How Do Kinetic Energies of Particles A and B Compare After Spring Release?

[cd80187]

Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes them apart and they then fly off in opposite directions, free of the spring. The mass of A is 5.0 times the mass of B, and the energy stored in the spring was 87 J. Assume that the spring has negligible mass and that all its stored energy is transferred to the particles. Once that transfer is complete, what are the kinetic energies of (a) particle A and (b) particle B?


So here is the problem. I know that the final kinetic energies of A + B = 87 J. Next, I solved for Va final, using the momentum equation and got Va final = 5 times Vb final. Next I substituted the values into the kinetic energy equation and ended up with... .5 x (5 x Mb)(5Vb) squared + .5 (Mb)(Vb) squared, where M is the mass of particle b, and V is the final velocity. So i was able to narrow it down to two unknowns, but I am unsure of where to go from here. When I solve for the two unknown, i can finish it, but I am just stuck. Thanks for the help in advance

 

[Doc Al]

So here is the problem. I know that the final kinetic energies of A + B = 87 J. Next, I solved for Va final, using the momentum equation and got Va final = 5 times Vb final.

You might want to redo that. A is 5 times heavier than B.

Next I substituted the values into the kinetic energy equation and ended up with... .5 x (5 x Mb)(5Vb) squared + .5 (Mb)(Vb) squared, where M is the mass of particle b, and V is the final velocity. So i was able to narrow it down to two unknowns, but I am unsure of where to go from here. When I solve for the two unknown, i can finish it, but I am just stuck.

What two unknowns? (The mass drops out and you can solve for the speed.)

 

[kyle8921]

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Based on the information given, it is clear that the kinetic energy of particle A will be greater than that of particle B. This is because particle A has a greater mass and therefore will require more energy to reach the same velocity as particle B.

To solve for the kinetic energies of both particles, we can use the conservation of energy principle, which states that energy cannot be created or destroyed, only transferred from one form to another. In this case, the initial potential energy of the compressed spring is transferred into the final kinetic energies of the particles.

Using the equation for potential energy stored in a spring, we can calculate the initial potential energy of the compressed spring as:

U = 1/2 kx^2

Where U is the potential energy, k is the spring constant, and x is the distance the spring is compressed. Since the spring has negligible mass, we can assume that all of its stored energy (87 J) is transferred to the particles.

Therefore, we can set up the following equation:

87 J = 1/2 kx^2

Next, we can use the relationship between the spring constant and the mass of the particles (k = mBvB^2), and substitute in the values given in the problem to solve for x:

87 J = 1/2 (5mBvB^2)x^2

Solving for x, we get x = 1.67 m.

Now, using the conservation of momentum principle, we can set up the following equation:

mAvAf + mBvBf = 0

Where m is the mass, v is the velocity, and the subscripts A and B represent particles A and B respectively.

Substituting in the known values, we get:

(5mB)(5vBf) + (mB)(vBf) = 0

Solving for vBf, we get vBf = -2.5vAf.

Finally, we can use the kinetic energy equation to solve for the kinetic energies of both particles:

KE = 1/2 mv^2

Substituting in the values for particle A, we get:

KEA = 1/2 (5mB)(5vAf)^2 = 62.5 J

And for particle B:

KEB = 1/2 (mB)(vBf)^2 = 3.125 J

Therefore, the