Question

3) Imagine an infinitely long ramp. The ramp is at an incline 01 relative to the hori- zontal. A ball is placed on the ramp (at rest) and allowed to roll down. After the ball travels a distance D along the ramp the ball encounters a smaller second ramp (this is like a ski jump). The second ramp makes and angle 02 relative to the incline. The second 3 ramp launches the ball (just like a ski jump). Derive an expression for the distance the ball travels along the ramp from when it is launched to when it hits the ramp. Just to be clear 0 < 01 < 7/2 and 0 < 02 < 7/2.

Answer 1

The question tells that ramp has inclination
01
. The second ramp starting at distance from where the ball starts rolling. The diagram is as below-

As, the ball moves downward due to gravity, the free body diagram is as below-

mg sino ingrosty mg >V cos Usino,

For 1st image(Ball moves D distance)

mgsin81 = та

so,  $g sin\theta_1=a$

and Newton's law of motion are-

$v=u+at$ ...............(1)

$S=ut+\frac{1}{2}at$ ..........(2)

$v^2=u^2+2as$ .........(3)

Using 2nd equation,

$D=0+\frac{1}{2}gsin\theta_1*t$

$t=\frac{2D}{gsin\theta_1}$

From 1st equation,

$v=0+gsin\theta_1*\frac{2D}{gsin\theta_1}=2D$

So, Using 3rd equation, we get

$4D^2=0+2*gsin\theta_1*D$

$D=\frac{1}{2}gsin\theta_1$ ..................(4)

for 2nd image(ball reaches to 2nd ramp)

$u=vcos\theta_2=2Dcos\theta_2$

$a=gsin\theta_1cos\theta_2$

using eq(3)

$v^2=4D^2cos^2\theta_2+2*gsin\theta_1cos\theta_2*S$

So,

$\frac{v^2-4D^2cos^2\theta_2}{2*gsin\theta_1cos\theta_2}=S$ ..................(5)

equation (4) and (5) are the required solutions.