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

Answer 1

To solve the problem, we can use the law of conservation of momentum. The total momentum before the collision must equal the total momentum after the collision.

Let's denote:

• \( m_1 = 0.35 \) kg (mass of the first clay ball)
• \( v_1 \) = initial velocity of the first ball (unknown)
• \( m_2 = 0.35 \) kg (mass of the second clay ball, which is at rest)
• \( v_2 = 0 \) m/s (initial velocity of the second ball)
• \( v_f = 2.1 \) m/s (final velocity of the combined balls after collision)

Using the conservation of momentum, we have:

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

Substituting the known values into the equation:

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

This simplifies to:

\[ 0.35 v_1 = 0.7 \cdot 2.1 \]

Calculating the right-hand side:

\[ 0.7 \cdot 2.1 = 1.47 \]

Now we have:

\[ 0.35 v_1 = 1.47 \]

To find \( v_1 \), we divide by 0.35:

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

Now, since the final velocity is directed north, and considering the momentum directions, the initial velocity of the first ball must have been south to cause this result. Therefore, the answer is:

a) 4.2 m/s to the south.