The Bullet-Block Mystery

Transcription

1 LivePhoto IVV Physics Activity 1 Name: Date: 1. Introduction The Bullet-Block Mystery Suppose a vertically mounted 22 Gauge rifle fires a bullet upwards into a block of wood (shown in Fig. 1a). If the bullet enters the block in line with the its center-of-mass (shown in Fig. 2a), it penetrates into the block which then rises up to a height of about 0.5 meters without spinning noticeably. 1 If the vertical shot is repeated with the bullet hitting a block off center (shown in Figs. 1b and 2b), would the block also travel straight up or off to one side? Would it be spinning? Would it rise to the same height? It turns out that the off-center bullet (Fig. 1b), causes an identical block to rise straight up to essentially the same height, as it spins counter-clockwise while rising and falling. This is puzzling. How can the bullet impart more kinetic energy to the block that s shot off-center and yet it rises to the about the same height in both situations? This doesn t seem to make sense! Maybe the off-center bullet doesn t penetrate as far into the block, so that more of its initial energy is available to start the block spinning. However, x-rays of the two blocks shown in Fig. 3 show no difference in the penetration depth of the bullet in each block! This is a mystery! Can this be explained? What if the energy lost as the bullet penetrates into the block is much greater than the kinetic energy the bullet imparts to the block? Then a tiny change in penetration distance could result in significant changes in a block s linear and rotational kinetic energy. In this assignment, you can use manufacturer s data on a bullet s mass and muzzle velocity to calculate the kinetic energy a typical bullet has before entering the block and how much of it s energy is lost plowing into the wood. If you find that the energy the bullet loses changing shape and breaking chemical bonds in the wood is very large compared to the linear and rotational kinetic energy acquired by the block, the height difference between a spinning block and a non-spinning block might not be measureable! To complete this assignment, you ll need to consider conservation of linear momentum as the bullet penetrates the block. You ll also need to calculate linear and the rotational kinetic energy of the block just as it starts rising during on center and off center collisions between the bullet and block. Fig.1: A rifle aimed at a 2 x4 block edges: [a] on center and [b] off center. Fig. 2: Edges of 5 lengths of 2x4 hit (top) on-center and (bottom) off center. 1 A video of this phenomenon can be seen on YouTube: or by completing the Bullet Block Interactive Video Vignette available at <WEB address here>. IVV Assignment: Bullet-Block

2 I. Preliminary Questions Note: You will receive full credit for any thoughtful answers you provide for the preliminary questions whether or not they match the analyses you actually perform in the two sections on Activities and Reflection that follow. As part of the learning process it is important for you to think about how you can perform an analysis before starting. Do not change your answers to the questions that follow based on later analyses you perform. 1. Suppose a bullet is shot upward toward the center of mass of a block as shown in Figure 1(a) and penetrates into it. According to the laws of physics, what quantities of this bullet-block system should be the same immediately before the bullet hits the block and immediately after the bullet has penetrated into the block and the block starts rising? Circle your answer and explain why you picked it in each case. Linear Momentum of the Block s CoM T F Angular Momentum of the Block about it s CoM T F Kinetic Energy of the Block s CoM T F 2. Suppose a bullet is shot upward to the right of the center of mass of a block i.e. off center as shown in Figure 1(b) and penetrates into it. According to the laws of physics, what quantities of this bullet-block system should be conserved immediately before the bullet hits the block and immediately after the bullet has penetrated into the block so that it has just started to rise? Circle your answer and explain why you picked it in each case. NOTE: Linear Momentum T F Angular Momentum T F Kinetic Energy of the Block s CoM T F 11-2 Physics with Video Analysis

3 The x-ray images in Figure 3 show what happens to a bullet when it penetrates a wooden block on center and off-center. Each piece of wood and the speed each bullet as it enters a block is probably slightly different. However observations show that blocks rise to more or less that same height in different trials whether they are hit on-center or off-center. Fig. 3: X-ray images of the right half of two 5 lengths of 2 by 4 pine boards shows the penetration when a bullet enters a block: (a) on-center (left image) or (b) off- center (right image). The small hole half-way up each block is where support nails were mounted at the center of each block s face. What is conserved in the collision? (a) When the bullet travels upward and hits the block in line with its center of mass and becomes embedded, the bullet transfers both momentum and kinetic energy to the block and bullet. What do the laws of conservation of momentum and energy predict will happen (b) (c) QuickTime movie entitled <Galileo'sProjectile.mov>, describe how you might use Logger Pro software to verify Galileo s hypothesis that the ball s horizontal and vertical motions are independent. ANSWER: Use the Logger Pro software and analyze the movie by clicking on the ball throughout its motion. Focus the analysis of the motion on the ball when the ball is in the air, after it has left the table. Plot position vs. time graphs separately for the horizontal (x-data) and vertical (y-data) motions of the projectile. The x vs. t graph should be linear and look as if the ball were still rolling along the table, showing that the motion is uniform. The y vs. t data should be parabolic indicating that the vertical motion is showing accelerated motion and has the same shape as if the ball were just dropped off the edge of the table without a horizontal velocity. Each motion ought to show characteristic behavior that is independent of the other. To show that the curve of the motion of the projectile is a parabola, plot a y vs. x graph. Do a curve fit and determine if the result is described by a quadratic equation. Such a result would indicate that the actual path of the ball is parabolic because the y vs. x data points are what we see when the ball is in the air. Physics with Video Analysis 11-3

