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
- 7 years agoFavorite Answer
Hey! here is the solution: http://urjnaswxkfjjknseochallenge.blogspot.com/201...
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.
- 7 years ago
v = (15-kg * 4.4-m/s) / (77-kg + 136-kg)
v = 0.3098591549295775 to the leftSource(s): physics hmwk