Question

1. a 2.0 kg object initially moving at 4.0 m/s experiences an impulse of 6.0 n·s in the direction of motion,

a. calculate the object’s change in momentum.
b. determine the object’s final velocity.

Explanation:

Part a

Step1: Recall impulse-momentum theorem

The impulse-momentum theorem states that the impulse (\(I\)) applied to an object is equal to the change in its momentum (\(\Delta p\)). So, \(\Delta p = I\).

Step2: Substitute the given value

We are given \(I = 6.0\space N\cdot s\). So, \(\Delta p=6.0\space kg\cdot m/s\) (since \(1\space N\cdot s = 1\space kg\cdot m/s\)).

Step1: Recall the formula for momentum change

The change in momentum \(\Delta p=m(v_f - v_i)\), where \(m\) is the mass, \(v_f\) is the final velocity and \(v_i\) is the initial velocity. We know \(\Delta p = 6.0\space kg\cdot m/s\), \(m = 2.0\space kg\) and \(v_i=4.0\space m/s\).

Step2: Rearrange the formula to solve for \(v_f\)

From \(\Delta p=m(v_f - v_i)\), we can rearrange it as \(v_f=\frac{\Delta p}{m}+v_i\)

Step3: Substitute the values

Substitute \(\Delta p = 6.0\space kg\cdot m/s\), \(m = 2.0\space kg\) and \(v_i = 4.0\space m/s\) into the formula:
\(v_f=\frac{6.0}{2.0}+ 4.0\)
\(v_f = 3.0+4.0=7.0\space m/s\)

Answer:

The object's change in momentum is \(\boldsymbol{6.0\space kg\cdot m/s}\)

Part b