Polly Ester and Ray Ahn are doing the Elastic Collision lab on a low-friction track. Cart A has a mass of 1.00 kg and is moving rightward at 27.6 cm/s prior to the collision with Cart B. Cart B has a mass of 0.50 kg and is moving leftward with a speed of 42.9 cm/s. After the magnetic repulsion of the two carts, Cart A is moving leftward at 10.1 cm/s. Determine the post-collision speed and direction of cart B. I don't get how to exactly solve this...i'm getting 2/3 of it, just need to know how to do the last equation/formula.
try showing a bit of you work, maybe just the general steps.
p1=m1v1 p1=(1.00kg)(27.6cm/s) p1=10.1kg*cm/s p2=m2v2 p2=(0.50kg)(42.9cm/s) p2=21.45kg*cm/s I don't know from there.
Polyester and Rayon., really? Why don't you conserve energy and check?
What do you mean conserve energy? This is the conservation of momentum...
magnetic repulsion implies that now b attains momentum of a and a attains momentum of b and both start moving in opposite direction so now P1=21.45 and P2=10.1 now divide P2 by its mass so as to calculate speed of b in opposite direction. . .
That gives me 20.2m/s and 21.45m/s..correct?
20.5 is direction b and is toward right. . .
but the correct answer is 32.5 cm/s right. http://www.physicsclassroom.com/calcpad/momentum/problems.cfm number 26.
ohk then there is a twist in the story i missed this line. . . " After the magnetic repulsion of the two carts, Cart A is moving leftward at 10.1 cm/s. " i have to make some changes now
@some1 p1=m1v1 p1=(1.00kg)(27.6cm/s)= 27.6 "p1=10.1kg*cm/s" <----- wrong p2=m2v2 p2=(0.50kg)(42.9cm/s) p2=21.45kg*cm/s I don't know from there.
Really sorry....I didnt read the question properly the first time. you have \[p _{Ai} + p _{Bi} = p_{Af} + p_{Bf}\] i: Initial; f: final And you know the values of pAi, pBi and pAf. pBf can subsequently be found.
hm. i think i'll try that. thanks aish and theyatin. :D
Join our real-time social learning platform and learn together with your friends!