3) a red bus (3.0 kg) moving at 5.0 m/s collides elastically with a blue bus (3.0 kg) that is not moving…

3) a red bus (3.0 kg) moving at 5.0 m/s collides elastically with a blue bus (3.0 kg) that is not moving. after the collision, the red bus moves backward at -2.0 m/s. what is the final velocity of the blue bus?

3) a red bus (3.0 kg) moving at 5.0 m/s collides elastically with a blue bus (3.0 kg) that is not moving. after the collision, the red bus moves backward at -2.0 m/s. what is the final velocity of the blue bus?

Answer

Explanation:

Step1: List given values

Red bus: $m_1=3.0\ \text{kg}$, $v_{1i}=5.0\ \text{m/s}$, $v_{1f}=-2.0\ \text{m/s}$ Blue bus: $m_2=3.0\ \text{kg}$, $v_{2i}=0\ \text{m/s}$

Step2: Apply conservation of momentum

For elastic collisions, total momentum is conserved: $$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$$ Substitute known values: $$(3.0)(5.0) + (3.0)(0) = (3.0)(-2.0) + (3.0)v_{2f}$$

Step3: Simplify and solve for $v_{2f}$

Divide all terms by 3.0: $$5.0 + 0 = -2.0 + v_{2f}$$ Rearrange to isolate $v_{2f}$: $$v_{2f} = 5.0 + 2.0$$

Answer:

$3.0\ \text{m/s}$ (in the original direction of the red bus)