When the 6-kg box reaches point A it has a speed of vA=2m/s. Determine the angle θ at which it leaves the smooth circular ramp and the distance s to where it falls into the cart. Neglect friction.

Check attachment for necessary information

Step-by-step explanation:

At point B. Check attachment for free body diagram.

Where an is the centripetal acceleration and it is given as

an = Vb²/r

Fn = m•Vb²/r

Fn = 6 × Vb²/1.2

Fn = 5Vb²

Applying Newton's second law along the y direction

ΣF = m•ay

ay = 0, since the body is not moving in y direction

N —W = 0

N — WCosβ = 0

N = Fn = 5Vb²

5Vb²—58.86Cosβ = 0

Divide through by 5

Vb² — 11.772Cosβ = 0

Vb² = 11.772Cosβ, equation 1

Applying conservation of energy.

∆K.E(A) + ∆P.E(A) = ∆K.E(B) +∆P.E(B)

½m•Va²— 0+ mg•Ha — 0 = ½m•Vb² — 0 + mg•Hb — 0

½m•Va²+mg•Ha = ½m•Vb² + mg•Hb

From attachment

Ha = 1.13m

Hb = 1.2Cosβ.

Va = 2m/s²

½m•Va²+mg•Ha = ½m•Vb² + mg•Hb

½×6×2² + 6×9.81×1.13 = ½×6×Vb²+6×9.81×1.2Cosβ

12 + 66.512 = 3Vb² + 70.632Cosβ

From equation,.Vb² = 11.772Cosβ

78.512 = 3×11.772Cosβ+70.632Cosβ

78.512 = 35.316Cosβ+70.632Cosβ

78.512 = 105.948Cosβ

Cosβ = 78.512/105.948

Cosβ = 0.7410

β = ArcCos(0.7410)

β = 42.18 °

β = 42.2°

From the attachment, it is notice that,

θ + 20° = β

θ = β — 20°

θ = 42.2 — 20°

θ = 22.2°

The angle θ at which the box

left the smooth circular ramp is 22.2°

b. Using equation free fall motion

∆y = Vby•t + ½gt²

Let get Vb first, from equation 1

Vb² = 11.772Cosβ

Vb² = 11.772Cos42.2

Vb² = 8.721

Vb = √8.722

Vb = 2.95m/s

Now, to get Vby

Vby = VbSinβ

Vby = 2.95Sin42.2

Vby = 1.984 m/s

Then, applying free fall equation at point B

∆y = Vby•t + ½gt²

Hb - 0 = 1.984t + ½ × 9.81t²

1.2Cosβ = 1.984t + 4.905t²

1.2Cos42.2 = 1.984t + 4.905t²

4.905t² + 1.984t —0.889 = 0

Using formula method

t = [-b±√(b²-4ac)]/2a

a = 4.905 b = 1.984 and c = -0.889

t =[-1.984±√(1.984²-4×4.905×-0.889)] / 2×4.905

t = (-1.984±√21.378)/9.81

t = (-1.984± 4.624)/9.81

So,

t = (—1.984 — 4.624)/9.81

t = -0.674s

Or

t = (-1.984 + 4.624)/9.81

t = 2.64/9.81

t = 0.269s

Since time cannot negative, then,

t = 0.269s

Now, we can find the distance "s" by applying range formula, the part of motion is parabola this allow us to use projectile motion

R = Ux • t

s= Vbx × t

s= VbCosβ × t

s= 2.95Cos42.2 × 0.269

s= 0.588m

So, the distance "s" where the box fall into the cart is 0.588m