A roller coaster car full of lawyers has total mass 600 kg and is initially at r
ID: 2258278 • Letter: A
Question
A roller coaster car full of lawyers has total mass 600 kg and is initially at rest a height of H = 106 m above ground level. After it slides down the track, it is launched from a ramp at 30?degrees to the horizontal. The top of the ramp and the subsequent plane are h = 6.0 m above the ground, as shown. Ignore all friction and air resistance.
(a) Explain how to solve this problem, then explain why this was so easy , then proceed to compute how far the car flies across the plane (R) before crashing . . . without doing any work whatsoever.
(b) how can we make this problem harder
*equations please despite the weird questions. Thanks ahead of time
Explanation / Answer
From conservation of energy:
the velocity of car at the ground level will be:
mgh = .5mv^2
g*106 = .5*v^2
v = 45.58 m/sec
now this kinetic energy will be used to climb the ramp,
so,
.5*m*(45.58)^2 = .5mv'^2 + mg*6
v' = 44.27m/sec
now we can find the distance before crash using projectile formula:
R = (v'^2*sin2(30))/g
R = 173.19 m
This is a simple problem of energy conservation, at two points
this problem can be made tough if the inclined was having friction, or put a block at ground and cart comes and crashes with the block and block travels along the incline then find R, so we have mixed concept of momentum and energy conservation
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