# Problem: A baseball of mass m1 = 0.39 kg is thrown at another ball hanging from the ceiling by a length of string L = 1.95 m. The second ball m2 = 0.85 kg is initially at rest while the baseball has an initial horizontal velocity of V1 = 3.5 m/s. After the collision, the first baseball falls straight down (no horizontal velocity.)What is the angle that the string makes with the vertical at the highest point of travel in degrees?

###### FREE Expert Solution

From the law of conservation of momentum:

$\overline{){{\mathbf{m}}}_{{\mathbf{1}}}{{\mathbf{v}}}_{{\mathbf{1}}}{\mathbf{=}}{{\mathbf{m}}}_{{\mathbf{2}}}{{\mathbf{v}}}_{{\mathbf{2}}}}$

Law of conservation of energy;

$\overline{){{\mathbf{K}}}_{{\mathbf{i}}}{\mathbf{+}}{{\mathbf{U}}}_{{\mathbf{i}}}{\mathbf{+}}{\mathbf{w}}{\mathbf{=}}{{\mathbf{K}}}_{{\mathbf{f}}}{\mathbf{+}}{{\mathbf{U}}}_{{\mathbf{f}}}}$

Let's consider the following diagram, after the collision:  ###### Problem Details

A baseball of mass m1 = 0.39 kg is thrown at another ball hanging from the ceiling by a length of string L = 1.95 m. The second ball m2 = 0.85 kg is initially at rest while the baseball has an initial horizontal velocity of V1 = 3.5 m/s. After the collision, the first baseball falls straight down (no horizontal velocity.) What is the angle that the string makes with the vertical at the highest point of travel in degrees?