Concept explainers
Check Your Understanding What happens to force and accleration as the vehicles fall together? What wil our estimate of the velocity at a collision higher or lower than the speed actually be? And finally, what would happen if the masses were not identical? Would the force on each be the same or different? How about their accelerations?
The effect on force and acceleration when the vehicles fall together.
The estimate of the velocity at the higher or collision than the actual speed.
The effect on force and acceleration if the masses were not identical.
Answer to Problem 13.1CYU
Gravitation force, F, between two bodies and acceleration, a, due to the force is inversely proportional to the square of the distance which infers the increase in F and a at higher rate due to which speed attained by body will be greater. Change in mass will change the acceleration linearly while force will follow the Newton’s third law, that is equal and opposite.
Explanation of Solution
Introduction:
Gravitational force exerted by one body of mass
Acceleration a of body of mass
- Determine force
From equation 2, it is observed that gravitational force is inversely proportional to square of the distance between the two bodies. Force between two bodies when they are at distance r is given below
If body is moved closer half of the initial distance, that is
Therefore, force increased to 4 times as distance is decreased.Hence, force increases at a faster rate as the two vehicles brought closer.
- Determine acceleration
Acceleration also follows the similarfashion force, as given by equation 2. The new acceleration
- Determine speed
Final speed after moving a distance of scan be used using the following equation
As the acceleration increases at higher rate, hence the speed of the body (payload) will also be greater.
- Effect mass on force and acceleration
If the mass is changed (increased or decreased), force would also be changed accordingly. But according to Newton’s third law, gravitational force acting on each other would always be equal and directing opposite. As
Conclusion:
Hence, gravitation force, F, between two bodies and acceleration, a, due to the force is inversely proportional to the square of the distance
Want to see more full solutions like this?
Chapter 13 Solutions
University Physics Volume 1
Additional Science Textbook Solutions
An Introduction to Thermal Physics
College Physics (10th Edition)
Physics for Scientists and Engineers: A Strategic Approach, Vol. 1 (Chs 1-21) (4th Edition)
Essential University Physics: Volume 1 (3rd Edition)
Physics: Principles with Applications
Essential University Physics (3rd Edition)
- (a) What is the momentum of a 2000 kg satellite orbiting at 4.00 km/s? (b) Find the ratio of this momentum to the classical momentum. (Hint: Use the approximation that =1+(1/2)v2/c2 at low velocities.)arrow_forwardCheck Your Understanding Galaxies are not single objects. How does the gravitiational force of one galaxy exerted on the “closer” stars of the other galaxy compare to those farther away? What effect would this have on the shape of the galaxies themselves?arrow_forwardOne particle has mass mand a second particle has mass 2m . The second particle is moving with speed vand the first with speed 2v . How do their kinetic energies compare?arrow_forward
- Check Your Understanding If we send a probe out of the solar system starting form Earth’s surface, do we only have to escape the Sun?arrow_forward( Done T A 10 kg bowling ball rolls down the lane with speed of 7.8 m/s and hits a 0.9 kg pin. The pin rolls away at 11.2 m/s. What is the new velocitv of the bowling ball after impact? Your Answer: Answer units Send Toarrow_forwardConsider a binary star system that has bright lines at 656.72 and 656.86 nm,56.46-nm hydrogen line in the rest frame, estimate the speed V of the center of mass of the binary system is 151km/s how do I determine the mass of each star, they orbit in a circular motion.arrow_forward
- a) find velocit r ņ afder cullision. is moving to Hhe right mass m = with velou'ty v= 1"s and it collide s wi th at rest , all in one dimension. After He colli sion, velocity f m, is O.2' b) is th collisin elasdie or inelastic? K =! mvarrow_forwardPlease answer the question and its subquestions entirely! This is one question with two subquestions. According to the official Bartleby guidelines, I am alowed to have up to two subquestion! 1) Which of the following are equivalent units for the rate of change of momentum with respect to time? kg × m/s2 N s kg × s/m N/s a) While moving in the positive-x direction, Mass 1 strikes Mass 2, which is initially at rest. After the collision Mass 2 moves off in the same direction and at the same speed Mass 1 had initially. Which of the following is true? Mass 2 is larger than Mass 1. Mass 2 is the same size as Mass 1. Mass 2 is smaller than Mass 1. b) A 75 kg swimmer dives horizontally off a 500 kg raft. If the diver's speed immediately after leaving the raft is 4.0 m/s, what is the corresponding raft speed? 0.20 m/s. 0.50 m/s. 0.60 m/s. 4.0 m/s.arrow_forward1) A 5-kg particle's momentum is (211, 329) kg*m/s. Calculate its kinetic energy. 2)Calculate the x-component of vSJ. vS = 6 km/s vJ = 9 km/s θ = 23o 3) In Jupiter's frame, the incoming spacecraft is moving rightward and upward, at 45o. The outgoing spacecraft, having already flown by, is going at the same relative speed, but leftward and upward, also at 45o. (See the figures.) You are going to transform back to the sun's frame and calculate the final outgoing speed. You need to follow several steps: Calculate the vector components of the outgoing vSJ. Add Jupiter's vector velocity to get the outgoing vS. Finally calculate the outgoing speed, vS = |vS| in km/s. Enter that as the final answer. use these parameters: vSJ = 4 km/s vJ = 8 km/sarrow_forward
- dont use AI A proton has mass 1.7x10-27 kg. What is the magnitude of the impulse required in the direction of motion to increase its speed from 0.991c to 0.994c?arrow_forwardA spaceship is cruising at constant speed v = 0.5e as seen from an inertial ob- server in S'. A pie of rest mass mg = 9kg is thrown at a speed ute = 0.8c as seen from the spaceship pilot and collides with a second pie (initially at rest) of rest mass 7kg, after which the two pies stick together. 1. What is the momentum of the first pie before collision (as seen by the spaceship pilot)? 2. What is the speed of the two pies together after collision according the spaceship pilot and the observer in S'? 3. If the pies are radioactive with rest half-lifetime 79 = 10“ years, how long will it take according to the spaceship pilot for half of the pie atoms to decay?arrow_forwardCalculate the momentum a 1500kg object with a velocity of 25m/s produces?arrow_forward
- University Physics Volume 1PhysicsISBN:9781938168277Author:William Moebs, Samuel J. Ling, Jeff SannyPublisher:OpenStax - Rice UniversityUniversity Physics Volume 3PhysicsISBN:9781938168185Author:William Moebs, Jeff SannyPublisher:OpenStaxClassical Dynamics of Particles and SystemsPhysicsISBN:9780534408961Author:Stephen T. Thornton, Jerry B. MarionPublisher:Cengage Learning
- College PhysicsPhysicsISBN:9781938168000Author:Paul Peter Urone, Roger HinrichsPublisher:OpenStax College