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Dec 6, 2023

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Part I-- Newton's 1st Law 1. Assume that you sitting in your stopped car with your seatbelt fastened waiting for a green light. Another car suddenly hits your car from behind. After recovering from the surprise, you notice a pain in your head and neck. What do you think happens to the head of a buckled-up driver when the car is hit from behind. 2. Now assume you are a passenger in a moving car and this car hits the back of a stopped car. What do you think happens to the head of a buckled-up passenger in a moving car when the car hits a stopped car. 3. Place a ball on a book that you hold out in front of you like a tray with one hand. Record what happens to the ball when you conduct the following three experiments. 1. From rest, walk quickly forward. From the rest, the ball rolls backward. 2. From rest, walk quickly backwards. From the rest, the ball rolls forward. 3. Walk forward at a steady pace, keeping the ball on the book with your other hand. Let go of the ball while walking steadily. Then stop suddenly. The ball was at rest, and then, after stopping suddenly, it rolled forward . The First Newton’s Law states that an object at rest will stay at rest unless another object acts upon it. If you are sitting in the stopped car and another car hits you from behind, your fastened body is suddenly propelled forward together with the car (as the external force applied) and then stopped by the seatbelt from moving forward. But since your head and neck are not fastened (not attached to the car), they remain in resting position. After a hit, your head is forced into the seat’s headrest - backward before quickly jerking forward to catch up with the car's speed. In this case, according to Newton’s first Law, an object in motion will continue in motion with the same velocity unless acted upon by unbalanced force. The buckled-up passenger is moving with constant velocity/in one direction. Once the moving car hits the back of a stopped car (external unbalanced force resulting in a change of velocity), the passenger body will continue to move with the same velocity/ same direction until it is forced to stop. The fastened body abruptly stops by seatbelt (or is forced to decelerate to rest,) but the head continues to move forward. So, the head jerks forward until it reaches the maximum range of motion that pulls the head backward.

Are your observations consistent with Newton's first law? Discuss! o Using Newton's First Law, predict what should happen to the head of the buckled-up driver in the stopped car. Where should the brain trauma occur in this type of accident? o The head of the buckled-up driver in the stopped car would have brain trauma in the o frontal lobe. They experience hyperextension. Yes, it's consistent with Newton's first Law because the ball stays at rest or moves with the constant velocity (inertia) until the applied external force causes it to change. In this experiment, the ball's velocity is changed by how I moved or as a result of unbalanced force. In experiment 1, when I moved forward quickly, the force began to act on the resting ball, changing the ball's state. The ball rolled backward as friction force between the book and the ball started working on the ball, forcing the ball to move in the opposite, backward direction. In experiment 2, when I moved backward quickly, the force began acting on the resting ball. Similarly, the friction force started operating on the resting balls to force the ball to move in the opposite, forward direction. In experiment 3, when I walked steadily, the ball was moving with constant velocity along with the book. But when I stopped abruptly, which represents an external force applied to the book, the ball continued to move forward until it skidded off the front of the book.

o The head of the buckled-up driver in the stopped car would have brain trauma in the o frontal lobe. They experience hyperextension. o The head of the buckled-up driver in the stopped car would have brain trauma in the o frontal lobe. They experience hyperextension. I think this rapid transition of the head from rest to sudden forward motion and back often results in whiplash. It would most likely cause hyperflexion and trauma in the brain's back part/occipital lobe.

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o Using Newton's First Law, predict what should happen to the head of the buckled-up passenger in the moving car. Where should the brain trauma occur in this type of accident? In this scenario, I think the motion of the head - first jerky forward and then pulling back would cause hyperextension, and brain damage would occur in the front part/frontal lobe of the brain. Part II-- Newton's 2 nd law 4. Print your graphs. 5. Refer to your graph and describe the relationship between force and acceleration using words. 6. Write down Newton's 2nd law in the form of an equation. Define any variables and/or constants. What is your best estimate for the mass of the cart and the sensors in the video clips? -2 -1.8 -1.6 -1.4 -1.2 -1 -0.8 -0.6 -1.2 -1 -0.8 -0.6 -0.4 -0.2 0 f(x) = 0.62 x Acceleration vs Force Acceleration (m/s2) Force (N) As we see from the graph the relationship between force and acceleration has linear graph meaning that force and acceleration is directly proportional to acceleration, for example if acceleration increases force increases proportionally. The slope on the graph represents the mass of the cart according to Newton’s Second Law: F=ma

Part III—Elevator Ride 0.00 5.00 10.00 15.00 20.00 25.00 -5 0 5 10 15 20 25 Velocity vs Time Time (s) Velocity (m/s) Newton’s Second Law F=ma, where F (N) is force, m (kg) is mass of the object, a is acceleration (m/s 2 ). According to Newton’s Second Law the equation of the current graph y=0.6188x can be described: y (variables) - force applying on the cart x (variables) – the cart’s acceleration slope = 0.6188 (constant) – is the mass of the cart. Best estimates for the mass of the cart is 0.6188 kg

0.00 5.00 10.00 15.00 20.00 25.00 -2.00 -1.50 -1.00 -0.50 0.00 0.50 1.00 1.50 2.00 Acceleration vs Time Time (s) Acceleration (m/s/s) 0.00 5.00 10.00 15.00 20.00 25.00 0 2 4 6 8 10 12 14 16 Position vs Time Teme (s) Position (m) 7. From the elevator ride, discuss what the graphs tell you.

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8. From the elevator ride, explain in detail how to relate the information in the graphs to a ride in an elevator. The Velocity vs Time ( ∆v = a*∆t ) shows that the speed/velocity of the elevator started from zero and gradually increased for 1.8 s until it became constant 0.831 m/s for about 16.0 s. At the end of the graph, velocity decreased until it reached the constant velocity -0.525 m/s to the stop in the next 3.30 s. The slope to this graph is acceleration, which changes at the beginning and the end of the graph, showing increasing and decreasing velocity and horizontal line represents the constant velocity – zero acceleration. The Acceleration vs Time graph ( a = ∆v /∆t ) corresponds to the Velocity vs Time and shows the relation between acceleration and velocity as acceleration is a change of velocity. The graph represents the acceleration of the elevator. At the beginning, for 1.8 s, acceleration increases to 1.5 m/s 2 and then decreases to zero, which means the velocity is changing. Then acceleration is zero (constant velocity) for the next about 16 s. At the end of the graph, it goes in the negative direction, meaning deceleration (decreasing velocity) with -1.5 m/s 2 and then increasing to reach zero in 1.3 s or velocity becomes constant for the last 3.30 s to the stop. The Position vs Time graph ( ∆d = v*∆t) shows the displacement of the elevator from zero to 14.56 m and its stop. For the first 1.8 s, the graph increases slowly, showing low speed; then, for the next 16 s, the graph is linear and increases in the positive direction, representing the constant speed until it reaches the final destination of about 14.56 m up, then slightly decreasing at the end meaning decreasing speed for the last 3.30 s to the stop. The elevator started from the ground floor (zero). It began to accelerate to 1.5 m/s 2 in the first 0.6 s and then decelerated with kind of jerky motion to reach the constant velocity 0.831 m/s, and it takes 1.8 s. Then, the elevator went up about 13 m with the same constant velocity (zero acceleration), about 16.0 s. After that, the elevator started decelerating at -1.5 m/s2 and then accelerated (in a similar jerky motion) in 1.3 s to reach the constant velocity -0.525 m/s and then stopped in 3.30 s. We can see from the graph that the total displacement/entire trip was about 14.56 m up, and that took 22.80 s.

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