An indestructible bullet 1.85 cm long is fired straight through a board that is 11.5 cm thick. The bullet strikes the board with a speed of 400 m/s and emerges with a speed of 285 m/s.
(a) What is the average acceleration of the bullet as it passes through the board?
Give your answer in in scientific notation rounded to three significant figures.
(b) What is the total time that the bullet is in contact with the board?
Round your answer to the nearest millionth, if necessary. Give your answer in seconds.
**This I have correctly calculated to be 0.000390seconds***
(c) What thickness of board (calculated to 0.1 cm) would it take to stop the bullet, assuming the bullet's acceleration through all boards is the same?
Give your answer in centimetres.
a=(V^2-Vo^2)/2•s
s=11.5+1.85=13.35 cm=0.1335 m
a =2.95•10^5 m/s
t=(V-Vo)/a =0.000390 s
s1 = V^2/2•a = 0.542 m=54.2 cm
To solve this problem, we can use the equations of motion to analyze the motion of the bullet as it passes through the board.
(a) Let's start by calculating the average acceleration of the bullet. The average acceleration can be found using the equation:
average acceleration = (final velocity - initial velocity) / time
In this case, the initial velocity is 400 m/s, the final velocity is 285 m/s, and we know the time from part (b) is 0.000390 seconds. Plugging these values into the equation, we get:
average acceleration = (285 m/s - 400 m/s) / 0.000390 seconds
Calculating this, we find that the average acceleration is approximately -7.82 x 10^5 m/s^2. (Note: The negative sign indicates that the acceleration is in the opposite direction of the initial velocity.)
(b) As you correctly calculated, the time that the bullet is in contact with the board is 0.000390 seconds.
(c) To find the thickness of the board needed to stop the bullet, we can use the equation of motion:
final velocity^2 = initial velocity^2 + 2 * acceleration * distance
Since the bullet is indestructible, it will stop completely at the end of the board, so the final velocity is 0 m/s. The initial velocity is 400 m/s (given in the problem). We know the acceleration from part (a), which is -7.82 x 10^5 m/s^2.
Rearranging the equation, we get:
distance = (final velocity^2 - initial velocity^2) / (2 * acceleration)
= (0 - (400 m/s)^2) / (2 * -7.82 x 10^5 m/s^2)
Calculating this, we find that the distance the bullet travels through the board is approximately 0.1025 m.
To find the thickness of the board needed to stop the bullet, we subtract the initial length of the bullet from the distance traveled through the board:
thickness of board = distance - initial length of bullet
= 0.1025 m - 0.0185 m
= 0.084 m
Converting this to centimeters (since the answer asks for the thickness in centimeters), we get approximately 8.4 cm. However, the answer asks for the thickness rounded to 0.1 cm, so the thickness of the board needed to stop the bullet is 8.4 cm.