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 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)

Question

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  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

2.3 m/s
2 point 3 meters per seconds

1.0 m/s
1 point 0 meters per seconds

3.0 m/s
3 point 0 meters per seconds

2.7 m/s

GPT-4o mini · Answer

To prove conservation of momentum, we need to calculate the total momentum before and after the collision and ensure they are equal.

Step 1: Calculate the total momentum before the collision.

The momentum $p$ is given by the formula: $p = m v$ where $m$ is mass and $v$ is velocity.

For the first bumper car: $p_{1} = m_{1} v_{1} = 120, kg \times 4.0, m/s = 480, kg m/s$

For the second bumper car: $p_{2} = m_{2} v_{2} = 90, kg \times (- 5.0, m/s) = - 450, kg m/s$

Now, calculate the total momentum before the collision: $p_{total before} = p_{1} + p_{2} = 480, kg m/s + (- 450, kg m/s) = 30, kg m/s$

Step 2: Calculate the total momentum after the collision.

Let $v_{2}^{'}$ be the velocity of the second bumper car after the collision. The momentum of the second bumper car after the collision is: $p_{2}^{'} = m_{2} v_{2}^{'} = 90, kg \times v_{2}^{'}$

The momentum of the first bumper car after the collision is: $p_{1}^{'} = m_{1} v_{1}^{'} = 120, kg \times (- 2.0, m/s) = - 240, kg m/s$

Now, calculate the total momentum after the collision: $p_{total after} = p_{1}^{'} + p_{2}^{'} = - 240, kg m/s + 90, kg \times v_{2}^{'}$

Step 3: Set the total momentum before equal to the total momentum after. $30 = - 240 + 90 v_{2}^{'}$

Step 4: Solve for $v_{2}^{'}$ . Add 240 to both sides: $30 + 240 = 90 v_{2}^{'}$ $270 = 90 v_{2}^{'}$

Divide both sides by 90: $v_{2}^{'} = \frac{270}{90} = 3.0, m/s$

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