Posted by **Anonymous** on Wednesday, July 21, 2010 at 12:20am.

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?

- physics -
**drwls**, Wednesday, July 21, 2010 at 1:43am
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)

- physics -
**David**, Sunday, April 14, 2013 at 12:27am
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)

- physics -
**David**, Sunday, April 14, 2013 at 8:07am
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)

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