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 = 17.5 kg and an initial velocity of = 8.50 m/s, due east. Object B, however, has a mass of mB = 28.3 kg and an initial velocity of = 5.32 m/s, due north. Find the (a) magnitude and (b) direction of the total momentum of the two-object system after the collision.
Ah, a physics question! Let's solve it with humor.
(a) To find the magnitude of the total momentum, we need to add up the momentum of both objects. It's like being the coach of a two-person relay team, except they don't pass the baton, they stick together like glue!
We start with object A, which has a mass of 17.5 kg and is heading east at 8.50 m/s. This gives us a momentum of mA * vA = 17.5 kg * 8.50 m/s = 148.75 kg⋅m/s in the east direction.
Now let's tackle object B. It has a mass of 28.3 kg and is moving north at 5.32 m/s. Its momentum is mB * vB = 28.3 kg * 5.32 m/s = 150.476 kg⋅m/s in the north direction.
Since it's a completely inelastic collision, the two objects stick together like best friends at a conjoined twins convention. You could call them Object AB. To get the magnitude of their total momentum, we add up their individual momenta: 148.75 kg⋅m/s + 150.476 kg⋅m/s ≈ 299.226 kg⋅m/s.
(b) Now let's find the direction of the total momentum of the two-object system. Picture a compass in your mind. Object A is moving east, which corresponds to 90 degrees on the compass, while Object B is moving north, which corresponds to 0 degrees. When they collide and become one, their total momentum can be represented as 299.226 kg⋅m/s at an angle of θ = arctan(vB/vA) from the positive x-axis.
So, plug in the values for vA and vB, find the arctan, and voila! You have your direction.
Remember, laughter may not always solve physics problems, but it sure makes them more enjoyable!
To solve this problem, we first need to find the total momentum of the system before the collision.
The momentum of an object is given by the product of its mass and velocity. The momentum of object A is given by:
pA = mA * vA
Where mA is the mass of object A and vA is its velocity. Similarly, the momentum of object B is given by:
pB = mB * vB
Where mB is the mass of object B and vB is its velocity.
In this problem, object A is moving due east, so its velocity vA only has an x-component. Similarly, object B is moving due north, so its velocity vB only has a y-component.
Given:
mA = 17.5 kg
vA = 8.50 m/s due east
mB = 28.3 kg
vB = 5.32 m/s due north
To find the total momentum of the system before the collision, we can use the Pythagorean theorem to combine the x and y components of the velocities.
The x-component of the total momentum is given by the sum of the x-components of the individual momenta:
px = mA * vA
The y-component of the total momentum is given by the sum of the y-components of the individual momenta:
py = mB * vB
Using the given values, we can calculate the x and y components:
px = 17.5 kg * 8.50 m/s = 148.75 kg·m/s (east)
py = 28.3 kg * 5.32 m/s = 150.476 kg·m/s (north)
Now, we can use the Pythagorean theorem to find the magnitude of the total momentum:
|p| = sqrt(px^2 + py^2)
Substituting the values:
|p| = sqrt((148.75 kg·m/s)^2 + (150.476 kg·m/s)^2)
|p| = sqrt(22156.5625 + 22642.155776)
|p| = sqrt(44798.718276)
|p| ≈ 211.696 kg·m/s
So the magnitude of the total momentum of the two-object system after the collision is approximately 211.696 kg·m/s.
To find the direction of the total momentum, we can use trigonometry. The direction can be found using the inverse tangent function:
θ = arctan(py/px)
Using the given values:
θ = arctan(150.476 kg·m/s / 148.75 kg·m/s)
θ ≈ 45.48° north of east
So the direction of the total momentum of the two-object system after the collision is approximately 45.48° north of east.
Therefore, the answers are:
(a) The magnitude of the total momentum is approximately 211.696 kg·m/s.
(b) The direction of the total momentum is approximately 45.48° north of east.
To find the total momentum of the two-object system after the collision, we need to first calculate the momentum of each object before the collision, and then add them together.
The momentum of an object is given by the formula p = m * v, where p is the momentum, m is the mass, and v is the velocity.
Let's calculate the momentum of each object before the collision:
For Object A:
Mass of Object A (mA) = 17.5 kg
Initial velocity of Object A (vA) = 8.50 m/s (due east)
Momentum of Object A (pA) = mA * vA
= 17.5 kg * 8.50 m/s
= 148.75 kg*m/s (due east)
For Object B:
Mass of Object B (mB) = 28.3 kg
Initial velocity of Object B (vB) = 5.32 m/s (due north)
Momentum of Object B (pB) = mB * vB
= 28.3 kg * 5.32 m/s
= 150.496 kg*m/s (due north)
Now, to find the total momentum of the two-object system, we need to add the individual momenta together.
Total momentum (pTotal) = pA + pB
In this case, since the objects collide and stick together, the final direction of the momentum will depend on the resultant vector of the individual momenta. We can use the Pythagorean theorem to find the magnitude of the resultant vector, and trigonometry to find its direction.
Magnitude of the total momentum (a):
|pTotal| = sqrt((pA)^2 + (pB)^2)
= sqrt((148.75)^2 + (150.496)^2)
≈ 211.15 kg*m/s
Direction of the total momentum (b):
To find the direction, we can use the inverse tangent function:
θ = arctan(pB / pA)
= arctan(150.496 kg*m/s / 148.75 kg*m/s)
Calculating this value will give you the direction of the total momentum.