Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic collision. Momentum is conserved. Object A has a mass of mA = 17.9 kg and an initial velocity of = 8.82 m/s, due east. Object B, however, has a mass of mB = 28.9 kg and an initial velocity of = 4.01 m/s, due north. Find the (a) magnitude and (b) direction of the total momentum of the two-object system after the collision.

To find the magnitude and direction of the total momentum of the two-object system after the collision, we can use the principles of conservation of momentum.

Let's start by finding the total momentum of object A before the collision. The momentum of an object is given by the product of its mass and velocity:

P(A) = m(A) * v(A)

Substituting the given values, we have:

P(A) = 17.9 kg * 8.82 m/s

P(A) = 157.98 kg·m/s

Now, let's find the total momentum of object B before the collision:

P(B) = m(B) * v(B)

Substituting the given values, we have:

P(B) = 28.9 kg * 4.01 m/s

P(B) = 116.389 kg·m/s

Since momentum is a vector quantity, we need to consider both the magnitudes and directions. Vector addition is used to determine the total momentum after the collision.

To add the two momenta together, we can use the Pythagorean theorem:

P(total) = √(P(A)^2 + P(B)^2)

Substituting the numbers:

P(total) = √((157.98)^2 + (116.389)^2)

P(total) = √(24958.0404 + 13545.1921)

P(total) = √38410.2325

P(total) ≈ 196.00 kg·m/s

So, the magnitude of the total momentum of the two-object system after the collision is approximately 196.00 kg·m/s.

Now let's find the direction of the total momentum. We can use trigonometry to determine the angle between the resultant momentum and the positive x-axis (east direction).

θ = arctan (P(B) / P(A))

Substituting the values:

θ = arctan (116.389 / 157.98)

θ ≈ arctan (0.7372)

θ ≈ 35.318°

Since object A was moving due east before the collision, the direction of the total momentum is 35.318° east of the positive x-axis.

Therefore, the (a) magnitude of the total momentum is approximately 196.00 kg·m/s and (b) the direction is approximately 35.318° east of the positive x-axis.

To find the total momentum of the two-object system after the collision, we first need to calculate the individual momenta of object A and object B before the collision.

For object A:
Momentum = mass × velocity
P(A) = m(A) × v(A)

Given that:
m(A) = 17.9 kg
v(A) = 8.82 m/s (due east)

P(A) = 17.9 kg × 8.82 m/s

Similarly, for object B:
P(B) = m(B) × v(B)
m(B) = 28.9 kg
v(B) = 4.01 m/s (due north)
P(B) = 28.9 kg × 4.01 m/s

Now, let's calculate the total momentum of the two-object system after the collision, which will be the sum of the individual momenta:

Total momentum = P(A) + P(B)

(a) To calculate the magnitude of the total momentum, we need to find the vector sum of P(A) and P(B). Since P(A) is in the east direction and P(B) is in the north direction, we can use vector addition to find the magnitude.

Magnitude of total momentum = √[(P(A) × cos(θ))^2 + (P(B) × sin(θ))^2]

where θ is the angle between the east and north directions, which is 90 degrees.

Magnitude of total momentum = √[(17.9 kg × 8.82 m/s × cos(90))^2 + (28.9 kg × 4.01 m/s × sin(90))^2]

(b) To find the direction of the total momentum, we can use the tangent function:

Direction of total momentum = tan^(-1)[(P(B) × sin(θ)) / (P(A) × cos(θ))]

Direction of total momentum = tan^(-1)[(28.9 kg × 4.01 m/s × sin(90)) / (17.9 kg × 8.82 m/s × cos(90))]

Calculating these values should give you the magnitude and direction of the total momentum.