Two vehicles are approaching an intersection. One is a 2500-
kg pickup traveling at 14.0 m/s from east to west (the negative x-direction),
and the other is a 1500-kg Mercedes going from south to north (the positive
y-direction at 23.0 m/s). (a) Find the x and y components of the net
momentum of this system. (b) What are the magnitude and direction of the
net momentum?
Problem
35000i
34500j
To find the x and y components of the net momentum, we first need to calculate the individual momenta of each vehicle.
The momentum of an object is given by the equation:
p = m * v
Where p is the momentum, m is the mass, and v is the velocity.
For the pickup truck:
Mass (m1) = 2500 kg
Velocity (v1) = -14.0 m/s (negative because it is traveling in the negative x-direction)
p1 = m1 * v1
p1 = 2500 kg * (-14.0 m/s)
p1 = -35,000 kg*m/s (momentum of the pickup truck)
For the Mercedes:
Mass (m2) = 1500 kg
Velocity (v2) = 23.0 m/s
p2 = m2 * v2
p2 = 1500 kg * 23.0 m/s
p2 = 34,500 kg*m/s (momentum of the Mercedes)
(a) To find the x-component of the net momentum, we add the individual x-components together:
Net Momentum (Px) = Px1 + Px2
Px1 = m1 * vx1
Px1 = 2500 kg * (-14.0 m/s)
Px1 = -35,000 kg*m/s
Px2 = m2 * vx2
Px2 = 1500 kg * 0 m/s (since the Mercedes is not traveling in the x-direction)
Px2 = 0 kg*m/s
Therefore, Px = -35,000 kg*m/s + 0 kg*m/s = -35,000 kg*m/s (negative because it is in the negative x-direction)
(b) To find the y-component of the net momentum, we add the individual y-components together:
Net Momentum (Py) = Py1 + Py2
Py1 = m1 * vy1
Py1 = 2500 kg * 0 m/s (since the pickup truck is not traveling in the y-direction)
Py1 = 0 kg*m/s
Py2 = m2 * vy2
Py2 = 1500 kg * 23.0 m/s
Py2 = 34,500 kg*m/s
Therefore, Py = 0 kg*m/s + 34,500 kg*m/s = 34,500 kg*m/s
To find the magnitude and direction of the net momentum, we use Pythagorean theorem:
|p| = sqrt(Px^2 + Py^2)
|p| = sqrt((-35,000 kg*m/s)^2 + (34,500 kg*m/s)^2)
|p| = sqrt(1,225,000,000 + 1,192,500,000)
|p| = sqrt(2,417,500,000)
|p| = 49,163.24 kg*m/s
The direction of the net momentum can be found using trigonometry:
direction = arctan(Py / Px)
direction = arctan(34,500 kg*m/s / -35,000 kg*m/s)
direction = arctan(-0.9857)
direction = -45.13° (rounded to two decimal places)
So, the magnitude of the net momentum is 49,163.24 kg*m/s and the direction is -45.13°.
To find the x and y components of the net momentum, we need to calculate the momentum of each vehicle separately and then add them together.
The momentum (p) of an object is given by the mass (m) times the velocity (v):
p = m * v
Let's start with the pickup truck:
Mass of pickup truck (m1) = 2500 kg
Velocity of pickup truck (v1) = -14.0 m/s (negative because it's traveling west)
p1 = m1 * v1
p1 = (2500 kg) * (-14.0 m/s)
p1 = -35000 kg*m/s (negative because the truck is moving in the negative x-direction)
Now let's calculate the momentum of the Mercedes:
Mass of Mercedes (m2) = 1500 kg
Velocity of Mercedes (v2) = 23.0 m/s (positive because it's traveling north)
p2 = m2 * v2
p2 = (1500 kg) * (23.0 m/s)
p2 = 34500 kg*m/s
(a) The x and y components of the net momentum are:
Net momentum in the x-direction = p1 = -35000 kg*m/s
Net momentum in the y-direction = p2 = 34500 kg*m/s
(b) To find the magnitude of the net momentum, we can use the Pythagorean theorem:
Magnitude of net momentum (p_net) = √(p1^2 + p2^2)
p_net = √((-35000 kg*m/s)^2 + (34500 kg*m/s)^2)
p_net = √(1225000000 + 1190250000) kg^2*m^2/s^2
p_net = √(2415250000) kg^2*m^2/s^2
p_net ≈ 49142.94 kg*m/s
To find the direction of the net momentum, we can use trigonometry. The direction is the angle (θ) that the net momentum vector makes with the positive x-axis:
θ = arctan(p2 / p1)
θ = arctan((34500 kg*m/s) / (-35000 kg*m/s))
θ ≈ -44.3° (negative because it is in the opposite direction from the positive x-axis)
So, the magnitude of the net momentum is approximately 49142.94 kg*m/s, and the direction is approximately -44.3° from the positive x-axis.