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Two skaters on in-line skates, Lisa and Bart, are initially at rest. They push apart. and start moving in opposite directions. If Lisa's speed just after they push apart is 2.0 m/s and her mass is 85% of Bart's mass, how fast is Bart moving at that time?

2006-10-12 17:50:38 · 8 answers · asked by bdbush007 3 in Science & Mathematics Physics

8 answers

Starting from rest, their resultant momentum should be zero at the time of impact.
Considering the magnitudes of momenta,
Lisa's speed X Lisa's mass = Bart's speed X Bart's mass
Thus Bart's speed = 2 X 0.85= 1.7 m/s

This can be done using vectors and velocity also.

2006-10-12 18:14:46 · answer #1 · answered by eternal_quest 2 · 1 0

its momentum!

mass of lisa x speed of lisa = mass of bart x speed of bart
0.85 x 2 = 0.15 x v
v = 1.7/0.15
v = 11.33 m/s

therefore speed of bart is 11.33 m/s

2006-10-13 04:19:44 · answer #2 · answered by Ir Jamie 2 · 0 1

they each had the same energy of acceleration applied. if E=rate x mass, and it does; and M(L) = .85M(B); then; R(L) x .85M(B) = R(B) x M(B), and; 2m/s x .85M(B) = R(B) x M(B).. so.. dividing by M(B) to isolate rate of bart begats Rate of Bart = 2m/s x .85.. therefore; On a frictionless plane his rate would be about 85% hers or = 1.7 m/s

2006-10-13 01:21:06 · answer #3 · answered by mr.phattphatt 5 · 0 2

physics is a lot like that u r now here means because of physics only.think about flight,car,computer or any other equipment today u r utilizing.since physics is the brother of maths & father of electronics,mechanics &.....................a reality for ur dreams comes true by physics. Once put u r all equipment aside & try to live for one day.then u can automatically find answer for u r question.

2006-10-13 02:49:05 · answer #4 · answered by G K 2 · 0 2

2.0 m/s

2006-10-13 00:53:00 · answer #5 · answered by s t 2 · 0 1

1.7 m/s

2006-10-13 01:09:14 · answer #6 · answered by ? 4 · 0 1

1.7 m/s

2006-10-13 00:55:06 · answer #7 · answered by S--slick 4 · 0 1

2.2m/s

2006-10-13 00:58:03 · answer #8 · answered by It Co$t To Be Around The Bo$$ 4 · 0 1

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