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Physics Help - Deceleration?

Ok. This is like the first class I have ever been completely stuck on and it is actually making me angry... like The Hulk angry. Anyways, here is my problem: A rifle bullet with a muzzle speed of 330 m/s is fired directly into a special dense material that stops the bullet in 25cm. Assuming the bullet's deceleration to be constant, what is its magnitude? I know that the answer is 2.2 x 10^5 m/s/s but I can not for the life of me figure out how to get that answer. I tried using the equation defined third on the list here: http://webphysics.iupui.edu/152/152Basics/kinematics/kinematics.html After playing with the equation, making sure all the units were the same (tried converting m to cm, leaving them alone, etc etc) and all I get is a = 66. This cant be right! Can anybody help?

Public Comments

1. s=0.25 u=330 v=0 a=? using v^2=u^2+2as, I get 0=108900+0.5a and a=-217800m/s/s There you go. The equation used was one of the SUVAT equations: http://en.wikipedia.org/wiki/Equations_of_motion
2. We are assuming that the bullet doesn't slow due to air friction. Therefore, the entire 330m/s must be decelerated in 0.25m . distance = 1/2 * acceleration * t^2 = 0.25 0.5 = acceleration * t^2 acceleration = 0.5 / t^2 dv/dt = 0.5/t^2 dv = 0.5/t^2 * dt v = 0.5 * integrate( 1/t^2 ) = 0.5 * integrate( t^(-2) ) v = 0.5 * ( t^(-1) / -1 ) + c v = -0.5/t + c v(0) = 330 330 = -0.5/0 + c, division by 0 is undefined, but lets continue c = 330 + 0.5/0 v(t) = -0.5/t + 330 + 0.5/0, still undefined, we can go no farther. In order to solve this problem you must subdue your hulkishness and ask for the extra piece of information which has been excluded. Either the deceleration time or the deceleration average magnitude.
3. Use the law of conservation of energy,i.e., kinetic energy of the bullet is equal to the energy absorbed by the special dense material in stopping the bullet. KE = (1/2)m(V^2) where KE = kinetic energy of the bullet m = mass of the bullet V = speed of the bullet = 330 m/sec (given) KE = (1/2)(m)(330)^2 = (54450)m Work absorbed by the special dense material, W = F(d) where F = stopping force d = distance travelled by the bullet in the special dense material before being stopped = 25 cm = 0.25 m NOTE that from Newton's 2nd law, F = ma where m = mass of the body a = acceleration of the body Substituting, F = ma in the equation for "W", W = (ma)(0.25) Since W =KE, (ma)(0.25) = (54450)m and since "m" appears in both sides of the equation, it will simply cancel out and the above is simplified to 0.25(a) = 54450 and solving for "a", a = 54450/0.25 = 217,800 a = 2.178 x 10^5 m/sec^2 (which is close to your answer of 2.2 x 10^5 m/sec^2)