# 0.5 6.6 inelastic collisions in one dimension  (Page 3/5)

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## Section summary

• An inelastic collision is one in which the internal kinetic energy changes (it is not conserved).
• A collision in which the objects stick together is sometimes called perfectly inelastic because it reduces internal kinetic energy more than does any other type of inelastic collision.
• Sports science and technologies also use physics concepts such as momentum and rotational motion and vibrations.

## Conceptual questions

What is an inelastic collision? What is a perfectly inelastic collision?

A small pickup truck that has a camper shell slowly coasts toward a red light with negligible friction. Two dogs in the back of the truck are moving and making various inelastic collisions with each other and the walls. What is the effect of the dogs on the motion of the center of mass of the system (truck plus entire load)? What is their effect on the motion of the truck?

## Problems&Exercises

A 0.240-kg billiard ball that is moving at 3.00 m/s strikes the bumper of a pool table and bounces straight back at 2.40 m/s (80% of its original speed). The collision lasts 0.0150 s. (a) Calculate the average force exerted on the ball by the bumper. (b) How much kinetic energy in joules is lost during the collision? (c) What percent of the original energy is left?

(a) 86.4 N perpendicularly away from the bumper

(b) 0.389 J

(c) 64.0%

Professional Application

Two manned satellites approaching one another, at a relative speed of 0.250 m/s, intending to dock. The first has a mass of $4\text{.}\text{00}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{kg}$ , and the second a mass of $\text{7.50}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{kg}$ . (a) Calculate the final velocity (after docking) by using the frame of reference in which the first satellite was originally at rest. (b) What is the loss of kinetic energy in this inelastic collision? (c) Repeat both parts by using the frame of reference in which the second satellite was originally at rest. Explain why the change in velocity is different in the two frames, whereas the change in kinetic energy is the same in both.

(a) 0.163 m/s in the direction of motion of the more massive satellite

(b) 81.6 J

(c) $8\text{.}\text{70}×{\text{10}}^{-2}\phantom{\rule{0.25em}{0ex}}\text{m/s}$ in the direction of motion of the less massive satellite, 81.5 J. Because there are no external forces, the velocity of the center of mass of the two-satellite system is unchanged by the collision. The two velocities calculated above are the velocity of the center of mass in each of the two different individual reference frames. The loss in KE is the same in both reference frames because the KE lost to internal forces (heat, friction, etc.) is the same regardless of the coordinate system chosen.

Professional Application

A 30,000-kg freight car is coasting at 0.850 m/s with negligible friction under a hopper that dumps 110,000 kg of scrap metal into it. (a) What is the final velocity of the loaded freight car? (b) How much kinetic energy is lost?

Professional Application

One of the waste products of a nuclear reactor is plutonium-239 $\left({}^{\text{239}}\text{Pu}\right)$ . This nucleus is radioactive and decays by splitting into a helium-4 nucleus and a uranium-235 nucleus $\left({}^{4}{\text{He}}^{}+{}^{\text{235}}U\right)$ , the latter of which is also radioactive and will itself decay some time later. The energy emitted in the plutonium decay is $\text{8.40}×{\text{10}}^{–\text{13}}\phantom{\rule{0.25em}{0ex}}J$ and is entirely converted to kinetic energy of the helium and uranium nuclei. The mass of the helium nucleus is $\text{6.68}×{\text{10}}^{–\text{27}}\phantom{\rule{0.25em}{0ex}}\text{kg}$ , while that of the uranium is $3\text{.}\text{92}×{\text{10}}^{–\text{25}}\phantom{\rule{0.25em}{0ex}}\text{kg}$ (note that the ratio of the masses is 4 to 235). (a) Calculate the velocities of the two nuclei, assuming the plutonium nucleus is originally at rest. (b) How much kinetic energy does each nucleus carry away? Note that the data given here are accurate to three digits only.

Professional Application

The Moon’s craters are remnants of meteorite collisions. Suppose a fairly large asteroid that has a mass of $5\text{.}\text{00}×{10}^{\text{12}}\phantom{\rule{0.25em}{0ex}}\text{kg}$ (about a kilometer across) strikes the Moon at a speed of 15.0 km/s. (a) At what speed does the Moon recoil after the perfectly inelastic collision (the mass of the Moon is $7\text{.}\text{36}×{10}^{\text{22}}\phantom{\rule{0.25em}{0ex}}\text{kg}$ ) ? (b) How much kinetic energy is lost in the collision? Such an event may have been observed by medieval English monks who reported observing a red glow and subsequent haze about the Moon.

(a) $1\text{.}\text{02}×{\text{10}}^{-6}\phantom{\rule{0.25em}{0ex}}\text{m/s}$

(b) $5\text{.}\text{63}×{\text{10}}^{\text{20}}\phantom{\rule{0.25em}{0ex}}J$ (almost all KE lost)

What is the speed of a garbage truck that is $1\text{.}\text{20}×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\text{kg}$ and is initially moving at 25.0 m/s just after it hits and adheres to a trash can that is 80.0 kg and is initially at rest?

24.8 m/s

(a) During an ice skating performance, an initially motionless 80.0-kg clown throws a fake barbell away. The clown’s ice skates allow her to recoil frictionlessly. If the clown recoils with a velocity of 0.500 m/s and the barbell is thrown with a velocity of 10.0 m/s, what is the mass of the barbell? (b) How much kinetic energy is gained by this maneuver? (c) Where does the kinetic energy come from?

(a) 4.00 kg

(b) 210 J

(c) The clown does work to throw the barbell, so the kinetic energy comes from the muscles of the clown. The muscles convert the chemical potential energy of ATP into kinetic energy.

show that the set of all natural number form semi group under the composition of addition
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The denominator of a certain fraction is 9 more than the numerator. If 6 is added to both terms of the fraction, the value of the fraction becomes 2/3. Find the original fraction. 2. The sum of the least and greatest of 3 consecutive integers is 60. What are the valu
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2. (x) + (x + 2) = 60 2x + 2 = 60 2x = 58 x = 29 29, 30, & 31
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4
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Solve for the first variable in one of the equations, then substitute the result into the other equation. Point For: (6111,4111,−411)(6111,4111,-411) Equation Form: x=6111,y=4111,z=−411x=6111,y=4111,z=-411
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(61/11,41/11,−4/11)
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x=61/11 y=41/11 z=−4/11 x=61/11 y=41/11 z=-4/11
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