Question

The average speed of a nitrogen molecule in air is about 833 m/s, and its mass is about 4.68 × 10−26 kg. If it takes 2.3 × 10−13 s for a nitrogen molecule to hit a wall and rebound with the same speed but in the opposite direction, what is the magnitude of the average acceleration of the molecule during this time interval? Answer in units of m/s².

Answer 1

The magnitude of the average acceleration of a nitrogen molecule that hits a wall and rebounds within 2.3 x 10^-13 s is 7.24 x 10^15 m/s². The acceleration is calculated by taking the change in velocity (which is twice the initial speed since the direction is reversed) and dividing it by the time interval of the event.

Step-by-step explanation:

The magnitude of the average acceleration of a nitrogen molecule during a collision with a wall can be calculated using the formula a = Δv / Δt, where Δv is the change in velocity and Δt is the change in time. The molecule hits the wall and rebounds with the same speed but in the opposite direction, which means the velocity changes by twice its original speed (because it goes from +833 m/s to -833 m/s).

First, we calculate the change in velocity (Δv):

• Δv = final velocity (after collision) - initial velocity (before collision) = (-833 m/s) - (833 m/s) = -1666 m/s

Since acceleration is a vector quantity, and we are interested in the magnitude, we take the absolute value of Δv:

• |Δv| = 1666 m/s

Now, we use the given time interval (2.3 x 10^-13 s):

• a = |Δv| / Δt = 1666 m/s / (2.3 x 10^-13 s) = 7.24 x 10^15 m/s²

Therefore, the magnitude of the average acceleration of the nitrogen molecule during the collision with the wall is 7.24 x 10^15 m/s².