A particle with an initial linear momentum of 2.28 kg · m/s directed along the positive x-axis collides with a second particle, which has an initial linear momentum of 4.56 kg · m/s, directed along the positive y-axis. The final momentum of the first particle is 3.42 kg · m/s, directed 45.0° above the positive x-axis. Find the final momentum of the second particle.

find the magnitude and direction above the x-axis

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To solve this problem, we can use the conservation of momentum principle, which states that the total momentum of an isolated system remains constant before and after a collision.

Let's analyze the momenta of the two particles separately before the collision:

1. Initial momentum of the first particle (P1_initial) = 2.28 kg · m/s in the positive x-axis direction.
2. Initial momentum of the second particle (P2_initial) = 4.56 kg · m/s in the positive y-axis direction.

After the collision, the final momentum of the first particle (P1_final) is given as 3.42 kg · m/s in a direction 45.0° above the positive x-axis.

To find the magnitude and direction of the final momentum of the second particle (P2_final), we can use vector decomposition and trigonometry.

First, let's find the x and y components of the final momentum of the first particle using its magnitude and direction:

P1_final_x = P1_final * cos(theta)
= 3.42 kg · m/s * cos(45°)
= 2.42 kg · m/s (rounded to two decimal places)

P1_final_y = P1_final * sin(theta)
= 3.42 kg · m/s * sin(45°)
= 2.42 kg · m/s (rounded to two decimal places)

Since momentum is conserved, the total momentum in the x-direction before the collision must be equal to the total momentum in the x-direction after the collision:

P1_initial_x + P2_initial_x = P1_final_x + P2_final_x

Since the second particle initially has no momentum in the x-direction (P2_initial_x = 0), we can simplify the equation to:

P1_initial_x = P1_final_x + P2_final_x

Substituting the known values, we have:

2.28 kg · m/s = 2.42 kg · m/s + P2_final_x

Rearranging the equation, we get:

P2_final_x = 2.28 kg · m/s - 2.42 kg · m/s
= -0.14 kg · m/s (rounded to two decimal places)

The negative sign indicates that the final momentum of the second particle is in the opposite direction along the x-axis compared to the initial momentum of the first particle.

To find the magnitude and direction of P2_final, we can use the Pythagorean theorem and trigonometry:

P2_final = sqrt(P2_final_x^2 + P2_final_y^2)

= sqrt((-0.14 kg · m/s)^2 + (4.56 kg · m/s)^2)

≈ 4.57 kg · m/s (rounded to two decimal places)

The angle (θ) above the positive x-axis can be found using the arctan function:

θ = arctan(P2_final_y / P2_final_x)

= arctan(4.56 kg · m/s / -0.14 kg · m/s)

≈ -86.67° (rounded to two decimal places)

The negative sign indicates that the direction is below the positive x-axis.

Therefore, the magnitude of the final momentum of the second particle is approximately 4.57 kg · m/s, and the direction is approximately -86.67° below the positive x-axis.