Question

Two carts collide and bounce apart. Cart 1 had a momentum of –6 kg • m/s before the collision. Cart 2 had a momentum of 10 kg • m/s before the collision.

What is the total momentum of the carts after the collision?

–16 kg • m/s
–10 kg • m/s
4 kg • m/s
10 kg • m/s

Answer 1

The total momentum of a system before a collision is equal to the total momentum after the collision, according to the law of conservation of momentum.

Before the collision, the momentum of Cart 1 is \(-6 \, \text{kg} \cdot \text{m/s}\) and the momentum of Cart 2 is \(10 \, \text{kg} \cdot \text{m/s}\).

To find the total momentum before the collision, we can add these two momenta together:

\[
\text{Total momentum before collision} = (-6 \, \text{kg} \cdot \text{m/s}) + (10 \, \text{kg} \cdot \text{m/s})
\]

Calculating this:

\[
-6 + 10 = 4 \, \text{kg} \cdot \text{m/s}
\]

Thus, the total momentum before the collision is \(4 \, \text{kg} \cdot \text{m/s}\).

Since momentum is conserved, the total momentum of the carts after the collision will also be \(4 \, \text{kg} \cdot \text{m/s}\).

So the correct answer is:

**4 kg • m/s**