An aerospace technician is testing an experimental probe with a weight of 50\text{ N} on E… — Armed Services Vocational Aptitude Battery (ASVAB)

An aerospace technician is testing an experimental probe with a weight of $50\text{ N}$ on Earth (where $g = 10\text{ m/s}^2$ ). The probe is released from rest inside an evacuated drop tube, ensuring there is no air resistance. It falls freely for $200\text{ milliseconds}$ . What is the probe's velocity right before impact, and how would this impact velocity change if a $100\text{ N}$ probe were tested under the exact same conditions?

A
$2000\text{ m/s}$ ; the impact velocity would remain exactly the same because free-fall acceleration is independent of an object's weight.
B
$1.0\text{ m/s}$ ; the impact velocity would increase to $2.0\text{ m/s}$ because the object's mass, rather than a constant gravitational rate, determines its falling speed.
$2.0\text{ m/s}$ ; the impact velocity would remain exactly the same because free-fall acceleration is independent of an object's weight.Answer
D
$2.0\text{ m/s}$ ; the impact velocity would increase to $4.0\text{ m/s}$ because a greater mass results in a proportionally greater rate of acceleration.

Answer

The probe's impact velocity is

2.0\text{ m/s}

, and the velocity would remain exactly the same for the heavier probe.

First, the fall time must be converted to standard units (

200\text{ ms} = 0.2\text{ s}

). Using the velocity formula

v = g \times t

, the speed is calculated as

10 \times 0.2 = 2.0\text{ m/s}

. Second, in the complete absence of air resistance, the acceleration due to gravity is a constant. Therefore, a heavier

100\text{ N}

probe would fall at the exact same rate and strike the ground with the exact same velocity.

Step-by-Step Solution

Convert the free-fall time from milliseconds to seconds.

200\text{ ms} = 0.2\text{ seconds}

Standard kinematic equations require time to be in standard SI units (seconds) to match the acceleration rate (

10\text{ m/s}^2

Calculate the impact velocity using the formula for an object falling from rest (

v = g \times t

v = 10\text{ m/s}^2 \times 0.2\text{ s} = 2.0\text{ m/s}

This determines the exact speed of the

50\text{ N}

probe right before it strikes the bottom of the drop tube.

Evaluate how doubling the probe's weight to

100\text{ N}

affects the velocity.

The impact velocity remains

2.0\text{ m/s}

In an evacuated space (a vacuum), there is no air resistance. All objects accelerate due to gravity at the same constant rate, regardless of their individual masses or weights.

Key Concept

Free Fall Velocity and Mass Independence

Estimated Time:1m 0s

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