Question

1. impulse and momentum a 0.174-kg softball is pitched horizontally at 26.0 m/s. the ball moves in the opposite direction at 38.0 m/s after it is hit by the bat.

a. draw arrows showing the ball’s momentum before and after the bat hits it
b. what is the change in momentum of the ball?
c. what is the impulse delivered by the bat?
d. if the bat and ball are in contact for 0.80 ms, what is the average force the bat exerts?

Explanation:

Step1: Define coordinate system

Let the initial pitch direction be negative, so $v_i = -26.0\ \text{m/s}$, final velocity $v_f = 38.0\ \text{m/s}$, mass $m=0.174\ \text{kg}$.

Step2: Calculate initial momentum

$p_i = m v_i = 0.174\ \text{kg} \times (-26.0\ \text{m/s}) = -4.524\ \text{kg·m/s}$

Step3: Calculate final momentum

$p_f = m v_f = 0.174\ \text{kg} \times 38.0\ \text{m/s} = 6.612\ \text{kg·m/s}$

Step4: Find change in momentum

$\Delta p = p_f - p_i = 6.612\ \text{kg·m/s} - (-4.524\ \text{kg·m/s}) = 11.136\ \text{kg·m/s}$

Step5: Relate impulse to momentum change

Impulse $J = \Delta p = 11.136\ \text{kg·m/s}$

Step6: Calculate average force

First convert time: $t=0.80\ \text{ms}=0.80\times10^{-3}\ \text{s}$. Use $J=F_{avg}t$, so $F_{avg}=\frac{J}{t}=\frac{11.136\ \text{kg·m/s}}{0.80\times10^{-3}\ \text{s}} = 13920\ \text{N}$

Answer:

a. (Arrow drawing): Draw a left-pointing arrow labeled $p_i$ (momentum before hit) and a longer right-pointing arrow labeled $p_f$ (momentum after hit), since the final speed is greater than the initial speed.
b. $\Delta p = 11.1\ \text{kg·m/s}$ (rounded to 3 significant figures)
c. $J = 11.1\ \text{kg·m/s}$ (rounded to 3 significant figures)
d. $F_{avg} = 1.4\times10^4\ \text{N}$ (rounded to 2 significant figures)