Question

A baseball of mass

m
1
=
0.39
kg is thrown at another ball hanging from the ceiling by a length of string L = 1.35 m. The second ball
m
2
=
0.72
is initially at rest while the baseball has an initial horizontal velocity of
V
1
=
2.5
m/s. After the collision the first baseball falls straight down (no horizontal velocity.)


Randomized Variables

m
1
=
0.39

m
2
=
0.72

L = 1.35 m

V
1
=
2.5
m/s

What is the angle that the string makes with the vertical at the highest point of travel in degrees?

θ
=
_____

Answer 1

θ = 21.44º

Step-by-step explanation:

Given

m₁ = 0.39 Kg

m₂ = 0.72 Kg

L = 1.35 m

u₁ = 2.5 m/s

u₂ = 0 m/s

v₁ = 0 m/s

It is an elastic collision. After the collision, we apply the following equation for the horizontal motion

⇒ v₂ = (m₁*u₁+ m₂*u₂- m₁*v₁)/(m₂)

⇒ v₂ = (0.39 Kg *2.5 m/s + 0.72 Kg*0 m/s - 0.39 Kg *0 m/s) / (0.72 Kg)

⇒ v₂ = 1.3542 m/s

Then we apply

Einitial = Efinal ⇒ Ki + Ui = Kf + Uf

Knowing that

At the lowest point of travel h = 0 m

At the highest point of travel v = 0 m/s

⇒ Ki + 0 = 0 + Uf ⇒ 0.5*m* v22 = m*g*h

⇒ h = 0.5*v₂²/g

⇒ h = 0.5*(1.3542 m/s)²/(9.8 m/s²) = 0.0935 m

The maximum angle the rope makes with the vertical is:

θ = ArcCos ((L-h)/L)

⇒ θ = ArcCos ((1.35 m -0.0935 m)/ 1.35 m)

⇒ θ = 21.44º