Question

a 0.15 - kilogram baseball moving at 20 m/s is stopped by a catcher in 0.010 seconds. the average force stopping the ball is (a) 3.0×10² n (b) 3.0×10⁰ n (c) 3.0×10¹ n (d) 3.0×10² n

Explanation:

Step1: Use the impulse - momentum theorem

The impulse - momentum theorem is \(J = \Delta p\), and \(J = F_{avg}\Delta t\), \(\Delta p=m(v - u)\). Here, \(m = 0.15\space kg\), \(u = 20\space m/s\), \(v=0\space m/s\), \(\Delta t = 0.010\space s\).

Step2: Calculate the change in momentum

\(\Delta p=m(v - u)=0.15\times(0 - 20)=- 3\space kg\cdot m/s\). The magnitude of \(\Delta p\) is \(3\space kg\cdot m/s\).

Step3: Calculate the average force

Since \(J = F_{avg}\Delta t=\Delta p\), then \(F_{avg}=\frac{\Delta p}{\Delta t}\). Substitute \(\Delta p = 3\space kg\cdot m/s\) and \(\Delta t=0.010\space s\) into the formula: \(F_{avg}=\frac{3}{0.010}=300\space N = 3.0\times10^{2}\space N\)

Answer:

D. \(3.0\times10^{2}\space N\)