A: A ball is released at the bottom of a container of liquid. The ball rises upw
ID: 2261343 • Letter: A
Question
A:
A ball is released at the bottom of a container of liquid. The ball rises upward, pops out of the liquid and reaches a height of 20 cm, above the liquid, moving through a near vacuum. The experiment is carried out on Earth, and the time from the ball's release to the moment when it reaches its maximum height is 0.70 seconds.
1. Calculate the velocity of the ball when it leaves the liquid.
2. Determine the average acceleration in the liquid, assuming it to be constant.
3. Determine the depth of the liquid from the motion of the ball and the time.
B:
Astronauts must train for docking maneuvers with computerized go-carts and diesel trucks. The carts are controlled by the driver using a computer console much like the one for the spaceship they are training for. The diesel truck moves at 16 m/s on the roadway and the go-cart STARTS out at 12 m/s AND 50 meters behind the diesel truck. The idea is to enter a program which will get the cart up the ramp and into the trailer to stop in the "docking zone" in 240 seconds. The program consists of a velocity or acceleration and a duration in seconds. The distance from the end of the ramp to the docking spot is 5 meters, and the carts velocity is relative to the surface of the wheels.
(example video: http://www.youtube.com/watch?v=dYlstdCWzCY)
C:
A person is on a bike is attempting to jump a 20 meter wide trench. The ramp on the far side is 30° above horizontal and he wants to land his motorcycle on the ramp with a velocity parallel to its surface. Determine the angle of the launch and initial velocity necessary for this (without air resistance).
If he then has to go through a vertical circular track (loop-the-loop), calculate the diameter of the largest loop he will make it through, assuming he coasts to the top and the average effective resistance force can be taken to be 180 N. Use a combined mass of 400kg for the person and his bike.
I only give points for correct answers as well as clear steps, and short explanations.
I will increase the reward to 2000 points if all these conditions are met.
Explanation / Answer
On coming out of liquid, energy =0.5 mv^2 =mgh
v=1.98 m/s
time in air = sqrt(2h/g) = 0.2 sec
time in liquid = 0.5 sec
accn = change in vel / time = 1.98 / 0.5 = 3.96 m/s^2
depth = 0.5accn x time^2 = 49.5 cm
B starting velocity = 12, ending velocity = 16, time = 240 sec
relative velocity = -4, separation = 55
55=-4(240)+0.5 accn 240^2
accn =0.0352 m/s^2
C One of the solutions is that he takes off at 30degrees and just clears the trench
therefore, v^2 sin(60)/g = 20 => v=15.04 m/s
At the top, resistance =mv^2/r - mg
r=22.07 m