# A 77-kg fisherman in a 136-kg boat throws a package of mass m = 15 kg horizontally toward the right with a spe?

A 77-kg fisherman in a 136-kg boat throws a package of mass m = 15 kg horizontally toward the right with a speed of vi = 4.4 m/s as in the figure below. Neglecting water resistance, and assuming the boat is at rest before the package is thrown, find the velocity of the boat after the package is thrown.

What is the magnitude and direction

Relevance

Hey! here is the solution: http://urjnaswxkfjjknseochallenge.blogspot.com/201...

Momentum Conservation:

Since the boat is at rest initially, the total momentum is zero. So it has to be zero after the mass is thrown.

Initial Momentum = 0

Final Momentum = (77 + 136) * v + (15 * 4.4) where v is the final velocity of the boat

=> (77 + 136) * v + (15 * 4.4) = 0

=> 213 * v + 66 = 0

=> v = -66 / 213 = -0.3098591549295775

The boat moves in the opposite direction to where the package was thrown with a velocity of -0.3098591549295775.

Good luck!

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• v = (15-kg * 4.4-m/s) / (77-kg + 136-kg)

v = 0.3098591549295775 to the left

Source(s): physics hmwk
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