4 1. Preliminary Questions Note: You will receive full credit for any thoughtful answer to provide for this question whether or not it matches the analyses you actually perform in the two sections on Activities and Reflection that follow. As part of the learning process it is important for you to think about how you can perform an analysis before starting. Please do not change your answer to the question that follows. Suppose a bullet is shot upwards toward the center of mass of a block as shown in Figure 1(a) and penetrates into a wood block. Before YouTube Video at you know that when the bullet is shot vertically in to the block in line with boc penetrates into the block in line with the blocks center of mass and the block-bullet system moves straight upward. What is conserved in the collision? (1) linear mo (d) (e) its the block in line with its center of mass and becomes embedded, the bullet transfers both momentum and kinetic energy to the block and bullet.. What do the laws of conservation of momentum and energy predict will happen We know that when a bullet penetrates into the block some of its energy is transformed into the kinetic energy of the bullet and block that causes it to rise about half a meter. Do you think it is reasonable to assume that most of the original bullet s energy goes into giving the bloc penetrating into the block that the amount that is (f) (g) QuickTime movie entitled <Galileo'sProjectile.mov>, describe how you might use Logger Pro software to verify Galileo s hypothesis that the ball s horizontal and vertical motions are independent. ANSWER: Use the Logger Pro software and analyze the movie by clicking on the ball throughout its motion. Focus the analysis of the motion on the ball when the ball is in the air, after it has left the table. Plot position vs. time graphs separately for the horizontal (x-data) and vertical (y-data) motions of the projectile. The x vs. t graph should be linear and look as if the ball were still rolling along the table, showing that the motion is uniform. The y vs. t data should be parabolic indicating that the vertical motion is showing accelerated motion and has the same shape as if the ball were just dropped off the edge of the table without a horizontal velocity. Each motion ought to show characteristic behavior that is independent of the other. To show that the curve of the motion of the projectile is a parabola, plot a y vs. x graph. Do a curve fit and determine if the result is described by a quadratic equation. Such a result would indicate that the actual path of the ball is parabolic because the y vs. x data points are what we see when the ball is in the air Physics with Video Analysis

5 2. Activity-Based Questions (a) Collect vertical and horizontal position data: Open the Logger Pro experiment file <GalileoNow.cmbl> to open a video analysis file with the Galileo s Projectile movie inserted. The movie is already scaled in meters using the information in the title frame. Obtain x, y and t data by using the Add Point tool ( ) near the top right of the movie window. Click on the middle of the ball in each frame to record its horizontal and vertical positions in meters. Note: If you mess up, you can close and re-open the file or start over by choosing Clear All Data from the Data menu. Alternatively, return to the frame with the badly located point on it, click the Select Point tool ( ), and drag the bad point to its proper location with the mouse. Is the x-component of velocity constant during the entire movie? Is the y-component of velocity constant during the entire movie? Explain your answer by drawing conclusions from both: (1) the shape of the x vs. t graph and (2) from the apparent changes in the ball s vertical and horizontal motion from frame to frame in the movie. Hint: Play the movie in the Logger Pro experiment file. ANSWER: (1) The x-component of the ball s velocity appears to be constant throughout its motion because the x vs. t graph is a straight line indicating that the ball moves with uniform motion in the horizontal direction. Also, by examining the movie, I see that the horizontal distance the ball moves from frame to frame seems constant even after the ball is falling. (2) The y-component of the ball s velocity appears to be constant until the ball rolls of the table because the y vs. t graph is a straight line indicating that the ball moves with uniform motion in the horizontal direction. (2) Also, by examining the movie, I see that the horizontal distance the ball moves from frame to frame also seems constant even after the ball is falling. (b) A Logger Pro Analysis of Horizontal Motion During the Fall: Consider the movie frames after the ball has left the edge of the table between t 0 = s and t 1 = s (the final time). Highlight these frames by selecting lines 20 through 33 in the data table. Select the x vs. t graph by clicking on it and then call up the Curve Fit feature in the Analyze menu. Click the Time Offset box. Note that the Logger Pro Time Offset is the initial time of interest, t 0, used track the falling motion. Your time offset should be s. Choose the simplest equation (Proportional, Linear, Quadratic, Cubic, etc.) that you think might match your x vs. t data. Write the x(t) equation that led to the best Curve Fit. Then list the values of the coefficients and time offset with appropriate units to three significant figures. Also report your uncertainty of fit (RMSE). Note: Use Logger Pro symbols for the variables x, t and coefficients including time offset (A, B, C, D, m, b, or t 0 etc.). Equation with Logger Pro symbols: x(t) = Coefficients & time offset with symbols, values & units: m =1.49 m/s b = 1.06m t 0 = s Uncertainty of Fit (RMSE) Physics with Video Analysis 11-5

