A boy whose mass is 35kg and a girl at 25kg face each other initially at rest on frictionless rollar skates. The boy pushes the girl, who moves away at a speed of 1.5 m/s. Whats the boys speed?
I used the elastic collison equation:
V1f = (35-25/35+25)0 + (2(25)/35+25)1.5
V1i = (50/60)1.5
V1i = (.83)1.5
V1i = 1.25 m/s
am I close??
2006-10-24 02:26:28 · 5 answers · asked by Haley 2 in Science & Mathematics ➔ Physics
No. The forces on each are equal, so from f = ma
35a = f = 25*1.5, solve for a
2006-10-24 02:53:26 · answer #1 · answered by sofarsogood 5 · 0⤊ 0⤋
No, your answer is not correct. To solve this problem set the momentum of the boy equal to the momentum of the girl: 25*1.5=35*v......v=(25/35)*1.5=1.07m/s.
2006-10-24 09:32:10 · answer #2 · answered by bruinfan 7 · 0⤊ 0⤋
mbvb=mgvg
by the principle of conservation of linear momentum, vb=mgvg/mb=25*1.5/35=1.07 since the boy is heaver than girl.
2006-10-24 09:37:00 · answer #3 · answered by Anonymous · 0⤊ 0⤋
You have to use the Newton's Second Law of Motion.
Equate the momentum.
2006-10-24 09:42:34 · answer #4 · answered by itsme 1 · 0⤊ 0⤋
your speed will be the speed of the police car that hauls you in for pushing a cute little girl
2006-10-24 12:41:11 · answer #5 · answered by hell oh 4 · 0⤊ 0⤋