, keeping the body relatively safe in comparison. A hard surface would simply crash against the body’s mass and bring it to a stop by applying a massive amount of force in a single second. Airbags allow a body in motion to stop gradually by absorbing the motion of the body’s mass. This is true whether or not an airbag is used, so why use an airbag? How does it reduce injuries?Īirbags reduce injury because they exert far less force on a person than would a dashboard or windshield. In a car collision, the drivers body must change speed from a high value to zero. Click OK.Purpose: “To collect force, velocity, and time data as a cart experiences different types of collisions and to determine an expression for the change in momentum, delta p, in terms of the force and duration of a collision.”Ĭ(1): Following directions and instructionsī(3): Mathematical modeling of linear relationship Choose Motion Sensor,and set the trigger rate to 100. In the Sensor Properties window, choose Measurement, then de- select Position and Acceleration, so only Velocity is selected. Connect the yellow motion sensor plug into digital channel 1, and the other plug into channel 2. Under Measurement, de-select all choices and select Force (only Force should be checked). In the Sensor Properties window, choose General, then select Fast Force Changes, and increase the sample rate to 1000 Hz. Connect the Force Sensor lead to Channel A on the Science Workshop Interface box. Launch the Data Studio program, choose Create Experiment, select Force Sensor, and Motion Sensor. Program Selection and Preparation Turn on the Pasco Science Workshop Interface, then the computer, and login. What was the total impulse on the ball? Apparatus Pasco 750 Interface 1.2 meter dynamics track Force sensor attached to the Force sensor bracket with attachments (hook, rigid spring, less rigid spring, rubber bumper, magnetic bumper) Dynamic cart (without plunger) and mass block Motion sensor (set on "narrow beam') Pulley and clamp Large table clamp Mass holder and 500 grams of masses Thread Mass block to elevate track Detailed Procedure and Analysis for the Impulse – Momentum Experiment I. A 4 kg ball moving at 3 m/s bounces off a wall and is observed to be moving at 3 m/s in the opposite direction. ![]() (Hint: For parts 1 to 4 above, graph the force vs. The force decreases linearly to 0 in 2 seconds. The force stays constant at 8 Newtons for 3 seconds. The force grows linearly from 0 to 8 Newtons in 2 seconds. A force in the positive x direction acts on the mass for 7 seconds as follows: a. A 4kg mass is initially moving in the x direction at 5 m/s. 12/04/20 1 Pre-Lab Assignment Complete the following problem, using the impulse-momentum relationship. In addition, the initial and final (maximum and minimum) velocities can be obtained, making it easy to calculate initial and final momentum, and test the impulse-momentum relation. A statistics package is used to integrate the force versus time curve to obtain the impulse. The computer graphs force versus time, and also the cart’s velocity versus time. A motion sensor detects the position of the cart versus time, enabling its velocity to be calculated as a function of time. In this experiment, a moving cart collides with a stationary “force sensor.” The force sensor measures the collision force as it varies with time throughout the collision. Integrating from time t1 to time t2, 2 2 1 1 2 1 t p res t p F dt dp p p p . Multiplying both sides of the equation by dt we obtain, resF dt mdv d mv dp . This can easily be derived from Newton’s second law res dv F ma m dt . 2 1 t res t F dt The impulse-momentum relationship states that if an object with mass m is acted on by a force over the time interval from t1 to t2, the impulse is equal to the change in momentum: 2 1 2 1 t res t F dt p p p . ![]() The impulse of a resultant force from time t1 to time t2 is When the force is plotted versus time, the impulse is the area under the curve between t1 and t2. The momentum of an object with mass m and velocity v is p mv . Download Impulse – Momentum Experiment - Lab Experiment #9 | PHYS 211 and more Physics Lab Reports in PDF only on Docsity!Physics 211 Experiment #9 Impulse – Momentum Experiment Discussion Impulse, momentum, and the impulse-momentum relationship are defined and discussed in the text. ![]()
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