To find the velocity of the ball after it is hit, we can use the conservation of momentum principle.
The total momentum before the collision must equal the total momentum after the collision. The momentum \( p \) can be calculated as:
\[ p = m \cdot v \]
Where \( m \) is mass and \( v \) is velocity.
Before the collision:
- The cue stick has a momentum of \( 0.5 , \text{kg} \times 2.5 , \text{m/s} = 1.25 , \text{kg m/s} \).
- The ball is initially at rest, so its momentum is \( 0.2 , \text{kg} \times 0 , \text{m/s} = 0 , \text{kg m/s} \).
Thus, the total initial momentum is:
\[ \text{Total Initial Momentum} = 1.25 , \text{kg m/s} + 0 = 1.25 , \text{kg m/s} \]
After the collision, let's say the velocity of the ball \( v_b \) is what we want to find, and the cue stick continues moving with some velocity \( v_c \). Assuming an elastic collision (for simplicity), we would typically apply conservation of momentum as follows:
\[ \text{Total Final Momentum} = (0.5 , \text{kg} \cdot v_c) + (0.2 , \text{kg} \cdot v_b) \]
Setting the total initial momentum equal to the total final momentum:
\[ 1.25 , \text{kg m/s} = (0.5 , \text{kg} \cdot v_c) + (0.2 , \text{kg} \cdot v_b) \]
Without additional information on the post-collision velocity of the cue stick \( v_c \), a direct calculation for \( v_b \) is not possible.
However, if we assume an elastic collision where both the cue stick and ball exchange velocities (which isn't typical in a practical scenario), we can apply a simple ratio derived from masses, yielding:
Using the idea of elastic collisions: \[ v_b = \frac{(m_1 - m_2)}{(m_1 + m_2)} \cdot v_a + \frac{2m_2}{(m_1 + m_2)} \cdot u_2 \] With:
- \( m_1 = 0.5, \text{kg} \) (mass of cue stick)
- \( m_2 = 0.2, \text{kg} \) (mass of ball)
- \( v_a = 2.5, \text{m/s} \) (velocity of cue stick)
- \( u_2 = 0, \text{m/s} \) (initial velocity of ball)
Plugging these into:
\[ v_b = \frac{(0.5 - 0.2)}{(0.5 + 0.2)} \cdot 2.5 + \frac{2 \cdot 0.2}{(0.5 + 0.2)} \cdot 0 \]
Calculating that gives us:
\[ v_b = \frac{0.3}{0.7} \cdot 2.5 = \frac{0.3 \times 2.5}{0.7} \approx 1.0714 , \text{m/s} \]
We realized we have mismatched elastic equations.
So generally, with normal impacts again for ball impacts:
Typically, we might conclude:
- Offers require exploration across elastic impacts plus conservation consideration but usually doesn't result in varying exact sums beyond given velocities.
From available options, 3.6 m/s would fit (as sounds close if interacting equation adds reasonably; beyond that crush ratio expands), but with conservation spawn of momentum all results towards velocity systems that seem fitting through normal impacts protocols across mass distribution.
Thus, none of the answers exactly fit direct outputs unless calibrated especially across input-output variations in collisions, plus supplementary collisions usually dictate considering alternates if deeper ranges flavor.
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