Question

1. a force of 186 n acts on a 7.3 kg bowling ball for 0.4 s. (a) what is the impulse on the bowling ball? (b) what is its change in velocity?

Explanation:

Part (a)

Step 1: Recall the formula for impulse

Impulse (\(J\)) is defined as the product of force (\(F\)) and the time (\(t\)) for which the force acts. The formula is \(J = F \times t\).

Step 2: Substitute the given values

We are given \(F = 186\space N\) and \(t = 0.4\space s\). Substituting these values into the formula:
\(J = 186\space N \times 0.4\space s\)

Step 3: Calculate the impulse

\(J = 74.4\space N\cdot s\) (or \(kg\cdot m/s\) since \(1\space N\cdot s = 1\space kg\cdot m/s\))

Part (b)

Step 1: Recall the impulse - momentum theorem

The impulse - momentum theorem states that the impulse on an object is equal to the change in its momentum (\(\Delta p\)). Momentum is given by \(p = m\times v\), so the change in momentum \(\Delta p=m\times\Delta v\), where \(\Delta v\) is the change in velocity. From the impulse - momentum theorem, \(J=\Delta p = m\times\Delta v\). We can solve for \(\Delta v\) as \(\Delta v=\frac{J}{m}\)

Step 2: Identify the values of \(J\) and \(m\)

We found \(J = 74.4\space kg\cdot m/s\) from part (a) and the mass \(m = 7.3\space kg\)

Step 3: Calculate the change in velocity

Substitute \(J = 74.4\space kg\cdot m/s\) and \(m = 7.3\space kg\) into the formula \(\Delta v=\frac{J}{m}\)
\(\Delta v=\frac{74.4}{7.3}\space m/s\)

Step 4: Perform the division

\(\Delta v\approx10.2\space m/s\) (rounded to one decimal place)

Answer:

s:
(a) The impulse on the bowling ball is \(\boldsymbol{74.4\space N\cdot s}\) (or \(\boldsymbol{74.4\space kg\cdot m/s}\))
(b) The change in velocity is approximately \(\boldsymbol{10.2\space m/s}\) (or more precisely \(\boldsymbol{\frac{74.4}{7.3}\approx10.2\space m/s}\))