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
Momentum and Projectile Question. PLEASE HELP!?
A 1.25kg block of wood sits at the edge of a table 0.752m above the floor. A 10.5g bullet moving horizontally with a speed of 725m/s hits the block and embeds itself within the block. How far from the edge of the table does the block and bullet land?
2 Answers
- Andrew SmithLv 73 years ago
All these problems ever are is a series of several simple steps.
1. Use conservation of momentum to find the velocity after the collision. ( mv1 = (M+m) v2 )
2. Use the vertical motion to find the time in the air ( h = 1/2 g t^2)
3. Use s = vt to find the distance moved in this time.
- electron1Lv 73 years ago
Initial momentum = 0.0105 * 725 = 7.6125
This is momentum after the collision.
Total mass = 0.0105 + 1.25 = 1.2605 kg
1.2605 * v = 7.6125
v = 7.6125 ÷ 1.2605
The horizontal velocity after the collision is approximately 6.04 m/s. To determine the horizontal distance its moves as it falls 0.752 meter, we need to determine the time for the fall. Let’s use the following equation.
d = vi * t + ½ * g * t^2, vi 0
0.752 = ½ * 9.8 * t^2
t = √(0.752 ÷ 4.9)
The time is approximately 0.39 second.
d = (7.6125 ÷ 1.2605) * √(0.752 ÷ 4.9)
This is approximately 2.37 meters.
