QuestionRex (m=81.5 kg) and Tex ( 94.3 kg ) board the bumper cars at the local carnival. Rex is moving at a full speed of $2.33 \mathrm{~m} / \mathrm{s}$ when he rear-ends Tex who is at rest in his path. Tex and his 125-kg car lunge forward at 1.16 $\mathrm{m} / \mathrm{s}$. Determine the post-collision speed of Rex and his $125-\mathrm{kg}$ car.

Studdy Solution
Solve for the post-collision speed $v$ of Rex and his car.
$$206.5 \times 2.33 = 219.3 \times 1.16 + 206.5 \times v$$
Calculate the left side: $$481.145 = 254.388 + 206.5 \times v$$
Subtract $254.388$ from both sides: $$481.145 - 254.388 = 206.5 \times v$$
$$226.757 = 206.5 \times v$$
Divide both sides by $206.5$: $$v = \frac{226.757}{206.5}$$
$$v \approx 1.098 \, \text{m/s}$$
The post-collision speed of Rex and his car is approximately:
$$\boxed{1.098 \, \text{m/s}}$$