A baseball of mass m1 = 0.26 kg is thrown at another ball hanging from the ceiling by a length of string L = 1.45 m. The second ball m2 = 0.64 kg is initially at rest while the baseball has an initial horizontal velocity of V1 = 2.5 m/s. After the collision the first baseball falls straight down (no horizontal velocity).

The angle is approximately 50°

Explanation:

We know,

h = L – L * cos θ

  = 1.55 – 1.55 (cos θ )

To determine the maximum height, we use conservation potential and kinetic energy. As 2nd ball rises to its maximum height, the increase of its potential energy is equal to the decrease of its kinetic energy. Below is the equation for determining the value of h from the 2nd ball’s initial velocity.

Potential energy, PE = mgh = m X 9.8 X h

Kinetic energy, KE =

Set PE equal to KE and solve for h.

h = v²/19.6  

To determine the initial kinetic energy of the second ball, we need to the velocity of the 2nd ball immediately after the collision. To do this we need to use conservation of momentum. Below is the equation for determining the value of h from the 2nd ball’s initial velocity.

For the 1st ball, horizontal momentum = 0.26 X 2.5 = 0.65

Since this ball falls straight down after the collision, its final horizontal momentum is 0.

For the 2nd ball, horizontal momentum = 0.61 X vf

0.64 * vf = 0.65

vf = 0.65/.0.64

vf = 0.02m/s

h = (0.65/0.64)²/19.6

h = 0.00002m

0.00002 = 1.45 – 1.45 * cos θ

Subtract 1.55 from both sides.

0.00002 – 1.45 = -1.45 * cos θ

θ = 50°