Recent content by ron360

yeah but change i get is F=6.006*10^8(kg*m/s^2) and its not the actual force

in know its change in momentum which is mv, but how would you apply to my force?

Pressure problem -- Impulse of object striking a head 1. A police crime lab is trying to determine whether someone was murdered or died as a result of an accident. He was struck in the temple by a 4.20 kg sculpture that is alleged to have fallen off a bookcase. The sculpture presumably fell a...

thanks a lot for all the responses, you are right a) is 4.66 m/s and b) i think is actually -2.32, correct me if I am wrong about b)

1/2*m*(vf^2-vi^2) + m*g*(hf-hi) = 0 m's cancel, vi = 0, and hf = 0 vf = sqrt(2*g*hi) vf = sqrt(2*9.81 m/s^2 * 2.5 m) = 7.00 m/s m1*v1i + m2*v2i = m1*v1f + m2*v2f KE: (with the 1/2's canceled) m1*v1i^2 + m2*v2i^2 = m1*v1f^2 + m2*v2f^2 To make equations a little simpler, m2 = 2*m1 and v2i...

sorry for being unclear here is my work for parts b) and c) i just need to double check i got the right answers, thanks a lot. since i got V2f=4.67 m/s from a) b) 1/2m1v1i^2 + 1/2 m2v2i^2 = 1/2m1v1f^2 + 1/2m2v2f^2 1/2(1)(7) + 1/2(2)(0)^2 = 1/2(1)V1f^2 + 1/2(2)(4.67)^2 v1f = 2.319 c)Sum of...

the formula is actually: [1/2m1v1i^2 + 1/2 m2v2i^2 = 1/2m1v1f^2 + 1/2m2v2f^2] for a) and b) ,for c) i used [m*a = -u*m*g] and [vf^2 = vi^2 + 2*a*d] ,i got 4.67m/s for a) 2.319m/s for b) and 4.82 m for c) , i know a) is right but i don't have the answers for b) and c) so i want to make sure i...

Collision -- a ball and string form a pendulum... 1. A 1.0-kg ball is attached to the end of a 2.5-m string to form a pendulum. This pendulum is released from rest with the string horizontal. At the lowest point in its swing when it is moving horizontally, the ball collides elastically with a...