The center of mass is the point at which all the mass is balanced.
The weighted average of all points of mass along each axis is what it is found by.
The center of mass is where our forces are applied in our diagrams.
The center of mass of an object is determined by the positions specified in our physics problems.
An unsymmetrical object may not appear to travel in a parabola when undergoing projectile motion as it twists and turns during its flight.
The center of mass is following the path.
The overall momentum of a system of interacting particles obeys all of the laws of nature.
A mass of gas consisting of many moving and interacting molecules can be modeled as being attracted to Earth by placing the total mass of the gas cloud at its center of mass.
The quantity is equal to mv-.
The area under a graph of force versus time is where the impulse is found.
The change in momentum is equal to the change in impulse.
The rate of change of momentum is equal to force.
In an isolated system, the total momentum is not changed.
The system's energy is conserved in an elastic collision.
In an inelastic collision where mass stick together, the energy loss transforms into heat.
In an isolated system, momentum is always maintained.
If you read a problem that doesn't explicitly state that momentum is involved, you can assume that total momentum is conserved, and you should write the equations for that.
Unless the collision is elastic, the energy is not necessarily conserved.
Decide if there is an impact or a collision.
If there is a collision, you should know whether it is elastic or inelastic.
If the collision is inelastic, be sure to determine the new combined mass.
If the collision is elastic, you should write the equations to conserve both momentum and energy.
The change in momentum of a mass is equal to the impulse given to it.
The change in momentum is the same as the net force.
Remember the sign conventions for left, right, up, and down motions, and be sure to take into account any reversal of directions.
If studied in the center-of-mass frame of reference, motions may be simpler.
The motions of mass particles relative to the center of mass are considered in the frame.
The center of mass will follow a smooth path after an internal explosion since the initial momentum in that frame was zero.
The center of mass of the debris still follows the original trajectory if a projectile is launched at an angle.
Two carts have masses of 1.5 and 0.7 kilogram and are held together by a massless spring.
The cart moves to the left with a speed of 7 m/s.
A ball with a mass of 0.15 kg has a speed of 5 m/s.
It bounces off the wall with a speed of 3 m/s.
A baseball is thrown at 35 m/s.
The batter hits it with a high speed.
An object is moving with a speed of 6 m/s to the right.
It collides and sticks to an object moving with a speed of 3 m/s in the same direction.
A mass moving with a speed of 7 m/s collides with a mass moving with a speed of 4 m/s.
The mass has a new speed of 3 m/s after it reverses direction.
A mass m is attached to a massless spring.
The mass rests on the floor.
The system is released after being compressed a distance from the spring's initial position.
Two blocks are moving on a horizontal surface.
The 1-kg block has a speed of 12 m/s, and the 4-kg block has a speed of 4 m/s, as shown in the diagram below.
A massless spring is attached to the end of the block.
The spring has a force constant of 1,000 N/m.
A disk is at rest on a horizontal surface.
It is hit by a disk with a speed of 4 m/s.
The disk has a speed of 2 m/s after the collision.
A girl is standing on a platform with wheels on a horizontal surface.
The platform is attached to a massless spring with a force constant of 1,000 N/m.
The girl throws a ball at an angle of 30 degrees to the horizontal.
In a recoil the mass go in opposite directions.
The rebound velocity is in the opposite direction.
Since both objects are moving in the same direction, we can write (1)(6) + (2) (3) v '.
The initial energy of the 1- kilo object is 18 J, while the initial energy of the 2- kilo mass is 9 J.
The total initial energy is 27 J.
After the collision, the object has a speed of 4 m/s and a final energy of 24 J.
3 J of energy has been lost.
We write (2)(7),(4),(4),(4),(4),(4),(4) and get v ', which is 1 m/s.
The mass is taken back under the radical sign so that we get p-xmk.
For an instant, we have an inelastic collision.
We get v-f=53.6 m/s when we solve for the final velocity.
The initial values for the velocities are 72 J and 32 J for the blocks.
The total initial energy is 104 J.
We get a final energy of 78.4 J using the final velocity of 5.6 m/s and the combined mass of 5 kg.
The spring is compressed by the difference in energy of the two objects.
The maximum compression of the spring after the collision is calculated using the formula for the work done against it.
If we treat the situation as elastic, we can use the initial velocities as a "before" condition for momentum and energy.
After the rebound has taken place, we want to solve for the two final velocities of the blocks.
To find both velocities, we need two equations, and we use the conserved of energy in this case to assist us.
The known numbers are 28 v 1 f + 4 v 2 f and 208 v1f2+4v2f2.
Only one of the two choices provides a meaningful set of solutions.
The first mass has to rebound and its final velocity has to be negative.
Our final answers are v 2 f and v 1 f.
In the x -direction, we only have the initial momentum of the 0.1- kilogram disk with a speed of 4 m/s.
The x-component of momentum given by (0.1)(2)cos 43 is the result of the collision.
The disk has an unknown angle to the x- axis and has zero momentum.
The x component will be positive and the y component negative if the angle is below the x axis.
We will get a positive answer if our assumption is correct.
A negative answer will let us know that the assumption is incorrect.
The y component of final momentum is given by (0.4) v 2 f cos th.
The ratio gives a tan of 0.535 and a th of 28deg.
The final velocity for the disk is 0.72 m/s.
The initial energy is found by using the data.
The final data gives us a total final energy of 0.2 J.
The amount of energy has been lost.
The ball's velocities are 35 cos 30 and 30.31 m/s.
(1)(30.31) is used to state that (1,050) v '.
The recoil velocity of the platform-girl system can be expressed as v'-0.0289 m/s.
The object's momentum is zero because it has zero velocity.
An upward force is given to an object when it bounces.
If it is large enough, it can be dangerous.
If the mass is placed vertically onto the object, the increase in mass appears to lower the velocity since no force was acting in the direction of motion.
Since the object did not have a horizontal motion, the energy from the moving object must act to accelerate the mass m. From the mass M frame of reference, it appears that the mass m is moving towards the object, slowing it down.