Two air carts of mass m1=0.84 kg and m2=0.42 kg are placed on a frictionless track. Cart 1 is at

Two air carts of mass m1=0.84 kg and m2=0.42 kg are placed...

Two air carts of mass $m_{1}=0.84 \mathrm{kg}$ and $m_{2}=0.42 \mathrm{kg}$ are placed on a frictionless track. Cart 1 is at rest initially, and has a spring bumper with a force constant of 690 $\mathrm{N} / \mathrm{m} .$ Cart 2 has a flat metal surface for a bumper, and moves toward the bumper of the stationary cart with an initial speed $v=0.68 \mathrm{m} / \mathrm{s}$ . (a) What is the speed of the two carts at the moment when their speeds are equal? (b) How much energy is stored in the spring bumper when the carts have the same speed? (c) What is the final speed of the carts after the collision?

Two air carts of mass $m_{1}=0.84 \mathrm{kg}$ and $m_{2}=0.42 \mathrm{kg}$ are placed on a frictionless track. Cart 1 is at rest initially, and has a spring bumper with a force constant of 690 $\mathrm{N} / \mathrm{m} .$ Cart 2 has a flat metal surface for a bumper, and moves toward the bumper of the stationary cart with an initial speed $v=0.68 \mathrm{m} / \mathrm{s}$ .
(a) What is the speed of the two carts at the moment when their speeds are equal?
(b) How much energy is stored in the spring bumper when the carts have the same speed? (c) What is the final speed of the carts after the collision?

Key Concepts

Elastic Collision Dynamics

An elastic collision is one where both momentum and kinetic energy are conserved. For systems that involve springs and collisions, assuming elasticity allows us to predict not only the speeds at maximum compression but also the final speeds of the objects after the spring decompresses, as the exchange of energy is idealized without losses.

Center-of-Mass Velocity

The center-of-mass velocity is the weighted average velocity of all the masses in the system and remains constant if there is no external force. In collision problems, especially at the moment of maximum compression when the objects move together, the common velocity matches the center-of-mass velocity, which simplifies the analysis of the system’s energy distribution.

Spring Potential Energy

Spring potential energy is the energy stored in a spring when it is deformed, described by Hooke's law through the spring constant and the degree of compression or extension. In collision problems, this concept explains the temporary storage of kinetic energy during the maximum compression phase and its subsequent release, influencing the motion of the colliding bodies.

Conservation of Momentum

In collision dynamics, the total momentum of a system remains constant if no external forces act on it. This principle is used to analyze how the individual momenta of bodies combine and redistribute during interactions, such as when two objects collide or compress a spring, allowing the determination of common velocities or final speeds after impact.

Energy Conservation and Transformation

Energy conservation states that the total energy in an isolated system remains constant, although it may change form. In collisions involving springs, kinetic energy can be temporarily converted into spring potential energy at the point of maximum compression, and then reconverted to kinetic energy, a process that is essential for analyzing the collision’s dynamics.

Transcript

00:01 Question 84 is kind of a fun problem that involves conservation of momentum and conservation of energy.

00:05 So we have two air carts, mass of 0 .84 and mass of 0 .42.

00:13 And the smaller cart is initially, so cart 1, the larger cart is at rest, so v is 0.

00:20 And then the other cart is initially pushed towards it with a speed of 0 .68 meters per second headed this direction.

00:28 There is a spring on this that's going to allow these to have an elastic collision, and that spring has a constant of 690 newtons per meter.

00:42 Question a says, what is the speed of the two carts at the moment when their speeds are equal? so, a, if they have equal velocities, what is that velocity? and so this is a conservation of momentum problem, and so we have 0 .68 times 0 .42 plus 0 .0, because that's the momentum, they have at the beginning is equal to the momentum they have at the end but this is not the end after the collision it's the end when they have the same velocity so i can go ahead and write 0 .42 the massive cart 1 plus 0 .84 the massive cart 2 times their common velocity v and so 0 .68 times 0 .42 is 0 .2856 and this has to equal 1 .26 times v, so if we divide by 1 .26, we get a final velocity of 0 .227 meters per second.

01:39 And that's the velocity the two carts have when they have the same velocity.

01:44 Part b of this question says, how much energy is stored in the spring when the carts have the same speed? so b, we want the elastic potential energy in the spring.

01:53 This one, we're actually going to rely on the law of conservation of energy four.

01:57 The carts started with some kinetic energy.

02:01 The gravitational potential never changes on this level track so we can ignore it.

02:05 So the kinetic energy we start with is constant, well, the mechanical energy we start with, is going to be converted into the kinetic energy of both of the carts plus whatever elastic potential energy is stored in the spring.

02:18 And so i'm going to have one half the mass of the velocity or one half the mass of 0 .42 times the velocity of 0 .680.

02:27 Squared that is the kinetic energy we started with has to equal half the mass well at the end of this problem we have both of these carts so 1 .26 kilograms with their final velocity of 0 .227 squared plus whatever elastic potential energy is stored in that spring so the energy we started with has got to be distributed everywhere and so 0 .5 times 0 .42 times 0 .68 squared is 0 .0 .6 is 0.

02:58 0 .07, 0971, 04.

03:04 This is equal to 0 .5 times 1 .26 times 0 .227 squared, or 0 .03246, plus the elastic potential energy in the spring...

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