A completely inelastic collision occurs between two balls of wet putty that move directly toward cach other along a vertical axis. Just before the collision, one ball, of mass 3.0 kg, is moving upward at 20 m/s and the other ball, of mass 2.0 kg, is moving downward at 10 m/s. How high do the combined two balls of putty rise above the collision point? (Neglect air drag.)

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the scenario
We have two balls of wet putty colliding inelastically. One ball has a mass of 3.0 kg moving upward at 20 m/s, and the other has a mass of 2.0 kg moving downward at 10 m/s. After the collision, they stick together and move with a common velocity.

### Step 2: Apply the conservation of momentum
The total momentum before the collision must equal the total momentum after the collision.

Let:
- \( m_1 = 3.0 \, \text{kg} \) (mass of the first ball)
- \( u_1 = 20 \, \text{m/s} \) (velocity of the first ball)
- \( m_2 = 2.0 \, \text{kg} \) (mass of the second ball)
- \( u_2 = -10 \, \text{m/s} \) (velocity of the second ball, negative because it is moving downward)

Using the conservation of momentum:
\[ m_1 u_1 + m_2 u_2 = (m_1 + m_2) v \]
Substituting the values:
\[ (3.0 \, \text{kg} \cdot 20 \, \text{m/s}) + (2.0 \, \text{kg} \cdot -10 \, \text{m/s}) = (3.0 \, \text{kg} + 2.0 \, \text{kg}) v \]
\[ 60 - 20 = 5v \]
\[ 40 = 5v \implies v = \frac{40}{5} = 8 \, \text{m/s} \]

### Step 3: Determine the maximum height reached after the collision
After the collision, the combined mass of the two balls is:
\[ m = m_1 + m_2 = 3.0 \, \text{kg} + 2.0 \, \text{kg} = 5.0 \, \text{kg} \]
The initial velocity \( v = 8 \, \text{m/s} \) (upward). We will use the kinematic equation to find the maximum height \( h \):
\[ v^2 = u^2 + 2as \]
Where:
- \( v = 0 \, \text{m/s} \) (final velocity at the maximum height)
- \( u = 8 \, \text{m/s} \) (initial velocity)
- \( a = -9.8 \, \text{m/s}^2 \) (acceleration due to gravity, negative because it acts downward)
- \( s = h \) (displacement, which is the height we want to find)

Substituting the values:
\[ 0 = (8)^2 + 2(-9.8)h \]
\[ 0 = 64 - 19.6h \]
\[ 19.6h = 64 \implies h = \frac{64}{19.6} \approx 3.27 \, \text{m} \]

### Step 4: Conclusion
The maximum height that the combined two balls of putty rise above the collision point is approximately **3.27 meters**.

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