Two boxes are tied together by a string and are sitting at rest on a frictionless surface

Two boxes are tied together by a string and are sitting at rest in the middle of a large frictionless surface.Between the two boxes is a massless compressed spring. The string tying the two boxes together is cutsuddenly and the spring expands, pushing the boxes apart. The box on the left has four times the mass ofthe box on the right.4MSpringStringMAt the instant (after the string is cut) that the boxes lose contact with the spring, will the speed ofthe box on the left begreater than, less than,orequal tothe speed of the box on the right?Explain.
Two boxes are tied together by a string and are sitting atrest in the middle of a large frictionless surface. Betweenthe two boxes is a massless compressed spring. Thestring tying the two boxes together is cut suddenly andthe spring expands, pushing the boxes apart. The box onthe left has four times the mass of the box on the right.At the instant (after the string is cut) that the boxes losecontact with the spring, the speed of the box on the leftwill be…A.)Greater than the right boxB.)Less than the right boxC.)Equal to the right boxD.)Not enough information providedi-ClickerIn Case A, a metal bullet penetrates a wooden block. InCase B, a rubber bullet with the same initial speed andmass bounces off of an identical wooden block.Will the speed of the wooden block after the collisionbegreater in Case A, greater in Case B,orthe same inboth cases?A.The speed will be greater in Case A because themetal bullet exerts a larger force on the block.B.The speed will be greater in Case B because thebullet changes direction.C.The speed will be the same in both cases becausethe bullets have the same mass and initial speedand give the block the same momentum.D.Cannot be determined.i-Clicker

Kristie F.

Physics 101 Mechanics 4 months, 4 weeks ago

B5-CT21: Two BOXES ON A FRICTIONLESS SURFACE _MOMENTUM AND SPEED Two boxes are tied together by a string and are ~sitting at rest on a frictionless surface. Between the two boxes is a massless compressed spring: The string tying the two boxes together is cut and the spring expands, pushing the boxes apart: The box On the left has four times the mass of the box On the right: String 4M M Spring (a) After the string is cut and the boxes lose contact with the spring, will the magnitude of the momentum o the box On the left be (i) greater than, (ii) less than, Or (iii) equal to the magnitude of the momentum of the box OH the right? Explain your reasoning: (b) At the instant (after the string is cut) that the boxes lose contact with the spring; will the speed of the box 0n the left be (i) greater than, (ii) less than, or (iii) equal to the speed of the box on the right? Explain your reasoning:

Two boxes are tied together by a string and are sitting at rest on a frictionless surface

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Two boxes are tied together by a string and are sitting at rest on a frictionless surface

Robert D.

University of Salford

Two blocks of masses and 3 are placed on a frictionless, horizontal surface. A light spring is attached to the more massive block, and the blocks are pushed together with the spring between them (Fig. P9.9). A cord initially holding the blocks together is burned; after that happens, the block of mass 3 moves to the right with a speed of 2.00 (a) What is the velocity of the block of mass ? (b) Find the system's original elastic potential energy, taking . (c) Is the original energy in the explain your answer cord? (d) Explain your answer to part (c). (e) Is the momentum of the system conserved in the how that is possible considering (f) there are large forces acting and (g) there is no motion beforehand and plenty of motion afterward?

Two boxes are tied together by a string and are sitting at rest on a frictionless surface

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Video Transcript

Okay, so here we have a situation whether two blocks connected by a spring on a small light string on that like string snaps, sending the two masses in opposite directions. Now, before anything happens, we have a total mental off zero. So it's right. Peanuts equals zero. So total months was going to be zero. Which means that in our second situation, um, one will be one final plus. And to be to final has to equal to zero. Now we know that the massive the larger object is three am in amassed. A small object is just so and the one plus three and me too, is equal to zero. The lodge, a block, has a mass of two meters per second going to the right. So now we know and the one plus 6 a.m. equals zero so that we could just rearrange remove the EMS. We can see that V one is equal to negative six. I had meters per second. It is moving to the left at six meters per second. All right, Poppy, I want to find the original amount of elastic potential energy. So we find the mass is 0.35 kilograms. Let's find the elastic potential energy. Knowing that energy is conserved, we should know that the final kinetic energy is equal to the initial elastic potential energy on the question friend. If a kinetic energy is 1/2 times mass times velocity squared. So substituting in numbers for both of our blocks. 1/2 times point 0.35 Terms of velocity of that first block, which is negative six squared plus 1/2 times three times 30.35 times a velocity of that one, which was two squat. Concluding that out, you'll find that the total kinetic energy and therefore the amount of elastic potential energy start in the spring is eight point full. Jules. So is the energy stored in the spring or the card. While the court doesn't still any energy, it doesn't have the ability to do work on the object. So the only thing able to store the energy is the spring. Unlike I said, work is done on the spring, which allows it to store the energy. The card exerts a force, but it doesn't exert that force over a specific distance. Now momentum here is conserved. Even though we have large forces acting in our system. They're all within the system. So the forces are all internal because all the forces air internal. That's not going to affect our overall momentum off the system. And again, there's plenty of motion after the card breaks in the spring pushes those blocks, and there's no momentum before while it's storing that energy. But the momentum is conserved because they go in opposite directions. The mass multiplied by the velocity is what we're looking at in terms of momentum. Now the small block moves to the left in a large block, moves to the rights, but they move with equal momenta and different velocities. It's equal in magnitude but opposites in direction, so the mountain is conserved. It's equal in magnitude, but opposite in direction. Even though the forces are all internal, that doesn't affect the momentum of our system.