The ball only loses energy during the collisions. Energy is conserved between collisions. We will denote E the totalenergy of the ball after the first collision, E2 the total energy of the ball after the second collision, and in general,En will represent the total energy of the ball after the nth collision.Consider the floor to be the reference for zero potential energy.Question 4a. Find the coefficient of restitution in terms of the ball's kineticenergy just before and just after the nth collision (n > 2).b. If the coefficient of restitution is e = 0, what does that implyabout the energy lost during the collision?c. If the coefficient of restitution is e = 1, what does that implyabout the energy lost during the collision?

Part A: The coefficient of restitution ε is defined as the ratio of relative velocity after the…

Answered: The ball only loses energy during the…

Similar questions

The following question concerns a collision in outer space, far enough from any planets so that the external forces can be considered as negligible. An unmanned spacecraft, having mass 1,500,000 kg and speed 25,000 km/hr collides with a small meteor, having mass 3,000 kg and speed 2,100 km/hr, in the Kuiper belt. a. Prior to the collision, what are the total kinetic energy and total momentum? b. Assuming a head-on collision, in which the meteor becomes lodged into the spacecraft’s body and the two objects move off together, what are the total kinetic energy and momentum after the collision? c. If instead of getting lodged into the spacecraft, the meteor bounced off, would the change in the space craft’s momentum be larger or smaller, or is there not enough information to tell? Explain.
Quesitos: The drawing shows a bullet passing through two blocks that rest on a horizontal, frictionless surface. Rule out air resistance. The bullet completely passes through the first block and is buried in the second block. Notice that after the collision, both blocks move. Can the Conservation Principle be applied from Linear Momentum to this three-body system? Justify your answer (Ignore any loss of mass from the first block). Problem: A 4.00-g bullet moves horizontally with velocity of + 355m / s. The mass of the first block is 1150 g, and its velocity after the bullet passes through it it is +0.550 m / s. The mass of the second block is 1530 g. (a) Obtain the speed of the bullet after passing through the first block, (b) obtain the velocity of the second block after the bullet is buried in it
A projectile has undergone an elastic one-dimensional collision with an initially stationary target, along an x axis. The figure below is a graph of position versus time for the projectile and target, before and after the collision. (Two line segments are parallel to the time axis.) Which is true about the masses of the projectile and target? 4 O the projectile's mass is greater O the target's mass is greater O the masses are equal
Marble A was released at a certain height of the ramp and hit Marble B on the horizontal end of the ramp. Both Marbles A and B weighs 5 grams. The horizontal distances covered by marbles A and B after the collision are 5 cm and 35 cm, respectively. The elevation from the edge of the ramp to the ground is 15cm. Assuming that energy and momentum are conserved, what is initial potential energy (in erg; erg = dyne.cm) of Marble A? Use g=980cm/s2. Round off your answer to two decimal places. Answer correctly, thank you.
A small block of mass M is released from rest at the top of the curved frictionless ramp shown in the figure. The block slides down the ramp and is moving with a speed of 3.50 m/s when it collides with a larger block mass 1.5M at rest at the bottom of the incline. The larger block moves to the right with a speed of 2m/s immediately after the collision. Express your answer in speed of the small block after the collision in terms of the given quantities and fundamental constants. •.. 1.5M 3.50 (m/s) M –1.50(m/s)M A V small block M 2 М — 1.50 V small block M 3.50(m/s) M- 3(m/s)M (c) v small block M ЗМ — 1.50М V small block M
A 83.2 g marble moving at 10.2 m/s approaches and collides head-on with a stationary 41.8 g marble. Assuming that the collision is elastic, what are the corresponding kinetic energy of each marble?
A 12 kg block, initially moving to the right, collides with a stationary 18 kg block, causing the 18 kg to move off to the right. The collision is perfectly elastic. We may conclude that after the collision, the 12 kg block will stop. move to the right with less speed than it had before the collision. move to the right with the same speed that it had before the collision. move to the left with less speed than it had before the collision. move to the left with the same speed that it had before the collision.
A pitcher claims he can throw a 0.148-kg baseball with as much momentum as a 3.28-g bullet moving with a speed of 1.50 x 10³ m/s. (a) What must the baseball's speed be if the pitcher's claim is valid? m/s (b) which has greater kinetic energy, the ball or the bullet? O The bullet has greater kinetic energy. The ball has greater kinetic energy. O Both have the same kinetic energy.
In the figure below, a 3.1 kg box of running shoes slides on a horizontal frictionless table and collides with a 1.8 kg box of ballet slippers initially at rest on the edge of the table, at height = 0.40 m. The speed of the 3.1 kg box is 5.0 m/s just h before the collision. If the two boxes stick together because of packing tape on their sides, what is their kinetic energy just before they strike the floor? ។ h
A small block of mass M is released from rest at the top of the curved frictionless ramp shown in the figure. The block slides down the ramp and is moving with a speed of 3.50 m/s when it collides with a larger block mass 1.5M at rest at the bottom of the incline. The larger block moves to the right with a speed of 2m/s immediately after the collision. Express your answer in speed of the small block after the collision in terms of the given quantities and fundamental constants. ... 1.5M 2 M - 1.50 A V small block M ЗМ— 1.50M в) V small block M 3.50 (m/s) M-3(m/s)M © v V small block M 3.50 (m/s) M- 1.50(m/s) M D V small block M
Two balls of mass m and M are attached by strings of length L. The two balls are initially at rest at an angle θ, and are then released. The balls undergo a totally inelastic collision at the bottom of their swings. Assume that m = 1.7kg, M = 2.8kg, L = 0.80m and θ = 67o.a) Calculate the speed of the balls immediately after the totally inelastic collision.b) To what maximum angle do the conjoined balls rise after the collision?
The ballistic pendulum is a device to measure the speed of a bullet using conservation of energy and momentum. A 20 g bullet is fired horizontally into a 1.1 kg wood block hanging from a 1.2-m-long string. The bullet embeds itself into the block, and the block then swings out to an angle of 35°. A. How high does the block of wood go relative to its initial height when its hanging straight down? B. What is the increase in gravitational potential energy of the wood block at its maximum height? C. Use conservation of energy to determine the speed of the block just after the bullet strikes it. D. Use conservation of momentum to determine the speed of the bullet right before it strikes the block.

SEE MORE QUESTIONS