Let's work through the problem step-by-step to determine which trial's cart has the greatest momentum at the bottom of the ramp.
1. Identify the relevant quantities: The momentum of an object is given by the product of its mass and its velocity. Therefore, for each trial, we need to calculate the momentum using the formula:
[tex]\[
\text{momentum} = \text{mass} \times \text{velocity}
\][/tex]
2. Calculate the momentum for each trial:
- Trial 1:
[tex]\[
\text{mass} = 200\, \text{kg}, \quad \text{velocity} = 6.5\, \text{m/s}
\][/tex]
[tex]\[
\text{momentum} = 200 \, \text{kg} \times 6.5 \, \text{m/s} = 1300.0 \, \text{kg}\cdot\text{m/s}
\][/tex]
- Trial 2:
[tex]\[
\text{mass} = 220\, \text{kg}, \quad \text{velocity} = 5.0\, \text{m/s}
\][/tex]
[tex]\[
\text{momentum} = 220 \, \text{kg} \times 5.0 \, \text{m/s} = 1100.0 \, \text{kg}\cdot\text{m/s}
\][/tex]
- Trial 3:
[tex]\[
\text{mass} = 240\, \text{kg}, \quad \text{velocity} = 6.4\, \text{m/s}
\][/tex]
[tex]\[
\text{momentum} = 240 \, \text{kg} \times 6.4 \, \text{m/s} = 1536.0 \, \text{kg}\cdot\text{m/s}
\][/tex]
- Trial 4:
[tex]\[
\text{mass} = 260\, \text{kg}, \quad \text{velocity} = 4.8\, \text{m/s}
\][/tex]
[tex]\[
\text{momentum} = 260 \, \text{kg} \times 4.8 \, \text{m/s} = 1248.0 \, \text{kg}\cdot\text{m/s}
\][/tex]
3. Compare the calculated momenta:
[tex]\[
\begin{align*}
\text{Momentum for Trial 1} & = 1300.0\, \text{kg}\cdot\text{m/s} \\
\text{Momentum for Trial 2} & = 1100.0\, \text{kg}\cdot\text{m/s} \\
\text{Momentum for Trial 3} & = 1536.0 \, \text{kg}\cdot\text{m/s} \\
\text{Momentum for Trial 4} & = 1248.0 \, \text{kg}\cdot\text{m/s}
\end{align*}
\][/tex]
4. Determine which trial has the greatest momentum:
By comparing the momenta, we see that Trial 3 has the greatest momentum at [tex]\( 1536.0 \, \text{kg}\cdot\text{m/s} \)[/tex].
Therefore, the cart in Trial 3 has the greatest momentum at the bottom of the ramp.
The correct answer would be:
Trial 3, because this trial has a large mass and a large velocity.
