Question

9 an object of mass 1.2kg is moving with a velocity of 2.0m/s when it is acted on by a force of 4.0n. the velocity of the object increases to 5.0m/s in the same direction. for which period of time does the force act on the object? a 0.90s b 1.1s c 1.5s d 3.6s

Explanation:

Step1: Calculate the change in momentum

The change in momentum \(\Delta p = m(v - u)\), where \(m = 1.2\space kg\), \(u = 2.0\space m/s\), \(v = 5.0\space m/s\).
\(\Delta p=1.2\times(5 - 2)=1.2\times3 = 3.6\space kg\cdot m/s\)

Step2: Use the impulse - momentum theorem \(F\Delta t=\Delta p\)

Given \(F = 4.0\space N\), from \(F\Delta t=\Delta p\), we can solve for \(\Delta t\).
\(\Delta t=\frac{\Delta p}{F}\)
Substitute \(\Delta p = 3.6\space kg\cdot m/s\) and \(F = 4.0\space N\) into the formula.
\(\Delta t=\frac{3.6}{4}=0.9\space s\)

Answer:

A. \(0.90s\)