Question

A toy car of mass 40g rides along a frictionless track of length 5m. One end of the track is held 2m above the floor. The end on the floor meets a tiny wedge inclined upward at 30 degrees, which acts as a ramp. If the car is let go from rest at the top of the incline, how far from [the ramp at the end of] the track does it land?

Answer 1

initial gravitational potential energy at the top

Ui = m*g*h

final kinetic energy at the end of the ramp Kf = (1/2)*m8v^2

from energy conservation

Kf = Ui

v = sqrt(2*g*h)

v = sqrt(2*9.8*2) = 6.26 m/s

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after leaving the ramp the car motion motion is a projectile

PROJECTILE

1)

along horizontal
________________

initial velocity vx = v*costheta

acceleration ax = 0

initial position = xo = 0

final position = x

time T = 1.52 s

displacement = x - x0

from equation of motion

x - x0 = vx*T+ 0.5*ax*T^2

x - xo = v*costheta*T

T = (x - xo)/(v*costheta)

along vertical
______________

initial velocity vy = v*sintheta

acceleration ay = -g = -9.8 m/s^2

initial position y0 = 0

final position y = 0

from equation of motion

y-y0 = vy*T + 0.5*ay*T^2 .........(2)

0 = v*sintheta*T - (1/2)*g*T^2

T = 2*v*sintheta/g

T = (x-xo)/(v*costheta)

(x-xo)/(v*costheta) = 2*v*sintheta/g

x - xo = 2*v*v*sintheta*costheta/g

x - xo = v^2*sin(2theta)/g

x - xo = 6.26^2*sin(2*30)/9.8

x - xo = 3.46 m <<<<---------ANSWER