A skydiver is falling straight downward (prior to having deployed his parachute) when he strikes a woman in a hang glider flying horizontally to the east. The skydiver lands on the glider and grabs hold of it. Just prior to this collision, the glider had a speed of 5ms east and the skydiver had a speed of 10ms downward. What is the magnitude of the velocity of the woman and glider immediately after the collision if the skydiver holds on to it? The combined mass of the woman and glider is 120kg and the mass of the skydiver is 80kg.

Question

A skydiver is falling straight downward (prior to having deployed his parachute) when he strikes a woman in a hang glider flying horizontally to the east. The skydiver lands on the glider and grabs hold of it. Just prior to this collision, the glider had a speed of 5ms east and the skydiver had a speed of 10ms downward. What is the magnitude of the velocity of the woman and glider immediately after the collision if the skydiver holds on to it? The combined mass of the woman and glider is 120kg and the mass of the skydiver is 80kg.

Answer 1

To solve this problem, we first need to understand the principle of conservation of momentum. According to this principle, the total momentum of a system remains constant before and after a collision, assuming no external forces are acting.

The momentum (p) of an object is given by the product of its mass (m) and velocity (v). Mathematically, it can be written as p = m * v.

Let's calculate the initial momentum of the system before the collision:

The momentum of the woman and glider is given by:
p1 = (mass of woman and glider) * (velocity of woman and glider)
= (120 kg) * (5 m/s east)
= 600 kg*m/s east

The momentum of the skydiver is given by:
p2 = (mass of skydiver) * (velocity of skydiver)
= (80 kg) * (10 m/s downward)
= 800 kg*m/s downward

Since the skydiver grabs hold of the glider and holds on to it, their combined mass remains the same, so the total mass of the system is still 200 kg (120 kg + 80 kg).

After the collision, the skydiver and the woman and glider move together as one object, so they have the same velocity. Let's call this velocity v_final.

The final momentum of the system is:
p_final = (total mass of the system) * (v_final)

Setting up the conservation of momentum equation, we can equate the initial momentum to the final momentum:

p1 + p2 = p_final

600 kg*m/s east + 800 kg*m/s downward = (200 kg) * (v_final)

Now, to find the magnitude of the velocity of the woman and glider immediately after the collision, we need to convert the eastward velocity component and the downward velocity component into a single vector. We can use the Pythagorean theorem to do this:

v_final^2 = (v_final east)^2 + (v_final downward)^2

We can substitute the given values into this equation:

v_final^2 = (v_final east)^2 + (v_final downward)^2
v_final^2 = (5 m/s)^2 + (-10 m/s)^2
v_final^2 = 25 m^2/s^2 + 100 m^2/s^2
v_final^2 = 125 m^2/s^2

To find the magnitude of the velocity, we take the square root of both sides:

v_final = √(125 m^2/s^2)
v_final ≈ 11.2 m/s

Therefore, the magnitude of the velocity of the woman and glider immediately after the collision, with the skydiver holding on, is approximately 11.2 m/s.