a baseball batter was able to impart to a 7.56 n-s impulse to a 0.144-kg baseball thrown at 67.3 mph, before receiving instruction. after improving his follow-through, the batter has been able to increase the impulse by 10.6%. what will be the new speed of the ball leaving the bat (in m/s) with this greater impulse? (use 1.0 m/s = 2.24 mi/hr)

Question

Accepted Answer

# Explanation:
## Step1: Convert initial speed to m/s
Given $ 1.0 \, \text{m/s} = 2.24 \, \text{mi/hr} $, initial speed $ v_i = 67.3 \, \text{mph} $. Convert to m/s: $ v_i=\frac{67.3}{2.24} \, \text{m/s} \approx 29.99 \, \text{m/s} \approx 30.0 \, \text{m/s} $
## Step2: Find initial momentum
Momentum $ p_i = m v_i $, mass $ m = 0.144 \, \text{kg} $, so $ p_i = 0.144 \times 30.0 = 4.32 \, \text{kg·m/s} $
## Step3: Calculate initial impulse (equal to initial momentum change, assuming initial momentum is $ p_i $ and final momentum before improvement is $ p_{f1} $, impulse $ J_1 = p_{f1}-p_i = 7.56 \, \text{N·s} $, so $ p_{f1}=J_1 + p_i = 7.56 + 4.32 = 11.88 \, \text{kg·m/s} $
## Step4: Find new impulse
Increase impulse by $ 10.6\% $, new impulse $ J_2 = J_1\times(1 + 0.106)=7.56\times1.106 \approx 8.361 \, \text{N·s} $
## Step5: Calculate new final momentum
New final momentum $ p_{f2}=J_2 + p_i = 8.361 + 4.32 = 12.681 \, \text{kg·m/s} $
## Step6: Find new speed
Speed $ v_{f2}=\frac{p_{f2}}{m}=\frac{12.681}{0.144} \approx 88.06 \, \text{m/s} $ (more accurately, let's recalculate each step with more precision)

### Recalculating with more precision:
Step1: $ v_i=\frac{67.3}{2.24} \approx 29.9910714 \, \text{m/s} $
Step2: $ p_i = 0.144\times29.9910714 \approx 4.3187143 \, \text{kg·m/s} $
Step3: $ p_{f1}=7.56 + 4.3187143 = 11.8787143 \, \text{kg·m/s} $
Step4: $ J_2 = 7.56\times1.106 = 7.56 + 7.56\times0.106 = 7.56 + 0.80136 = 8.36136 \, \text{N·s} $
Step5: $ p_{f2}=8.36136 + 4.3187143 = 12.6800743 \, \text{kg·m/s} $
Step6: $ v_{f2}=\frac{12.6800743}{0.144} \approx 88.056 \, \text{m/s} \approx 88.1 \, \text{m/s} $ (or more accurately, let's check the impulse - momentum relation again. Wait, actually, impulse is the change in momentum. So initial momentum is $ p_i = m v_i $, impulse $ J = \Delta p = p_f - p_i $, so $ p_f = J + p_i $. So when we calculate the new impulse, we should consider that the initial momentum is $ p_i $, so first, let's do it correctly:

Alternative approach:

Impulse is change in momentum, so $ J = m(v_f - v_i) $, so $ v_f=\frac{J}{m}+v_i $

First, find $ v_i $ precisely: $ v_i = 67.3 / 2.24 = 29.9910714286 \, \text{m/s} $

Initial impulse $ J_1 = 7.56 = 0.144(v_{f1}-v_i) $, so $ v_{f1}=\frac{7.56}{0.144}+v_i = 52.5 + 29.9910714286 = 82.4910714286 \, \text{m/s} $ (Wait, I made a mistake earlier! Oh no, I confused momentum and impulse. Impulse is $ J = \Delta p = m(v_f - v_i) $, so $ v_f = v_i+\frac{J}{m} $. That's a critical error. Let's correct this.)

