19. The tiger leaps with a velocity in the x-direction only, so velocity in y-direction = 0 m/s. As a result, we will use the change in y equation to find time first, then plug it into our change in x = vt equation to find the distance the tiger moves.
20. Another horizontally launced projectile problem! We know vxi = 1.6 m/s, vyi = 0 m/s, so we can use our y-direction and x-direction equations to find both of their respective displacements. Remember that there is no acceleration in the x-direction, and g acts in the y-direction
22. This is a tricky one. We have initial velocity in both the x and y-directions. The problems asks for the height where there is only a velocity in the x-direction. At it's maximum height, the pebble has a velocity only in the x-direction, and its y-velocity is zero. We know the change in y = 8.0 m, and change in x = 9.0m. We will use our final location as the maximum height, so vyf = 0m/s. First, solve for vyi using the vyf2 = vyi2 +2gy equation. Plug vyi into the vyf = vyi +gt equation to find time. Finally, use x = vt to find vx.
24. This problem is similar to 19 and 20 - vxi only, while vyi = 0 m/s. Again, use y-direction equations to solve for time, then plug into x = vt equation to find vx!!
26. First step ALWAYS: break the initial velocity into its horizontal and vertical components using vi, the angle, and trigonometry. Plug vyi into a y-direction equation to find time. Remember that change in y = 0m, since the projectile lands at same elevation as it is thrown. So, you can rearrange and simplify the y = vyit+0.5gt2 equation to find t.
27. The ball is thrown HORIZONTALLY, so we only have a vxi, while, vyi = 0 m/s. vxi = 22m/s, and we know x, so this time use the x equations to find time, then plug into y equations to find change in y.
4. use pythagorean theorem & inverse tangent to find resultant and its direction.
5. Solve this mathematically, similar to how you did it with the road trip activity. First, draw out what each of the vectors looks like. Then, break each vector into its x and y-components. Set up a table (like we did in class), so that you can add the x-components and y-components. Remember to use correct signs (North = +, South = -, etc.). Then, create a new triangle using tip-to-tail method. Find the resultant using pythagorean theorem, and inverse tangent function to find the angle.
7(a). Pay attention to the figure shown in the book! Look at the direction of each vector.
8. Draw each of the vectors, and follow the steps for #5 - break into components, find resultant with its magnitude and direction.
** in all of these problems, there is more than one way to solve the problem - either by using just one equation, or multiple equations!**
18. Using equations given, plug in an initial velocity = 0 m/s, xi = 0m and see how much the equations are simplified.
20. We are given initial velocity, final velocity (slows to rest, so vf = 0m/s), and distance of 85 m. Select appropriate equation, and plug in to solve for acceleration.
21. We are given final velocity and acceleration, and we want to find displacement. Assume that initial velocity is 0 m/s for this problem - that it is starting from rest, and that xi=0. knowing vf, a, vi,xi and wanting to find xf, select correct equation. remember that we do not know time! can use one equation or more than one.
22. We are given final velocity, final displacement (xf), and we also know that the runner starts from rest (vi=0m/s), and from xi = 0m. From here, select an equation to solve for a using xi, xf, vi, vf - rearrange it to solve for a. After solving this equation, you can really rearrange any equation to solve for t, and plug in!
23. We know vi, vf = 0m/s (slows to rest), time, and we want to find displacement (change in x). Find an equation to solve for t, or use multiple equations.
5. You already did this one the night before! Skip it - see other blog from previous HW if you have questions.
9. You are given the speed of each locomotive, and the distance between them. Since they are both going the same speed, we know they will meet at the half way mark. Therefore, you know the speed of one locomotive, you know the distance they will travel before they meet (8.5 km / 2), now you just need to solve for time using s = d/t.
13. Use the problem solving methodology outlined on pg. 29, and be aware of units in the problem - time is given in seconds, but final velocity is given in km/hr, so convert one of these units. Since it wants your final answer in m/s/s, I would convert km/hr to m/s. Then simply use acceleration = change in velocity/time to solve for acceleration.
14. Use problem solving methodology. You are given acceleration, initial velocity, and final velocity, and you want to find time (how long it takes). Write down all information given and see your units - once again, you must convert to make time units consistent. Then, rearragen a = change in v/time equation to solve for time. Solve for time.
