Well, we have a quiz tomorrow and we reviewed for that. I just realized that I'm posting a little late for you to look at this and learn more from it. Woops.
We basically went over a worksheet that looks strikingly similar to this one that I found on google images on the first try:
a. So we've gone over this twenty million times, but in case your memory is a bit foggy, the marbles motion is parabolic because the ball's horizontal velocity, due to negligible friction and inertia, will be constant. However, after it leaves the table, since the only vertical force acting on it is gravity, it will accelerate towards the floor at 9.8m/s/s.b.
While it is on the rail, we know that the ball is moving horizontally at a constant velocity of 10 m/s. This means no unbalanced forces, so the force diagram will have two arrows: an up arrow for the normal force the rail is exerting on the ball, and a down arrow for gravity. These two arrows are equal in length because I'm pretty sure the ball isn't accelerating upwards or downwards anytime soon.
When it leaves the table, since we are ignoring air resistance and inertia is not a force, there will only be one down arrow on your force diagram for gravity. (This computer doesn't have paint or anything similar to it, so no pretty drawings today :( ) It's horizontal movement is still constant because an outside force hasn't prevented the ball's inertia to keep it moving forward. As for vertical, it is accelerating downward.
c. Once the ball leaves the table, calculate how long it will take for the ball to hit the floor.
Okay, so here we use the equation: change in position = 1/2at^2+Vot
We are only talking about vertical movement. We know that the table is 1.5m tall, it's acceleration is 9.8 (gravity), and that it's initial velocity is 0, because the ball wasn't moving vertically at all before it left the table. So we plug this information in.
- 1.5m = 1/2(9.8m/s/s)t^2 + 0t (Then divide by 4.9 to get t^2 alone.)
- t^2 = 3.06 (Find the square root of both sides)
- t = .55s*
Alright, so now we're talking about horizontal movement, but now we have t, which we found in the last problem. We also know that initial velocity is 10m/s, because that was the velocity left the table, and will always be the horizontal velocity until an outside force acts on it. Also, since velocity is constant, we know that acceleration is 0. So plug that in to our equation again:
- changeX=1/2(0)(3.06) + 10*.55s
- X = 5.5m*
i. Determine the horizontal range of the marble as it falls to the floor.
ii. What effect does doubling the height have on the range of the marble?
iii. What other factors affect the range of the sphere?
i. Okay, this is basically combining c and d, except we have a new vertical height. We need to find how long it takes for the marble to hit the floor so that we can plug that time into another equation to see where, in regards to horizontal direction, the marble will be.
So, once again, thinking about vertical movement, we plug in our facts into the equation. (See c for more explanation if necessary.)
- 3m = 1/2(9.8)t^2 = 0t
- t^2 = .612
- t = .782s
- changeX = 1/2(0)t^2 + 10*.782
- x = 7.82m*
iii. Other factors that could affect the range of the sphere:
- Horizontal velocity, because, given the same amount of time it takes to hit the floor, the marble would move faster to go farther or slower to come up short.
- Increased friction or air resistance, since it would slow the ball down and change its range
- *** One thing that would NOT change it's horizontal range would be the weight or mass of the ball, as long as the velocity was still the same because, as we've seen and tested ourselves, if you drop two objects varying in weight at the same height, they will always land at the same time, meaning none of of the variables have changed, producing the same horizontal range.
So yep. That's all. I would do more, like go to the back of the worksheet, but I'm
-Norman
SCIENCE!!!!
