physics
posted by Anonymous on .
A 300kg motorboat is turned off as it approaches a dock and it coasts in toward the dock at .5 m/s. Isaac, whose mass is 62kg, jumps off the front of the boat with a speed of 3m/s relative to the boat. what is the velocity of the boat after Isaac jumps?

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)