Posted by Anonymous on Wednesday, July 21, 2010 at 12:20am.
The combined momentum of motorboat and Isaac remains the same. Choose a coordinate system based on land. The momentum conservation equation can then be written
0.5 m/s*(Mb + Mi) = Mb*Vb + Mi*(Vi+0.5)
Solve for the unknown velocity of the boat afterwards, Vb.
0.5*362 = 300 Vb + 62*3.5
300 Vb = 181 - 217 = -36
Vb = -0.12 m/s (backwards)
boat is m1 = 300 kg
Isaac is m2 = 62 kg
towards dock is +
away from dock is -
original velocity of both is
v = +0.50 m/s
v2' = final velocity of Isaac
v1' = final velocity of boat
v2'-v1' = +3.0 m/s (because Isaac jumps off the front of the boat & is moving toward the dock; this will make v2' positive and v1' negative.
(m1+m2)v = m1v1' + m2v2'
v1' = 0.014 m/s away from the dock (boat) and v2' - 2.986 m/s towards the dock (Isaac)
boat is m1 = 300 kg
Isaac is m2 = 62 kg
towards dock is +
away from dock is -
original velocity of both is
v = +0.50 m/s
v2' = final velocity of Isaac
v1' = final velocity of boat
v2'-v1' = +3.0 m/s (because Isaac jumps off the front of the boat & is moving toward the dock; this will make v2' positive and v1' negative.
(m1+m2)v = m1v1' + m2v2'
v1' = 0.014 m/s away from the dock (boat) and v2' = 2.986 m/s towards the dock (Isaac)