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

STEP 1

1. Rex's mass is $81.5 \, \text{kg}$ .
2. Tex's mass is $94.3 \, \text{kg}$ .
3. Rex's car mass is $125 \, \text{kg}$ .
4. Tex's car mass is $125 \, \text{kg}$ .
5. Rex is moving at $2.33 \, \text{m/s}$ before the collision.
6. Tex is initially at rest.
7. After the collision, Tex and his car move at $1.16 \, \text{m/s}$ .
8. We are to find the post-collision speed of Rex and his car.

STEP 2

1. Calculate the total mass of Rex and his car.
2. Calculate the total mass of Tex and his car.
3. Apply the law of conservation of momentum.
4. Solve for the post-collision speed of Rex and his car.

STEP 3

Calculate the total mass of Rex and his car.
$\text{Total mass of Rex and his car} = 81.5 \, \text{kg} + 125 \, \text{kg} = 206.5 \, \text{kg}$

STEP 4

Calculate the total mass of Tex and his car.
$\text{Total mass of Tex and his car} = 94.3 \, \text{kg} + 125 \, \text{kg} = 219.3 \, \text{kg}$

STEP 5

Apply the law of conservation of momentum. The total momentum before the collision equals the total momentum after the collision.
Before the collision: $\text{Momentum of Rex and his car} = 206.5 \, \text{kg} \times 2.33 \, \text{m/s}$ $\text{Momentum of Tex and his car} = 219.3 \, \text{kg} \times 0 \, \text{m/s} = 0$
Total initial momentum: $206.5 \, \text{kg} \times 2.33 \, \text{m/s} + 0$
After the collision: $\text{Momentum of Tex and his car} = 219.3 \, \text{kg} \times 1.16 \, \text{m/s}$ $\text{Momentum of Rex and his car} = 206.5 \, \text{kg} \times v$
Total final momentum: $219.3 \, \text{kg} \times 1.16 \, \text{m/s} + 206.5 \, \text{kg} \times v$
Set the total initial momentum equal to the total final momentum: $206.5 \, \text{kg} \times 2.33 \, \text{m/s} = 219.3 \, \text{kg} \times 1.16 \, \text{m/s} + 206.5 \, \text{kg} \times v$

STEP 6

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:
$\boxed{1.098 \, \text{m/s}}$

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