AP Physics 1 Kinematics Free Response 8
TLDRIn this AP Physics 1 video, Alan from Bottle Stem Coach tackles challenging kinematic problems, focusing on a two-part projectile motion scenario involving a ball thrown onto a flat horizontal roof. He guides viewers through the process of calculating the time it takes for the ball to land on the far side, emphasizing the importance of understanding the motion in both the vertical and horizontal directions.
Takeaways
- ๐งโ๐ซ Alan from Bottle Stem Coach is presenting AP Physics 1 kinematic problems.
- ๐ The focus is on advanced kinematic problems for thorough practice.
- โธ๏ธ Alan encourages viewers to pause and attempt solving the problem themselves before watching the solution.
- โฝ The problem involves a two-part projectile motion of a ball thrown onto a flat horizontal roof.
- ๐ข The ball lands on the roof at the highest point, rolls across the roof, and falls off the other side.
- ๐ Variables used: initial speed (v-not), angle (theta), and length of the building (L).
- ๐ข Time to maximum height is determined using the vertical motion equations.
- โ๏ธ Time to rise to maximum height equals V naught sine theta over G, and the time to fall is the same.
- โก๏ธ Time to slide across the roof is L over V naught cosine theta, considering horizontal motion with no acceleration.
- ๐ Total time is the sum of the times for rising, falling, and sliding across the roof: 2 V naught sine theta over G + L over V naught cosine theta.
Q & A
What is the context of the problem being discussed in the video?
-The problem involves a ball being thrown onto a flat horizontal roof, landing at the highest point of its path, rolling across the roof, and then falling off the other side. The task is to find the total time from when the ball is thrown to when it lands on the far side, using given variables.
Which variables are given to solve the problem?
-The given variables are the initial speed (v-not), the angle of the throw (theta), and the length of the building (L).
How is the time to reach the maximum height calculated?
-The time to reach the maximum height is calculated using the equation t_{rise} = \frac{v_{not} \sin(\theta)}{g}, where g is the acceleration due to gravity.
Why is the time to fall the same as the time to rise?
-The time to fall is the same as the time to rise due to the symmetry of projectile motion in the absence of air resistance.
What equation is used to calculate the time the ball spends sliding across the roof?
-The equation used is L = v_{not} \cos(\theta) \cdot t, which solves to t_{slide} = \frac{L}{v_{not} \cos(\theta)}.
How is the total time calculated?
-The total time is calculated by summing the time to rise, the time to fall, and the time to slide across the roof: t_{total} = 2 \frac{v_{not} \sin(\theta)}{g} + \frac{L}{v_{not} \cos(\theta)}.
What is the significance of ignoring friction and rotational energy in this problem?
-Ignoring friction and rotational energy simplifies the problem to pure translational motion, making the calculations straightforward and focusing only on the kinematic aspects.
Why is the vertical direction chosen to find the maximum height?
-The vertical direction is chosen because the maximum height is reached when the vertical component of the velocity (V_y) is zero.
How is the initial vertical velocity represented in the equations?
-The initial vertical velocity is represented as v_{not} \sin(\theta).
What assumptions are made to solve this problem?
-The assumptions include ignoring air resistance, friction, and rotational energy, and assuming constant acceleration due to gravity.
Outlines
๐ AP Physics 1 Kinematics Challenge
In this educational video, Alan from Bottle Stem Coach introduces a challenging AP Physics 1 kinematics problem involving projectile motion. The problem describes a ball thrown onto a flat horizontal roof, which lands at the highest point, rolls across, and falls off the other side without friction. Alan encourages viewers to attempt the problem before revealing the solution. The problem is broken down into three parts: the ball's ascent and descent, and its horizontal motion across the roof. The solution involves using kinematic equations to find the time it takes for the ball to complete its journey, considering the initial velocity, angle of projection, and the length of the building.
๐ Detailed Solution to the Projectile Motion Problem
Alan proceeds to solve the problem by first addressing the vertical motion, focusing on the time to reach the maximum height, which is determined by setting the final vertical velocity to zero and using the kinematic equation involving initial velocity, acceleration due to gravity, and time. He then calculates the time for the ball to slide horizontally across the roof, using the horizontal distance and the initial horizontal velocity component. The total time for the ball's journey is found by summing the time to rise, the time to fall symmetrically, and the time to slide across the roof. The units are checked for consistency, ensuring the solution is expressed in the correct units of time and distance.
Mindmap
Keywords
๐กKinematics
๐กProjectile motion
๐กInitial speed (v-not)
๐กAngle (theta)
๐กMaximum height
๐กSymmetry
๐กHorizontal distance (L)
๐กVertical velocity
๐กAcceleration due to gravity (g)
๐กFrictionless
Highlights
Introduction to a continuation of AP Physics 1 kinematic problems.
Emphasis on tackling challenging kinematic problems beyond simple AP Physics responses.
Encouragement for viewers to attempt problems before continuing with the video.
Description of a two-part projectile motion scenario involving a ball thrown onto a flat horizontal roof.
Assumption of no friction and the use of rotational energy concepts.
Introduction of variables v-not for initial speed, theta for angle, and L for the length of the building.
Objective to find the total time from when the ball is thrown to when it lands on the far side.
Breakdown of the problem into three distinct motion parts for analysis.
Explanation of the symmetry in the time to rise and fall due to the vertical motion.
Use of the equation Vf^2 = Vi^2 + 2ad to find the time to reach maximum height.
Identification of the initial vertical velocity as v-not sine theta.
Calculation of the time to reach the maximum height using the equation Vf = Vi + at.
Determination of the time spent sliding across the roof using horizontal motion equations.
Solution for the time spent sliding across the roof using the equation ฮx = Vโt + 1/2at^2.
Final calculation of total time by summing the rise, fall, and slide times.
Verification of units to ensure the correctness of the time calculation.
Conclusion and sign-off for the next video in the series.
Transcripts
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