A completely inelastic collision occurs between two balls of wet putty that move directly toward each 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 12 m/s. How high do the combined two balls of putty rise above the collision point? (Neglect air drag.)

Options

A. 2.3 m
B. 2.0 m
C. 2.6 m
D. 2.9 m

Answer

To solve the problem step by step, we will follow these main steps:

### Step 1: Calculate the initial momentum of both balls before the collision.
The momentum of an object is given by the formula:
\[ p = m \cdot v \]
Where:
- \( p \) is momentum,
- \( m \) is mass,
- \( v \) is velocity.

For the first ball (mass = 3.0 kg, velocity = 20 m/s upward):
\[ p_1 = 3.0 \, \text{kg} \cdot 20 \, \text{m/s} = 60 \, \text{kg m/s} \]

For the second ball (mass = 2.0 kg, velocity = -12 m/s downward):
\[ p_2 = 2.0 \, \text{kg} \cdot (-12 \, \text{m/s}) = -24 \, \text{kg m/s} \]

### Step 2: Calculate the total initial momentum before the collision.
The total initial momentum \( p_{initial} \) is the sum of the momenta of both balls:
\[ p_{initial} = p_1 + p_2 = 60 \, \text{kg m/s} - 24 \, \text{kg m/s} = 36 \, \text{kg m/s} \]

### Step 3: Calculate the combined mass after the collision.
After a completely inelastic collision, the two balls stick together. The total mass \( m_{total} \) is:
\[ m_{total} = m_1 + m_2 = 3.0 \, \text{kg} + 2.0 \, \text{kg} = 5.0 \, \text{kg} \]

### Step 4: Calculate the velocity of the combined mass after the collision using conservation of momentum.
Using the conservation of momentum:
\[ p_{initial} = p_{final} \]
Where:
\[ p_{final} = m_{total} \cdot V \]
Thus,
\[ 36 \, \text{kg m/s} = 5.0 \, \text{kg} \cdot V \]
Solving for \( V \):
\[ V = \frac{36 \, \text{kg m/s}}{5.0 \, \text{kg}} = 7.2 \, \text{m/s} \]

### Step 5: Calculate the maximum height reached after the collision.
To find the height \( h \) that the combined mass rises, we can use the kinematic equation:
\[ V^2 = U^2 + 2a h \]
Where:
- \( V = 0 \) (final velocity at the maximum height),
- \( U = 7.2 \, \text{m/s} \) (initial velocity after collision),
- \( a = -9.8 \, \text{m/s}^2 \) (acceleration due to gravity, negative as it acts downward).

Rearranging the equation to solve for \( h \):
\[ 0 = (7.2)^2 + 2(-9.8)h \]
\[ 0 = 51.84 - 19.6h \]
\[ 19.6h = 51.84 \]
\[ h = \frac{51.84}{19.6} \approx 2.64 \, \text{m} \]

### Final Answer:
The combined two balls of putty rise approximately **2.64 meters** above the collision point.

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