Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
Physics Help????!!!?????!!!!!?
A hockey player with a mass of 51.5 kg is traveling due east with a speed of 3.15 m/s. A second hockey player with a mass of 67.0 kg is moving due south with a speed of 6.65 m/s. They collide and hold on to each other after the collision, managing to move off at an angle θ south of east, with a speed of vf. Assume friction may be ignored and determine the following.
a) the angle θ __________ ° south of east
b) the speed vf ________m/s
1 Answer
- 8 years agoFavorite Answer
First of all, this is a conservation of momentum problem. Also, the collision is inelastic.
a) Using p=mv, calculate the momentum in each direction.
1st player: 51.5 x 3.15 = 162.225 newtons east
2nd player: 67.0 x 6.65 = 445.55 newtons south
When the players collide, they'll go south and east like this (their collision is represented by the hypotenuse):
http://i.imgur.com/GV2ieIK.png
This triangle has a base of 162.225 and a height of 445.55. From this, we can derive the angle. We have the opposite and adjacent sides, so we use tangent:
Theta = tan(-1) (162.225/445.55) = 20.0 degrees
b) In the triangle above, we can use the pythagorean theorem to get the hypotenuse length of 474.16. This represents the combined momentum force. Because the collision is inelastic, we can find the velocity by taking the momentum and dividing it by the combined masses.
p = mv, or v = p/m
v = 474.16/(51.5+67.0) = 4.00 m/s
