In a perfectly elastic collision, both momentum and kinetic energy are conserved.
Let's start by applying the conservation of momentum:
\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]
where
- \(m_1\) = mass of bumper car 1 = 50 kg,
- \(u_1\) = initial velocity of bumper car 1 = 1.4 m/s,
- \(m_2\) = mass of bumper car 2 = 50 kg,
- \(u_2\) = initial velocity of bumper car 2 = -2.1 m/s,
- \(v_1\) = final velocity of bumper car 1 = -1.9 m/s,
- \(v_2\) = final velocity of bumper car 2 (unknown).
Plugging in the values into the momentum conservation equation:
\[ 50 \cdot 1.4 + 50 \cdot (-2.1) = 50 \cdot (-1.9) + 50 \cdot v_2 \]
Calculating the left side:
\[ 50 \cdot 1.4 = 70 \] \[ 50 \cdot (-2.1) = -105 \] So: \[ 70 - 105 = -35 \]
Now substituting in the momentum conservation equation:
\[ -35 = 50 \cdot (-1.9) + 50 \cdot v_2 \]
Calculating \(50 \cdot (-1.9)\):
\[ 50 \cdot (-1.9) = -95 \]
Now substituting that back in:
\[ -35 = -95 + 50 v_2 \]
Adding 95 to both sides:
\[ -35 + 95 = 50 v_2 \] \[ 60 = 50 v_2 \]
Solving for \(v_2\):
\[ v_2 = \frac{60}{50} = 1.2 \text{ m/s} \]
So, the velocity of bumper car 2 after the collision is 1.2 m/s.
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