Two students in bumper cars want to prove conservation of momentum. They collide their bumper cars. The first bumper car has a mass of 120 kg

Question

and was moving with a velocity of 4.0 m/s
before the collision and with a velocity of −2.0 m/s
after the collision. The second bumper car has a mass of 90 kg
and was moving at a velocity of −5.0 m/s
before the collision. To prove conservation of momentum, what must the velocity of the second bumper car be after the collision?(1 point)
Responses

2.7 m/s
2 point 7 meters per seconds

2.3 m/s
2 point 3 meters per seconds

3.0 m/s
3 point 0 meters per seconds

1.0 m/s

Answer 1

To prove conservation of momentum, we need to calculate the total momentum before and after the collision, and set them equal to each other.

Before the collision:

1. Momentum of the first bumper car: \[ p_1 = m_1 \cdot v_1 = 120 , \text{kg} \cdot 4.0 , \text{m/s} = 480 , \text{kg} \cdot \text{m/s} \]

2. Momentum of the second bumper car: \[ p_2 = m_2 \cdot v_2 = 90 , \text{kg} \cdot (-5.0 , \text{m/s}) = -450 , \text{kg} \cdot \text{m/s} \]

Total momentum before the collision: \[ p_{\text{before}} = p_1 + p_2 = 480 , \text{kg} \cdot \text{m/s} - 450 , \text{kg} \cdot \text{m/s} = 30 , \text{kg} \cdot \text{m/s} \]

After the collision:

1. Momentum of the first bumper car (after collision): \[ p_1' = m_1 \cdot v_1' = 120 , \text{kg} \cdot (-2.0 , \text{m/s}) = -240 , \text{kg} \cdot \text{m/s} \]

2. Momentum of the second bumper car (after collision): Let the final velocity of the second bumper car be \( v_2' \): \[ p_2' = m_2 \cdot v_2' = 90 , \text{kg} \cdot v_2' \]

Total momentum after the collision: \[ p_{\text{after}} = p_1' + p_2' = -240 , \text{kg} \cdot \text{m/s} + 90 , \text{kg} \cdot v_2' \]

Setting before and after momentum equal (conservation of momentum): \[ p_{\text{before}} = p_{\text{after}} \] \[ 30 , \text{kg} \cdot \text{m/s} = -240 , \text{kg} \cdot \text{m/s} + 90 , \text{kg} \cdot v_2' \]

Solving for \( v_2' \): \[ 30 + 240 = 90 v_2' \] \[ 270 = 90 v_2' \] \[ v_2' = \frac{270}{90} = 3.0 , \text{m/s} \]

Therefore, the velocity of the second bumper car after the collision must be 3.0 m/s.