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$\bullet$ On a frictionless air track, a 0.150 $\mathrm{kg}$ glider moving at1.20 $\mathrm{m} / \mathrm{s}$ to the right collides with and sticks to a stationary0.250 $\mathrm{kg}$ glider. (a) What is the net momentum of this two-glider system before the collision? (b) What must be the net momentum of this system after the collision? Why? (c) Useyour answers in parts (a) and (b) to find the speed of the glid-ers after the collision. (d) Is kinetic energy conserved duringthe collision?

a) 0.18 $\mathrm{kg} \mathrm{m} / \mathrm{s}$b) Therefore, the momentum of the glider system after the collision is 0.18 $\mathrm{kg} \mathrm{m} / \mathrm{s}$c) 0.45 $\mathrm{m} / \mathrm{s}$d) since they stick together, the collision is inelastic, so kinetic energy is not conserved during the collision.

Physics 101 Mechanics

Chapter 8

Momentum

Physics Basics

Kinetic Energy

Potential Energy

Energy Conservation

Moment, Impulse, and Collisions

Fatma L.

December 13, 2021

On a frictionless, horizontal air table, puck A (with mass 0.25kg), which is moving to the right toward puck B (with mass 0.35kg), which is initially at rest. After the collision, puck A has a velocity of 0.12m/s to the left, and puck B has a velocity of

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On a frictionless air trac…

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So in this question, we have to players one ofthe mass syrup on 150 programs. Well, when you get the velocity off 1.20 meter per second and the other one off master, 0.250 feet hams. But the stationery after the moving lighter hits the stationary later, the state together before my larger later off zero point four kilograms, adding the two masses and they move. Let's say with the velocity V now for party. If we want to find out the initial off the system, the initial boredom is simply mass times velocity, which is zero for the second object. But for the first object, it's 0.15 kilograms multiplied by 1.2 meters per second, which is its transport. Toby. 01180 Hey, did you meet over second nor the three significant digits? Because all are all our numbers are given up to three things now for R V being asked what the final one will be off the system now simply because there are no other cars on the forces on the glider because there is no friction. The horizontal went there most weekends over, which means the initial mountain horizontally must be equal to the final warning horizontally and the final warned them will also be They don't want 180 It's you or the gliders. You know, for Parsi using this information, we want to find out the velocity V. So we know that the final award of this simply the mask which is 014 kilogram multiply by the velocity. This will be the final one. We know that this is already 0.18 kilogram meters per second. Therefore, the velocity is 0118 you're debasing one for which is 01450 Mr. President, Now we want to see if kind of energy is conserved in this collision. Now the initial kind of energy. In the initial case, the second letter is not moving. So the kind of country is only in the first lighter. So that would be half as masters. The las T squared on my excuse. You don't want 15 kilograms and the velocity square would be one point of squared. That turns out to be a 0.108 Jules. But finally the kind of energy would be the kind of energy off the full system, which is the kind of energy off the combined lighter moving just half times mask, which will be there about four kilogram modified by the velocity squared, which would be see the 145 meters per second squared. It's Chancellor Toby. 010405 Jules, notice the drop in the kind of energy this means the kind of energy is not concerned in this collision. In fact, any collision where objects stick together does not going so chaotic energy. That's because some work is done. Sticking them together on that kind of work usually turns into heat, and hence the kinetic energy will not be controlled.

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