Body A of mass 3kg has a velocity of +4ms^-1 and collides head on with body B, which has a mass of 2kg and a velocity of -2ms^-1. After the collision the velocity of B is found to be +3ms^-1.
What is the velocity of A after the collision?
What is the loss in kinetic energy of the bodies as a result of the collision?
2007-09-04 09:13:57 · 2 answers · asked by Alexander M 1 in Science & Mathematics ➔ Physics
mava1 +mbvb1 = mava2 + mbv2b2
unknow here is va2 so solve
va2 = (mvava1 + mbvb1 - mbvb2)/ma
Then compute KE before and KE after
KE1 = 1/2(mava1^2 +mbvb1^2)
KE2 - 1/2(mava2^2+mbvb2^2)
KE lost = KE1 - KE2
2007-09-04 09:21:20 · answer #1 · answered by nyphdinmd 7 · 0⤊ 0⤋
3*4 + (2*-2) = 3v + 2*3
12-4 = 3v + 6
3v =2 , v=2/3 ms^-1
KE before => 0.5*3*4^2 = 24J + 0.5*2*(-2)^2 = 4J = 28J
KE after => 0.5*3*(2/3)^2 = 2/3J + 0.5*2*3^2 = 9J = 29/3J
loss in KE = 28-29/3 = 55/3J = 18.3J
2007-09-04 18:57:41 · answer #2 · answered by SS4 7 · 0⤊ 0⤋