15. This one is simple, just use problem solving methodology. You know initial velocity, final velocity, and time, so solve for acceleration for (a), then use FLM for (b).
16. Use problem solving methodology. You know initial velocity, final velocity, and time, so solve for acceleration for the first part. When it asks many g's this is, you are just comparing your found acceleration to that of the acceleration due to gravity (9.80 m/s/s). Just divide: calculated a/9.80.
19. Use problem solving methodology. We know initial velocity, final velocity, and time. For the first part, just plug into a = change in v/time equation. For part 2, you will need to use an equation that we introduced earlier last week: average velocity = change in displacement/time. We don't know average velocity now, but can easily calculate it by adding intial and final velocities, and dividing by 2. Then, we can plug it into the average v = change in x/t equation to solve for xf. S
1, 2: just use s = d/t formula! may have to rearrange.
3. You are given distance - must convert seconds to hours to get distance.
4. Use factor label method!
5. Think of this problem as two segments of a trip:
segment 1: given speed, distance
segment 2: given speed
We also know total time, so you can solve for time for segment 1 using s = d/t. Then, solve for time for segment 2 by subtracting segment 1 time from total time. Finally, solve for distance of segment 2.
(a) add distances for segment 1 + segment 2
(b) average speed = total distance / total time
17. (a) For this one, you are just using factor label method to convert from meters to inches.
(b) In this one, you are actually looking for the number of atoms per centimeter. We know that 1 atom = 1.0x10^-10 m, so you can use this information as a conversion factor to ultimately get atoms/cm.
18. Convert each number to a common unit (I'd convert all to meters), then add them together. Remember that rule for sig figs when adding is that the least precise number determines the precision of your final number. Once all are converted, determine number with least precision, and round from there!
19. For each, you are calculating the conversion factors. In other words, for (a), you want to find out how many mi/hr are in 1.00 km/hr. Do the same for each problem!
21. The wording is tricky here!! The key is to know that a light-year is a unit of distance!
(a) In this one, calculate m/s and convert to m/year ---this is the number of meters in one light-year. Now you know that 1 light-year = xxxx meters
(b) We have a new conversion factor: 1.50x10^8 km = 1 AU. Use this information, and your answer for (a) to calculate AU/light-year.
(c) We know that 2.998x10^8 m/s. Start with this and convert it to m/hour, then use conversion factor 1.50x10^8 km = 1 AU to finally find AU/hr. NOTE: you will have to convert km --> m also.
pg. 234 Problems 1-4, 6-8, 20, 21
1. Just convert using a pi radians = 360 degrees
2. Drawing a picture of earth, and then the sun from someone's perspective on earth will help clarify what information is given here. Draw yourself on earth as a spot, and the sun from your perspective as a smaller circle. The angle is 0.5 degrees coming from your perspective looking at the sun, which is the arc length s. Radius r is the distance from earth to sun. You are solving for arc length but don't forget to divide by 2 when getting your final answer since youa re looking for radius. Use s = r*theta
3. Same as #2 but this time you are finding the theta - use book to give you earth-sun and earth-moon distances (this is your radius), and the earth and sun radiis x 2 to get s. Plug into s = r* theta
4. Draw a picture, where you are a distance from the Eiffel tower (this is radius), the eiffel tower is your s (300 m), and your theta = 6 degrees.
6. You are given diameter so remember to divide by 2 to get radius! BUT you actually don't need radius for this part - not until #7. convert revs/min to radians per second using pi radians = 180 degrees.
7. For this one, you'll use your answer in #6 and the radius from #6, and the equation v = rw.
8. Convert rev/min to rad/second. This will give you final angular velocity. Initial angular velocity = 0 (wi). Then use time and the equation angular acc = change in w divided by time to find angular acceleration.
20. Convert from rev/min to rad/ sec to get wf. wi = 0 rad/s. Convert the # of revs to radians using 2pi rad = 1 rev. Wo = 0 rad/s. Find equation to solve for angular acceleration.
21. Convert the rev/min to rad/sec to get wf. Wi = 0 rad/s. Using the time given, you can solve for angular acceleration. Then solve for theta (angular displacement) using a different equation- this will give you theta in radians so you'll need to convert back to revolutions.
