Yasmin
2021-03-11
An indestructible bullet 2.00 cm long is fired straight through a board that is 10.0 cm thick. The bullet strikes the boardwith a speed of 470 m/s and emerges with a speed of 250 m/s.
(a) What is the average acceleration of the bullet through the board?
(b) What is the total time that the bullet is in contact with the board?
(c) What thickness of board (calculated to 0.1 cm) would it take to stop the bullet, assuming that the acceleration through all boards the same?
Talisha
Skilled2021-03-12Added 93 answers
This is an exercise in the 3 equations you should know for solving this type of problems:
1) The correct equation here is
Solve this for a to get the acceleration - note that this will be a negative quantity.
2) Now you have the value for a and you can use v(final)=v(initial)+at
Solve this for t and get the time the bullet was in the board.
Note that v and a are vector quantities and are in opposite direction to each other.
3) You cannot again use the equation used in part 1 only now you know the deceleration produced by theboards and the final speed zero must be zero so you can solve for s the required board thickness.
Jeffrey Jordon
Expert2021-10-12Added 2605 answers
a)
We know that
b)
v= u + a t
c)
To stop the bullet it means that the final velocity will be zero.
s=13.94 cm
22+64
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