Question

section a

1. a 3.0 kg ball is struck with a bat with a force of 160 n. if the bat was in contact with the ball for 0.030 s, the balls velocity is

(a) 1.0 m/s
(b) 0.1 m/s
(c) 2.5 m/s
(d) 10 m/s
(e) none of the above

Explanation:

Step1: Recall impulse - momentum theorem

Impulse $J = F\Delta t$ and $J=\Delta p = m\Delta v$. Since the ball starts from rest (assumed), $\Delta v = v - 0=v$. So $F\Delta t=mv$.

Step2: Solve for velocity $v$

We know $F = 160\ N$, $\Delta t=0.030\ s$ and $m = 3.0\ kg$. Rearranging $F\Delta t=mv$ for $v$ gives $v=\frac{F\Delta t}{m}$.
Substitute the values: $v=\frac{160\times0.030}{3.0}$.
$v=\frac{4.8}{3.0}=1.6\ m/s$.

Answer:

(e) none of the above