Question

⑤ a constant force of 45 n acts on a 72 - kg object for 10.0 s. what is the change in the object’s momentum? ⑥ a football player gets ready for a field goal and kicks the ball with an impulse of 200 ns due to a force of 250 n. what is the time the ball is in contact with the kicker’s foot?

Explanation:

Question ⑤

Step1: Recall the impulse - momentum theorem

The impulse - momentum theorem states that the impulse \(J\) acting on an object is equal to the change in the object's momentum \(\Delta p\). The formula for impulse is \(J = F\times\Delta t\), and by the impulse - momentum theorem, \(\Delta p=J = F\times\Delta t\).

Step2: Substitute the given values

We are given that the force \(F = 45\space N\) and the time \(\Delta t=10.0\space s\). Substituting these values into the formula \(\Delta p=F\times\Delta t\), we get \(\Delta p = 45\space N\times10.0\space s\).

Step3: Calculate the result

\(45\times10.0 = 450\space N\cdot s\) (Since the unit of momentum change is the same as the unit of impulse, which is \(N\cdot s\) or \(kg\cdot m/s\)).

Step1: Recall the formula for impulse

The impulse \(J\) is given by the formula \(J = F\times\Delta t\), where \(F\) is the force and \(\Delta t\) is the time of contact. We know that impulse \(J = 200\space N\cdot s\) (assuming the unit of impulse is \(N\cdot s\)) and the force \(F = 250\space N\). We need to find the time \(\Delta t\). Rearranging the formula for impulse to solve for time, we get \(\Delta t=\frac{J}{F}\).

Step2: Substitute the given values

Substitute \(J = 200\space N\cdot s\) and \(F = 250\space N\) into the formula \(\Delta t=\frac{J}{F}\). So \(\Delta t=\frac{200\space N\cdot s}{250\space N}\).

Step3: Calculate the result

\(\frac{200}{250}=0.8\space s\)

Answer:

The change in the object's momentum is \(450\space kg\cdot m/s\) (or \(450\space N\cdot s\))

Question ⑥