PHY-105-101 Summer 2016 Jaquin Exam #2 1. Rotten Rodney, the brilliant but misch

Question

PHY-105-101 Summer 2016 Jaquin Exam #2 1. Rotten Rodney, the brilliant but mischievous high school physics student, has created a small cannon-like device from Pringles containers that can propel arawegg at 22 m/s (about 50 mph). He fires offone egg on a vertical trajectory and hen 3.00 seconds later he fires a second egg, also vertically along the same path as the first egg. The two eggs collide in mid-air a. At what height do they collide and what are their respective velocities when they collide? b. Sketch out the motion curves ofeach egg on a v t and plot. (Both eggs on the same v t and same x-t plots. Fulh annotate the plots. your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be launched from rest from the Earth's surface and is to reach a maximum height of 1500 meters. The rocket engines give rocket an acceleration of 140 m/s during the time Tthat they fire. After the engines shut of.the rocket is in free fall. What must be the value ofTin order for the rocket to reach its required altitude? ng the idea of sending another rover to Mars that would explore the polar regions of the are 3. NASA is covered in ice. Newly acquired high-resolution images from Mars orbit suggest that the ice is frequently fractured by crevasses that can be up to 5-m wide. NASA engineers are experimenting with a "pop-up" thruster to be installed on the bottom of the new rover that would provide a briefimpulse and suddenly accelerate the rover vertically giving the rover an initial vertical velocity of 10 m/s. The is tat the would proceed toward the crevasse edge at its top horizontal speed and "pop-up" just it approaches the edge of the crevasse. Its combined horizontal and vertical velocities would then at carry it over the crevasse landing safely on the opposite side on Mars? Note a. What required horizontal velocity of this new rover be ifitneeded to jump a 5-meter crevasse the acceleration due to gravity on Mars is 3.71 it cross a crevasse on b. If the rover is designed to have the maximum horizontal velocity you derived in part (a could Mars that was only 4-meters wide but whose far side was 5-meters above the launch point? 4. While walking through a park a physics student sees a water fountain installed in the center of a roughly circular pond similar to that shown in the image. The fountain sprays water upward at an angel of65 relative to the horizontal and the circular splash ring on of the surface of the pond (ie. where the fountain spay hits the water surface) has a radius 5.0 meters, Assume free fall. fountain? a. What is the initial speed ofthe water as it exits the b. Suppose the water pressure suddenly increased, doubling the initial velocity of the ter. How would the maximum height of the water fountain change compared to its normal height? How would the radius of the splash pattern change compared to its normal 5 meter radius? and notices that the circular splash ring of the fountain is On a windy day the physics student passes the same fountain head. Assuming that th no longer centered on the fountain itself, but is displaced horizontally 2.5 m from the fountain shift the splash wind is pushing all the fountain droplets at the speed of the wind, estimate the wind speed that would pattern by 50 cm. map in the equally old beach house you are staying at in Key Largo, Florida. The map giv code starting from the front porch of the beach house. Since you are a brilliant Math/Scienc ou are able after a time to read the coded directions. The directions tell you to do the following t follow the directions but find your path blocked by the new highw ue north from the

Explanation / Answer

a) in vertical, water came to samelevel.

y = uy * t + ay * t^2 /2

0 = (u sin65) - 9.81t^2/2

t = 2usin65/9.81

in horizontal, water travels 5 meters.

horizontal distance = horizontal speed * time

5 = vx * t

5 = (u cos65) ( 2usin65/9.81)

u = 8 m/s

b) maximum height = u^2 (sin65)^2 / 2g

now u is doubled hence maximum height will become 4 times.

Radius = 2 u^2 sin65 cos65 / 9.81

so radius will also become 4 times.

new R = 20 m

c.

time will not change.

t = 2 u sin65 / g = 1.48 sec

in this time horizontal shift have to be 50cm

hence d = vt

0.50 m = v ( 1.48 s)
v = 0.34 m/s

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