One way to approach this is to look at the balance of energies. First we can work out how far the ball drops vertically. We know the slope length is 4m and the angle of the incline is 30°. So the initial height of the ball is 4sin30=2m. We need this height to determine the potential energy due to gravity.
(1) Initial energy = final energy considering just what happens as the ball rolls from the top of the incline to the bottom.
Potential energy initially=mgh is all the initial energy, where m is the mass of the ball and g is the acceleration of gravity=9.81m/s² and h=height=2m. The final energy is the kinetic energy at the end of the incline=½mv² where v is the ball’s velocity at the end of the incline PLUS the rotational energy=½I⍵², where I is the moment of inertia of the ball and ⍵=angular velocity. Without deriving it we write I=⅖mr² where r=ball’s radius, and we assume a solid ball of uniform density. The ball is rotating at ⍵ radians per second, which is related to the translation velocity v, ⍵=v/r. So the rotational energy is ½(⅖mr²)(v²/r²)=⅕mv².
So the final energy is ½mv²+⅕mv²=0.7mv² and mgh=0.7mv². The mass cancels out and gh=0.7v², making v=√gh/0.7=√(9.81×2/0.7)=5.2942 m/s.
(2) Once the ball leaves the end of the incline it is in free fall. We can split its initial velocity at this point on its journey into a horizontal component and a vertical component, vᵪ=vsin30 and vᵧ=vcos30. We can’t use an energy equation this time because the final energy is not definable. When the ball falls into the cart it may bounce or the energy would be dissipated as sound. Also, we don’t know the radius of the ball. Therefore we’ll consider the ball as a point mass. In this case, we just use the equation sᵧ=vᵧt+½gt² where sᵧ=3m is the distance the ball drops and t is the time to fall that distance. Also, we can write sᵪ=vᵪt where sᵪ is the horizontal displacement.
We have a quadratic in t: ½gt²+vᵧt−sᵧ=0, t=2(-vᵧ+√(vᵧ²+2gsᵧ))/g=0.8874 seconds, when we plug in sᵧ=3 and vᵧ=vcos30=4.5849m/s. Note that the other root of the quadratic is negative which can be rejected.
Now we can find sᵪ=vᵪt=2.6471×0.8874=2.35m approx.