a) an 800kg car is at the top of Mount wellington (at a height of 1270m). if the car were allowed to role down the mountain without using any brakes, how fast would the car be travelling when it reaches sea level? Assume that there is no friction. b) after reaching sea level the car hits a ramp that sends it into the air at an angle of 30 degrees. how far would the car fly before hitting the ground? c) if the impact of the car and ground takes 0.1 second, what would be the force on the passengers in the car? 10 points for whoever can ive the best answer and show their working out...Good luck!!
Physics - 8 Answers
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1 :
whoa blew my mind, and i took this last year
2 :
What angle is the car travelling down the mountain?
3 :
a. The weight of the car is irrelevant. The potential energy at the top of the mountain equals the kinetic energy at the bottom. This will let you solve for the velocity. b. Assume that the velocity does not change while passing the ramp. Resolve the velocity into horizontal and vertical components. Noting that at the top of its arc the vertical velocity is zero, determine the distance that it must fall to re-gain the vertical velocity that it had originally. From this, compute the time, and double that since the car goes both up and down. Use this with the horizontal velocity to get the distance. c. Use the velocity from part a for this. a = dv/dt, and v and t are known.
4 :
1:fast work: because 2:far work: because also 3: to high to survive Because ppl r not indestructible
5 :
Conservation of energy: mgh = (1/2)mv^2 sqrt[2gh] = v = 157.77 m/s Plug this into the range equation and solve for the distance. Then use the kinematics equation and time = 0.1 to solve for the acceleration of the car on impact, with the acceleration, F=ma where m is the mass of the car.
6 :
You would need to know the angle of the mountain.
7 :
ha! nice try! trick question if i ever saw one. so much info missing, a)= what is the angle of the car?also, does the weight of the 800kg car reflect also the weight of the passengers? b)=what is the weight distribution? the angle of the fall? if it were front heavy, it would fall shorter, if it were back heavy, further away. is there any wind resistance?? again, what about the weight of the passengers?? and c)= what is the weight of the passengers? the impact resistance of the ground? of the car? is it nose diving or landing on the tires (shocks and rubber absorb alot of the impact)? are the passengers wearing seat belts (if so, what is their resistance)? these are just some of my concerns that lead me to conclude that this is a trick question.... nice try though... :-)
8 :
Thanks for such a good question!!... this re-kindled my rusted physics aptitude after a span of almost 7 years!! coming to the solution...., I have made quite a few assumptions, which I think are valid due to their details unavailable in the question! Assumptions: 1. The mountain has a 45degree straight frictionless slope. 2. Weight of the passenger can either be neglected or be assumed. I have chosen to 'neglect' the passenger's mass. Approach: I) To calculate the velocity, time & acc. components 1. Calculate the distance from the top of the mountain to it's base using the formula: sin45 = 1270 (Ht.)/ (Slope length) The slope lenght will come to: 1796.05 Mts 2. Calculate the force imparted by the car during acceleration of decscent from rest. (this is reqd. to calculate the impact force when the car lands on the ramp at the base!) F=5549.37 N (consider the sine component of 'g' - gravity for slope) 3. Calculate the time taken for the car to reach the slope with uniform acceleration from 'rest' using: S=ut+1/2gt2 t=22.75 secs is the time taken by the car to hit the ramp at the bottom. 4. Now, calculate the vel. at the ramp when car hits the bottom using v2=u2+2as. V=157.89 m/s. II) To calculte the impact force: Q=Fxt' (t'=0.1 sec) Q=554.94 N-s is the force of impact or the "moment of impact ". III) For calculation projectiles: 1. Vo=157.89 m/s will be initial vel. at the time of projectile calculate the Vx and Vy (Vel. components across X and Y axis) Use Vx=VoCos30 and etc... Vx= 137.73 m/s Vy=78.945 m/s 2. Time to attain max. hieght during projectile: Th=(Vy - Vo)/ay = (157.89-78.945)/9.81 Th=8.05 secs 3. Horizontal dist. traversed by the car: Sh=Vx X th = 136.78x8.05x2 Sh= 2201.35 Mts. Am i right?? :)) regards, Vinay
