A clay ball with a mass of 0.35 kg strikes another 0.35 kg clay ball at rest, and the two balls stick together. The final velocity of the balls is 2.1 m/s north. What was the first ball’s initial velocity?

Question

a
2.1 m/s to the north
b
4.2 m/s to the south
c
4.2 m/s to the north
d
2.1 m/s to the south

Answer 1

To solve for the initial velocity of the first ball, we will use the principle of conservation of momentum. The momentum before the collision should equal the momentum after the collision.

Let's define:

• \( m_1 = 0.35 , \text{kg} \) (mass of the first ball)
• \( m_2 = 0.35 , \text{kg} \) (mass of the second ball)
• \( v_1 \) = initial velocity of the first ball (unknown)
• \( v_2 = 0 , \text{m/s} \) (initial velocity of the second ball, at rest)
• \( v_f = 2.1 , \text{m/s} \) (final velocity of both balls after they stick together)

Using the conservation of momentum:

\[ m_1 \cdot v_1 + m_2 \cdot v_2 = (m_1 + m_2) \cdot v_f \]

Substituting the known values:

\[ (0.35) \cdot v_1 + (0.35) \cdot 0 = (0.35 + 0.35) \cdot 2.1 \]

This simplifies to:

\[ 0.35 v_1 = 0.70 \cdot 2.1 \]

Calculating the right side:

\[ 0.70 \cdot 2.1 = 1.47 \]

Now we have:

\[ 0.35 v_1 = 1.47 \]

Solving for \( v_1 \):

\[ v_1 = \frac{1.47}{0.35} \approx 4.2 , \text{m/s} \]

Since the final velocity of the combined mass is 2.1 m/s to the north and we know momentum was conserved, the first ball must have been moving in the opposite direction to achieve this (the second ball was at rest). Therefore, the first ball's initial velocity must be:

\[ 4.2 , \text{m/s} , \text{to the south}. \]

So, the answer is:

b) 4.2 m/s to the south.