Question

studied the momentum and velocity of two objects x and y. the student collected the data in the table below. what can you determine about the change
of object x compared to the change in velocity of object y?
the change in velocity of object x was greater than the change in velocity of object y.
the change in velocity of object x was less than the change in velocity of object y.
the change in velocity of object x was equal to the change in velocity of object y.
not enough information

Explanation:

Step1: Recall Impulse-Momentum Theorem

The impulse-momentum theorem states that the impulse (\(J\)) applied to an object is equal to the change in its momentum (\(\Delta p\)), and impulse is also given by \(J = F \cdot \Delta t\), where \(F\) is the net force and \(\Delta t\) is the time the force is applied. Also, momentum \(p = m \cdot v\), so \(\Delta p = m \cdot \Delta v\), which means \(\Delta v=\frac{\Delta p}{m}=\frac{F \cdot \Delta t}{m}\).

Step2: Calculate \(\Delta v\) for Object X

For object X: \(m_X = 10\space kg\), \(F_X = 15\space N\), \(\Delta t_X = 2\space s\).
Using \(\Delta v_X=\frac{F_X \cdot \Delta t_X}{m_X}\), substitute the values:
\(\Delta v_X=\frac{15 \times 2}{10}=\frac{30}{10} = 3\space m/s\).

Step3: Calculate \(\Delta v\) for Object Y

For object Y: \(m_Y = 6\space kg\), \(F_Y = 8\space N\), \(\Delta t_Y = 3\space s\).
Using \(\Delta v_Y=\frac{F_Y \cdot \Delta t_Y}{m_Y}\), substitute the values:
\(\Delta v_Y=\frac{8 \times 3}{6}=\frac{24}{6}=4\space m/s\).

Step4: Compare \(\Delta v_X\) and \(\Delta v_Y\)

We found \(\Delta v_X = 3\space m/s\) and \(\Delta v_Y = 4\space m/s\). So, \(\Delta v_X<\Delta v_Y\).

Answer:

The change in velocity of object X was less than the change in velocity of object Y.