A 45.0 kg girl is standing on a 160 kgplank. The plank, originally at rest, is f
ID: 1722107 • Letter: A
Question
A 45.0 kg girl is standing on a 160 kgplank. The plank, originally at rest, is free to slide on a frozenlake, which is a flat, frictionless surface. The girl begins towalk along the plank at a constant velocity of 1.55 m/s relative to the plank. (a) What is her velocity relative to thesurface of ice?1 m/s
(b) What is the velocity of the plank relative to the surface ofice?
2 m/s (a) What is her velocity relative to thesurface of ice?
1 m/s
(b) What is the velocity of the plank relative to the surface ofice?
2 m/s
Explanation / Answer
Givne that the mass of the girl is Mg = 45 kg mass of the plank is Mp = 160 kg Thevelocity of the girl relative to the plank is Vgp = V1 + V2 =1.55 -------(1) ------------------------------------------------------------------- From the conservation ofmomentum intiial momentum = final momentum Mg*V1 = Mp*V2 Mg*(1.55m/s - V2) = Mp*V2 then we get V2 = --------- m/s This is th eplankvelocity relative to the ice . substitude the value of the V2 in the equation (1) we get the valueof velocity fo the girl relative to the ice V1 = --------- m/sRelated Questions
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.