Our Schedule for this unit will be as follows:
Day 1 Topics: Center of Gravity
Center of gravity (Ch. 7 Section 7-8 pgs. 194-196)
Torque (Ch. 8 Section 8-4 pgs. 217-220)
Day 2 Topics: Torque Lab on Rotational Equilibrium
Day 3 Topics: Rotational Motion
Angular quantities θ, ω, α (Ch. 8 Section 8-1 pgs. 209-214)
Kinematic equations for rotational motion (Ch. 8 Section 8-2 pg 214-215)
Day 4 Topics: Rotational Dynamics & Rotational Analogies to Translational Motion (Conceptual – no problems)
Rotational Inertia & Newton’s First Law (Ch. 8 Section 8-5 pgs. 220-224)
Rotational Kinetic Energy (Ch. 8 Sec 8-7 pgs. 225-227)
Angular Momentum (Ch. 8 Sec 8-8 pgs. 228-229)
pg. 202-203 problems:
1. This is simple momentum problem using p = mv but don't forget to convert to kg!
2. total momentum initial = total momentum final. Initially, the momentum is zero since th boat and child are both at rest. In the final position, the masses of child + boat * velocity + mass of rock * its velocity = 0. Velocity Plug in #s, solve for velocity of boat.
4. Initial momentum = mass halfback*velocity halfback + mass corner back*velocity cornerback. For the final momentum, the two masses should be added. Solve for velocity final of halfback and cornerback.
5. In this problem, mass changes. Initial momentum = initial mass*velocity = final mass(load+mass of car)* velocity to solve for final velocity.
6. There are two objects initially: boxcar and second car, but only boxcar has velocity so initial momentum of second car is zero. Final velocity is given, so mass of second car can be solved for.
7. In this problem, there are two objects initially, and one object finally (just add masses when bullet is embedded). Using information, you can calculate the final velocity of wood with bullet in it using C of M. This will become your initial velocity on the wood block's flight upward. Since no other force is acting on the wood block, the mass will slow down due to gravity. So, you have Vi (from C of M), and a = -9.81 m/s2, solve for distance.
10. One object (atomic nucleus) initially becomes two objects final as an atomic nucleus emits an alpha particle. Velocities and masses are given, so final velocity of nucleus can be found.
14-19 - Use Impulse-momentum theorem
(Ft = m*change in velocity)
14. The tennis ball is beginning at rest when it first gets into contact with the racket. The time of contact is given, and it leaves the racket at 65 m/s. Using impulse = change in momentum, you can solve for the Force (Ft = m*(change in velocity)). Once you find the force in Newtons, campare this to the force in Newtons of a 60-kg person - is it larger?
15. You are given the initial velocity TOWARDS the bat (39 m/s), but then it is hit back AWAY from the bat towards the pitcher (at 52 m/s). Since velocity is a vector, it has direction - so you must designate a direction as being positive (towards or away) and use this to find change in velocity in order to solve the problem (so either initial velocity or final velocity will be negative). Then just plug in time, mass to find force.
16. You are looking at initial velocity when force is first applied (0 m/s) until it reaches its final velocity (45 m/s) at which point the force is no longer in contact with the golfball. (a) Impulse = m*change in v, (b) Ft = impulse to solve for F
17. This one is tricky - we know that momentum is a vector so for this problem, we have to break velocity into its x and y components since the velocity is at an angle. We can see that the components of velocity that are parallel to the wall (y-components) will remain the same, but the x-components will be equal and opposite. As a result, you must designate a positive direction for velocity (towards wall?) and a negative direction. Use these values to plug into your momentum-impulse equation.
18. Again, you must designate a direction as being positive (I like East). (a) use p = mv and plug infullbacker mass, velocity. (b) impulse = change in momentum, where your final velocity is 0 m/s since the fullbacker comes to rest, (c) the tackler will have the same force applied but in opposite direction, and it will be over the same time period.
(d) Set answer in (c) equal to Ft to solve for F.
19. We know that Ft = impulse, and we can find Ft by finding the area under the graph shown. (b) Initial velocity is zero since tennis ball is originally at rest before it is hit. Use solution from (a) and set it equal to m*(change in velocity) to find vf.
Due tomorrow: problems from book and pendulum lab questions and calculations ONLY - Power lab questions and calcs will be due on Wednesday before the test!
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