6 (c) Comment on how well the equation you chose matches the data. ANSWER: If the student answer to part (b) is the choice of a linear equation the student should report a close match., The higher order polynomials will give a good fits too, but these are not the simplest equation. Other choices will lead to bad matches of equation to data. (d) Is there a horizontal acceleration? Is the ball s velocity in the x-direction changing while it is falling? Why or why not? What is the value of a x while the ball is falling? ANSWER: a x = 0.00 m/s 2. Because the curve-fit equation is linear, this indicates that the horizontal motion of the ball is uniform and therefore the ball is not experiencing a horizontal acceleration. Also the horizontal distance the ball moves from frame to frame is not changing. (e) Which of the coefficients in equation in 2(b) represents the horizontal velocity component, v 0x, at the moment the ball leaves the table. What is the value of v 0x? ANSWER: At time t 0 = s, v 0x = m = 1.49 m/s. This is how fast the ball is rolling in the positive x-direction when it leaves the table (as well as before and after). (f) Which of the coefficients in equation in 2(b) represents the ball s horizontal position, x 0, at the time t 0 = s when it leaves the table? ANSWER: At time t 0 = s, b = x 0 = 1.06 m (g) Vertical Motion: Is there any difference in the shape of the graph of y vs. t before and after the ball leaves the table? What does your observation tell you about the nature of vertical motion for a projectile before and after it is falling? ANSWER: The slope of the graph, representing the vertical velocity of the before it falls off the table zero, After the ball leave s the table its vertical slope starts becoming more and more negative. This indicates that the ball has no vertical acceleration at first and then undergoes is a downward or negative vertical acceleration after leaving the table. (h) A Logger Pro Analysis of Vertical Motion During the Fall: Once again you need to use only the lines of data after the ball has left the edge of the track (t 0 = s to t 1 = 1.067s). Select the y vs. t graph by clicking on it and then choose Curve Fit from the Analyze menu. Click the Time Offset box and make sure t 0 is s. Choose the simplest equation (Proportional, Linear, Quadratic, Cubic, etc.) that can fit the data. Hint: It should be related to a kinematic equation! Write the y(t) equation that led to the best Curve Fit. Then list the values of the coefficients (including uncertainties reported as RSME values and units) and the time offset (with units) using three significant figures. Note: Use Logger Pro symbols for variables (y, t) and coefficients (A, B, C, D, m, b, or t 0 ). Equation with Logger Pro symbols: y(t) = Coefficients w/ uncertainties. All with symbols, values & A(t-t units: 0 ) 2 + B(t-t 0 ) + C A =( 4.93 ± 0.01)m/s 2 B = ( ± 0.005)m/s C=(1.11± 0.00)m Time Offset: t 0 = s Note: The values for A, B, and C should be within + 5 % of those listed. And students should comment on the goodness of fit they obtained Physics with Video Analysis

7 (i) According to your best fit equation and its related kinematic equation, what is the vertical component of the acceleration of the ball, a y, when it is falling? Use three significant figures and explain how you determined a y from your equation. How close is this value to the expected value for a y? Hint: The ball is in freefall after it rolls off the table. ANSWER: A = 4.93 m/s 2 which is half the acceleration of the ball. Therefore a y = m/s 2. This is within + 0.6% of the expected result of 9.8 m/s 2 for a y. The minus sign indicates that the ball is accelerating in the downward direction. (j) Again, refer to your equation to find the initial vertical velocity component, v 0y, at time t 0 = s and describe what this variable tells you about the ball s motion it the first frame of the movie showing the ball when it is no longer in contact with the table. ANSWER: B = v 0y = m/s. This value represents the initial velocity of the ball in the vertical direction. It is not zero in this case as would be expected because the ball has left the table and has already moved in the downward direction when the analysis begins. The minus sign indicates that the ball s initial vertical velocity points toward the floor. (k) Finally, use your equation to find the initial vertical position, y 0, at time t 0 = s and describe what this variable tells you about its motion. Report your result to three significant figures. ANSWER: C= y 0 = 1.1 m. This value represents the initial vertical position of the ball above the y=0m reference line defining the coordinate axes. Since that line is along the floor close to where the ball hits, this distance is greater than the distance the ball actually fell, 1.08 m, by about 4 cm, which is close to the 3 cm radius of the pool ball. (l) What can you conclude about the nature of vertical acceleration for a freely falling projectile? ANSWER: This analysis shows that the vertical motion of the projectile is a constant acceleration very close to the expected value of 9.8 m/s 2 due to the gravitational force acting in the vertical direction. Also, the vertical motion is also independent of the horizontal motion. 3. Reflections on Your Findings (a) Summarize what you learned about the nature of projectile motion from this assignment. ANSWER: Hopefully students will comment on the three aspects of the analysis: (1) the horizontal motion of the projectile shows that the object moves with a constant velocity as supported by the x vs. t graph or the function; (2) the vertical motion shows that the projectile moves with a constant acceleration based on the graph and/or the function; and (3) the motions are independent of each other. Physics with Video Analysis 11-7

8 Final Note: Physics teachers have done video explanations of the bullet block phenomen and posted them, include Rhett Alain Chad Orzel Physics with Video Analysis