## Correct Step2: Initial speed conversion is correct: $ v_i = 67.3 / 2.24 \approx 29.991 \, \text{m/s} $
## Correct Step3: Initial impulse $ J_1 = 7.56 = m(v_{f1}-v_i) $, so $ v_{f1}=v_i+\frac{J_1}{m}=29.991+\frac{7.56}{0.144}=29.991 + 52.5 = 82.491 \, \text{m/s} $
## Correct Step4: New impulse $ J_2 = J_1\times1.106 = 7.56\times1.106 = 8.36136 \, \text{N·s} $
## Correct Step5: New final speed $ v_{f2}=v_i+\frac{J_2}{m}=29.991+\frac{8.36136}{0.144}=29.991 + 58.065 = 88.056 \, \text{m/s} \approx 88.1 \, \text{m/s} $ (Wait, $ 8.36136 / 0.144 = 58.065 $, then $ 29.991+58.065 = 88.056 \approx 88.1 \, \text{m/s} $)

Wait, where was the mistake earlier? I thought impulse was equal to final momentum, but no, impulse is change in momentum, so $ J = p_f - p_i = m(v_f - v_i) $, so $ v_f = v_i + J/m $. That's the correct formula. So initial $ v_f1 = v_i + J1/m = (67.3/2.24)+(7.56/0.144) $

Calculating $ 67.3/2.24 \approx 29.991 $, $ 7.56/0.144 = 52.5 $, so $ 29.991 + 52.5 = 82.491 \, \text{m/s} $ (before improvement)

After improvement, $ J2 = 7.56\times1.106 = 8.36136 $, so $ v_f2 = 29.991+(8.36136/0.144)=29.991 + 58.065 = 88.056 \approx 88.1 \, \text{m/s} $ (or more precisely, let's compute $ 8.36136\div0.144 = 58.065 $, then $ 29.991+58.065 = 88.056 \approx 88.1 \, \text{m/s} $)

# Answer:
\boxed{88.1} (or more accurately, if we calculate with more precision, let's do:

$ v_i = 67.3 / 2.24 = 29.9910714286 $

$ J_1 = 7.56 $

$ v_{f1} = v_i + J_1/m = 29.9910714286 + 7.56/0.144 = 29.9910714286 + 52.5 = 82.4910714286 $

$ J_2 = 7.56 * 1.106 = 7.56 + 7.56*0.106 = 7.56 + 0.80136 = 8.36136 $

$ v_{f2} = v_i + J_2/m = 29.9910714286 + 8.36136/0.144 = 29.9910714286 + 58.065 = 88.0560714286 \approx 88.1 \, \text{m/s} $)

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step1: Convert initial speed to m/s

Step2: Find initial momentum

Step3: Calculate initial impulse (equal to initial momentum change, assuming initial momentum is \( p_i \) and final momentum before improvement is \( p_{f1} \), impulse \( J_1 = p_{f1}-p_i = 7.56 \, \text{N·s} \), so \( p_{f1}=J_1 + p_i = 7.56 + 4.32 = 11.88 \, \text{kg·m/s} \)

Step4: Find new impulse

Step5: Calculate new final momentum

Step6: Find new speed

Recalculating with more precision:

Correct Step2: Initial speed conversion is correct: \( v_i = 67.3 / 2.24 \approx 29.991 \, \text{m/s} \)

Correct Step3: Initial impulse \( J_1 = 7.56 = m(v_{f1}-v_i) \), so \( v_{f1}=v_i+\frac{J_1}{m}=29.991+\frac{7.56}{0.144}=29.991 + 52.5 = 82.491 \, \text{m/s} \)

Correct Step4: New impulse \( J_2 = J_1\times1.106 = 7.56\times1.106 = 8.36136 \, \text{N·s} \)

Correct Step5: New final speed \( v_{f2}=v_i+\frac{J_2}{m}=29.991+\frac{8.36136}{0.144}=29.991 + 58.065 = 88.056 \, \text{m/s} \approx 88.1 \, \text{m/s} \) (Wait, \( 8.36136 / 0.144 = 58.065 \), then \( 29.991+58.065 = 88.056 \approx 88.1 \, \text{m/s} \))

Answer: