Question

Three skydivers are plummeting earthward. They are initially holding onto each other, but then push apart. Two skydivers of mass 70 and 80 kg gain horizontal velocities of 1.2 m/s east and 1.4 m/s southeast, respectively. What is the horizontal velocity of the third skydiver, whose mass is 55 kg?

a) 1.5 m/s northeast
b) 1.3 m/s northwest
c) 0.2 m/s southeast
d) 1.1 m/s east

Answer 1

To find the horizontal velocity of the third skydiver, we need to apply the principle of conservation of momentum and separate the velocities into their x and y-components. Using the given masses and velocities, we can solve for the x-component velocity of the third skydiver.

Step-by-step explanation:

To find the horizontal velocity of the third skydiver, we need to apply the principle of conservation of momentum. The momentum before the push apart is equal to the momentum after the push apart.

Momentum is a vector quantity, which means that it has both magnitude and direction. The momentum of an object can be calculated by multiplying its mass by its velocity.

Let's assign positive x-axis to the east and positive y-axis to the north. We can separate the velocities of the two skydivers into their x and y-components using trigonometry.

The first skydiver has a velocity of 1.2 m/s east, which means its x-component velocity is 1.2 m/s and its y-component velocity is 0 m/s.

The second skydiver has a velocity of 1.4 m/s southeast, which can be separated into an x-component velocity of 1.4 * cos(45°) m/s and a y-component velocity of 1.4 * sin(45°) m/s.

Using the principle of conservation of momentum, we can now calculate the x-component velocity of the third skydiver:

(70 * 1.2 + 80 * (1.4 * cos(45°))) / 55 = x-component velocity

Solving this equation gives us approximately 1.11 m/s.