Question

a bat hits a 0.150 kg baseball for 0.0880 s. the balls velocity changes from -41.0 m/s to +37.0 m/s. how much force was applied to the ball? ? n

Explanation:

Step1: Calculate the change in momentum

The formula for momentum is $p = mv$. The change in momentum $\Delta p=m(v_f - v_i)$. Here, $m = 0.150\ kg$, $v_i=- 41.0\ m/s$, and $v_f = 37.0\ m/s$. So, $\Delta p=0.150\times(37.0-(-41.0))=0.150\times(37.0 + 41.0)=0.150\times78.0 = 11.7\ kg\cdot m/s$.

Step2: Calculate the force

According to the impulse - momentum theorem $F\Delta t=\Delta p$, so $F=\frac{\Delta p}{\Delta t}$. Given $\Delta t = 0.0880\ s$ and $\Delta p=11.7\ kg\cdot m/s$. Then $F=\frac{11.7}{0.0880}\approx133\ N$.

Answer:

$133$