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A 50.0 -g superball traveling at 25.0 \(\mathrm{m} / \mathrm{s}\) bounces off a brick wall and rebounds at 22.0 \(\mathrm{m} / \mathrm{s}\) . A high-speed camera records this event. If the ball is in contact with the wall for 3.50 \(\mathrm{ms}\) , what is the magnitude of the average acceleration the ball during this time interval? (Note: \(1 \mathrm{ms}=10^{-3} \mathrm{s} .\)

Short Answer

Expert verified
The ball's magnitude of average acceleration is 13428.6 \(\mathrm{m/s^2}\).

Step by step solution

01

Convert Time from Milliseconds to Seconds

The time of contact given is in milliseconds (ms). Convert it to seconds (s) because the standard unit for time in physics is seconds (s). Use the conversion factor where 1 ms equals to 0.001 s (or to \(10^{-3}\) s). Thus, conversion is: 3.5 ms = 3.5 \times 10^{-3} s.
02

Calculate the Change in Velocity

Determine the velocity change by subtracting the final velocity from the initial velocity. The ball starts with a velocity of 25.0 m/s and rebounds with a velocity of -22.0 m/s (the negative sign indicates a change in direction). Change in velocity \(\Delta v\) is: \(\Delta v = v_{final} - v_{initial} = -22.0 \mathrm{m/s} - 25.0 \mathrm{m/s} = -47.0 \mathrm{m/s}\).
03

Calculate the Average Acceleration

The average acceleration can be found using the formula \(a = \frac{\Delta v}{\Delta t}\), where \(\Delta v\) is the change in velocity and \(\Delta t\) is the time interval. Plug the values obtained from Step 1 and 2 into the acceleration formula to find the average acceleration.
04

Calculate the Magnitude of the Average Acceleration

Since the question asks for magnitude, the direction is not important, and only the absolute value of the acceleration is considered. Whether the acceleration sign is positive or negative, take the absolute value to report the final answer in magnitude form.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Change in Velocity
The concept of 'change in velocity' is fundamental when discussing motion in physics. It refers to the difference between the initial and final velocity. This can be understood as how much the velocity of an object has increased or decreased over a period of time.

Using the given superball example, the change in velocity occurred when the ball bounced off the wall and its direction reversed. The initial velocity was 25.0 m/s, and after the impact, it reversed at a speed of 22.0 m/s. Since the ball changed direction, the final velocity is considered to be negative in this context. Mathematically, the change in velocity \( \Delta v \) is represented by the equation:\[\Delta v = v_{final} - v_{initial}\]In our case:\[\Delta v = -22.0 \ m/s - 25.0 \ m/s = -47.0 \ m/s\]
Time Interval Conversion
Time interval conversion is essential when dealing with physical equations, as consistent units must be used. In many situations, like the one with our superball, time might be given in milliseconds (ms), but equations in physics typically require time in seconds (s).

The conversion is straightforward once you know the relationship between the units. There are 1000 milliseconds in one second, or:
\[1 \mathrm{ms} = 10^{-3} \mathrm{s}\]
In the example given, the ball is in contact with the wall for 3.5 ms. Converted to seconds, the time interval \( \Delta t \) is:\[3.5 \ ms = 3.5 \times 10^{-3} \ s\]This conversion allows us to accurately calculate phenomena like acceleration, especially when using a high-speed camera to capture events occurring over short time intervals.
High-Speed Camera Analysis
High-speed camera analysis is a vital tool in physics for capturing and analyzing events that happen too quickly for the human eye to see. A high-speed camera records motion at rates that are much faster than standard video cameras, allowing for the detailed examination of transient or sudden phenomena.

In this scenario, the high-speed camera was used to determine the exact time the superball was in contact with the wall, which is crucial for calculating the ball's average acceleration. By providing precise timing measurements, typically to the millisecond or beyond, these cameras ensure accurate data collection for time interval conversion in physics calculations.
Velocity and Acceleration
Velocity and acceleration are key concepts in physics that describe motion. Velocity is a vector quantity that includes the speed and direction of an object's motion, while acceleration is the rate at which an object's velocity changes over time.

In the example given, the superball changed its velocity from 25.0 m/s to -22.0 m/s due to its bounce off the wall. The average acceleration is determined by how quickly this change in velocity occurred, which is expressed by the formula:\[a = \frac{\Delta v}{\Delta t}\]Applying the formula to the problem, we take the change in velocity (-47.0 m/s) and divide it by the time interval (3.5 ms converted to seconds). The negative sign in the change in velocity value typically indicates the direction of acceleration, but since we are interested in the magnitude in this case, we disregard the direction and focus on the absolute value for the final answer.

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Most popular questions from this chapter

A ball is thrown directly downward, with an initial speed of 8.00 \(\mathrm{m} / \mathrm{s}\) , from a height of 30.0 \(\mathrm{m}\) . After what time interval does the ball strike the ground?

Automotive engineers refer to the time rate of change of acceleration as the "jerk." If an object moves in one dimension such that its jerk \(J\) is constant, (a) determine expressions for its acceleration \(a_{x}(t),\) velocity \(v_{x}(t),\) and position \(x(t),\) given that its initial acceleration, velocity, and position are \(a_{x i}, v_{x i},\) and \(x_{i},\) respectively. (b) Show that \(a_{x}^{2}=\) \(a_{x i}^{2}+2 J\left(v_{x}-v_{x i}\right) .\)

A test rocket is fired vertically upward from a well. A catapult gives it an initial speed of 80.0 \(\mathrm{m} / \mathrm{s}\) at ground level. Its engines then fire and it accelerates upward at 4.00 \(\mathrm{m} / \mathrm{s}^{2}\) until it reaches an altitude of 1000 \(\mathrm{m}\) . At that point its engines fail and the rocket goes into free fall, with an acceleration of \(-9.80 \mathrm{m} / \mathrm{s}^{2}\) . (a) How long is the rocket in motion above the ground? (b) What is its maximum altitude? (c) What is its velocity just before it collides with the Earth? (You will need to consider the motion while the engine is operating separate from the free-fall motion.)

Speedy Sue, driving at 30.0 \(\mathrm{m} / \mathrm{s}\) , enters a one-lane tunnel. She then observes a slow-moving van 155 \(\mathrm{m}\) ahead traveling at 5.00 \(\mathrm{m} / \mathrm{s}\) . Sue applies her brakes but can accelerate only at \(-2.00 \mathrm{m} / \mathrm{s}^{2}\) because the road is wet. Will there be a collision? If yes, determine how far into the tunnel and at what time the collision occurs. If no, determine the distance of closest approach between Sue's car and the van.

The position of a pinewood derby car was observed at various times; the results are summarized in the following table. Find the average velocity of the car for (a) the first second, \((b)\) the last 3 s, and \((c)\) the entire period of observation. $$\begin{array}{lllllll}\hline t(s) & {0} & {1.0} & {2.0} & {3.0} & {4.0} & {5.0} \\ \hline x(\mathrm{m}) & {0} & {2.3} & {9.2} & {20.7} & {36.8} & {57.5} \\ \hline\end{array}$$

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