A particle of mass m1 = 2.5 kg moving along the x axis collides with a particle of mass m2 = 4.7 kg initially at rest. The incoming particle is deflected in the direction 24 degrees above the x axis, whereas the target particle moves off at 14 m/s at 39 degrees below the x axis. What are the initial and final speeds of the 2.5 kg particle?

use the eqn v2'=v1(2*m1/(m1+m2))

This eqn is specially derived fro elastic collisions where v2=0.
The speeds should be components; north, south, east, west or up, down, leftm, right etc.
Anyways, find the components and then use pythagorean theroem.
(once you have v1, you can use several eqna, like m1v1+m2v2=m1v1'+m2v2')

Andrew: Whether a theory is siftneciic or not is a property of the theory, and has nothing to do with how good our particle accelerators are.We clearly disagree here, the problem with your definition is that it opens the door for a lot of pseudoscience which can be easily recast in such a way so as to meet all the other requirements including mathematical consistency mentioned by Bee. (It's quite easy to build a mathematically consistent theory when you don't have to worry about experimental results). But this is mostly an issue of semantics and I don't find it particularly important, so probably won't argue about it any longer.Andrew: Yeah, I think it becomes hard to justify spending so much money on a theory which we are unable to test due to our current technological restrictions. But like I say, you can never say a theory is wrong using that criteria. All you can do is put the theory in mothballs for a couple of thousand years (which is perhaps what we should do with string theory).Here we are in full agreement as I've never claimed it makes the theory wrong, only that it should be shelved until the time comes when it can be experimentally tested. This is the issue of optimal allocation of limited funds.Andrew: Eek, you're getting your theories mixed-up here in a way which is going to cause horrible confusion. Quantum mechanics deals with the behaviour of particles, and predicts - as you say - the well-established wave/particle dual behaviour of de Broglie (lambda = h/p). But string theory attempts to describe the structure of particles - it's a completely different theory.I am not confusing them, your original comment didn't mention the structure, it mentioned behavior:Andrew: Apparently, it appears to be the case that elementary particles behave more as if they are vibrations of a particular frequency - which seems quite a plausible and attractive suggestion.This is why I pointed out it's already a part of QM. But even particle structure is to some extent part of QM if you accept zitterbewegung interpretation of it.Bee: There are reasons to dislike string theory, you name some of them, but there are also good reasons why this theory has attracted many researchers that shouldn't be dismissed either.I am not dismissing them while I personally don't find them very appealing I agree it was worth a try. But 40 years passed and nothing came out of it despite thousands of papers published. Enough is enough, the funds are limited and should be reallocated to other approaches or just general exploration. Bee: "The point I was trying to make above is that in the absence of experimental guidance it is foolish to abandon either approach as long as they fulfill the quality standards, read: they should be expressed in consistent maths.Yes, it should not be completely abandoned, it should be shelved or at least greatly scaled back until time comes when it can be tested. No one advocates burning of papers and books, the studies can always be resumed if there are valid reasons to do so. In the meantime however funding should be redirected to exploration of different known approaches and search for completely novel ones.

To solve this problem, we can use the principles of conservation of momentum and conservation of kinetic energy.

Step 1: Calculate initial momentum

The initial momentum of the system is given by:
p_initial = m1 * v1_initial + m2 * v2_initial

Since the second particle is initially at rest, v2_initial = 0:
p_initial = m1 * v1_initial + m2 * 0
p_initial = m1 * v1_initial

Step 2: Calculate final momentum

The final momentum of the system is given by:
p_final = m1 * v1_final + m2 * v2_final

Step 3: Apply conservation of momentum

According to the conservation of momentum, the initial and final momenta are equal:
p_initial = p_final
m1 * v1_initial = m1 * v1_final + m2 * v2_final

Step 4: Apply conservation of kinetic energy

According to the conservation of kinetic energy, the initial and final kinetic energies are equal:
(1/2) * m1 * (v1_initial)^2 = (1/2) * m1 * (v1_final)^2 + (1/2) * m2 * (v2_final)^2

Step 5: Solve the system of equations

From Step 3, we have:
v1_initial = v1_final + (m2/m1) * v2_final (Equation 1)

From Step 4, we have:
(v1_initial)^2 = (v1_final)^2 + (m2/m1) * (v2_final)^2 (Equation 2)

Substitute Equation 1 into Equation 2:
(v1_final + (m2/m1) * v2_final)^2 = (v1_final)^2 + (m2/m1) * (v2_final)^2

Expand and simplify the equation:
(v1_final)^2 + 2 * (m2/m1) * v1_final * v2_final + ((m2/m1) * v2_final)^2 = (v1_final)^2 + (m2/m1) * (v2_final)^2

Simplify further by canceling out (v1_final)^2:
2 * (m2/m1) * v1_final * v2_final + ((m2/m1) * v2_final)^2 = 0

Factor out (m2/m1) * v2_final:
((m2/m1) * v2_final) * (2 * v1_final + (m2/m1) * v2_final) = 0

Since (m2/m1) * v2_final will not be zero, we can set the term in parentheses equal to zero to find v1_final:
2 * v1_final + (m2/m1) * v2_final = 0

Solving for v1_final:
v1_final = -(m2/m1) * v2_final / 2
= -(4.7 kg / 2.5 kg) * 14 m/s / 2
= -3.544 m/s

Step 6: Calculate v1_initial

Substitute v1_final into Equation 1:
v1_initial = v1_final + (m2/m1) * v2_final
= -3.544 m/s + (4.7 kg / 2.5 kg) * 14 m/s
= -3.544 m/s + 10.64 m/s
= 7.096 m/s (rounded to three decimal places)

Therefore, the initial speed of the 2.5 kg particle is 7.096 m/s and the final speed is -3.544 m/s.

To find the initial and final speeds of the 2.5 kg particle, we need to use the principles of conservation of momentum and conservation of kinetic energy.

First, let's calculate the initial momentum of the system before the collision. The momentum of an object is given by the product of its mass and velocity. Since the second particle is initially at rest, its momentum is zero. Therefore, the initial momentum of the system is equal to the momentum of the first particle:

Initial Momentum = m1 * v1

Next, let's consider the final momentum of the system after the collision. The momentum of an object after a collision is given by the product of its mass and velocity. In this case, we have two particles with different velocities.

The final momentum in the x-direction can be calculated by considering the x-components of the velocities of both particles:

Final Momentum in x-direction = m1 * v1x + m2 * v2x

The x-component of the velocity can be found by multiplying the magnitude of the velocity by the cosine of the angle:

v1x = v1 * cos(24°)
v2x = v2 * cos(39°)

Now, let's consider the final momentum in the y-direction:

Final Momentum in y-direction = m1 * v1y + m2 * v2y

The y-component of the velocity can be found by multiplying the magnitude of the velocity by the sine of the angle:

v1y = v1 * sin(24°)
v2y = v2 * sin(39°)

According to the principle of conservation of momentum, the initial momentum is equal to the final momentum:

Initial Momentum = Final Momentum in x-direction + Final Momentum in y-direction

m1 * v1 = m1 * v1x + m2 * v2x + m1 * v1y + m2 * v2y

Now, we have two equations to solve for the unknowns v1 and v2:

1. Equation for the x-component of momentum:
m1 * v1 = m1 * v1x + m2 * v2x

2. Equation for the y-component of momentum:
0 = m1 * v1y + m2 * v2y

Now we can plug in the given values for m1, m2, and the known angles, and solve the system of equations to find v1 and v2.