Math / Algebra

QuestionA baseball bat contacts a $0.145-\mathrm{kg}$ baseball for $1.3 \times 10^{-3} \mathrm{~s}$ The average force exerted by the bat on the ball is 8900 N .
Part A
If the ball has an initial velocity of $44 \mathrm{~m} / \mathrm{s}$ toward the bat and the force of the bat causes the ball's motion to reverse direction, what is the ball's speed as it leaves the bat? Express your answer with the appropriate units. $v_{\mathrm{f}}=\text { Valuet }_{\mathrm{s}}^{\mathrm{s}}$ Submit Previous Answers Request Answer Incorrect; Try Again; 9 attempts remaining

Studdy Solution

STEP 1

What is this asking? How fast is a baseball going after it's hit by a bat, if we know how fast it was going before it was hit, how heavy it is, how hard it was hit, and for how long? Watch out! Velocity has direction!
Don't forget that the initial velocity is towards the bat, meaning in the opposite direction of the final velocity.

STEP 2

1. Impulse-Momentum Theorem
2. Calculate the Impulse
3. Find the Final Velocity

STEP 3

Alright, let's start with the impulse-momentum theorem!
This theorem tells us that the change in momentum of an object is equal to the impulse applied to it.
Momentum is how hard it is to stop something, and impulse is a force acting over time.
It's like, the longer you push something, the more its momentum changes!

STEP 4

Mathematically, the impulse-momentum theorem is written as: $J = \Delta p$ Where $J$ is the impulse and $\Delta p$ is the change in momentum.

STEP 5

Momentum is simply mass times velocity.
So, the change in momentum is: $\Delta p = m \cdot v_f - m \cdot v_i$ Where $m$ is the mass, $v_f$ is the final velocity, and $v_i$ is the initial velocity.

STEP 6

Combining these, we get: $J = m \cdot v_f - m \cdot v_i$

STEP 7

Impulse is the average force multiplied by the time the force acts: $J = F \cdot t$ Where $F$ is the average force and $t$ is the time.

STEP 8

We're given $F =$ 8900 N and $t =$ 1.3 × 10⁻³ s.
Let's plug those in: $J = 8900 \text{ N} \cdot 1.3 \times 10^{-3} \text{ s}$ $J = 11.57 \text{ kg} \cdot \text{m/s}$ So, our impulse is 11.57 kg⋅m/s.

STEP 9

Now, let's go back to our impulse-momentum equation: $J = m \cdot v_f - m \cdot v_i$

STEP 10

We know $J =$ 11.57 kg⋅m/s, $m =$ 0.145 kg, and $v_i =$ -44 m/s (negative because it's going towards the bat).
We want to find $v_f$ .

STEP 11

Let's plug in our values: $11.57 \text{ kg} \cdot \text{m/s} = 0.145 \text{ kg} \cdot v_f - 0.145 \text{ kg} \cdot (-44 \text{ m/s})$

STEP 12

Simplify and solve for $v_f$ : $11.57 \text{ kg} \cdot \text{m/s} = 0.145 \text{ kg} \cdot v_f + 6.38 \text{ kg} \cdot \text{m/s}$ $11.57 \text{ kg} \cdot \text{m/s} - 6.38 \text{ kg} \cdot \text{m/s} = 0.145 \text{ kg} \cdot v_f$ $5.19 \text{ kg} \cdot \text{m/s} = 0.145 \text{ kg} \cdot v_f$ $v_f = \frac{5.19 \text{ kg} \cdot \text{m/s}}{0.145 \text{ kg}}$ $v_f = 35.79 \text{ m/s}$

STEP 13

The ball's speed as it leaves the bat is 35.79 m/s.

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Studdy Solution

STEP 1

STEP 2

1. Impulse-Momentum Theorem
2. Calculate the Impulse
3. Find the Final Velocity

STEP 3

STEP 4

Mathematically, the impulse-momentum theorem is written as: $J = \Delta p$ Where $J$ is the impulse and $\Delta p$ is the change in momentum.

STEP 5

Momentum is simply mass times velocity.
So, the change in momentum is: $\Delta p = m \cdot v_f - m \cdot v_i$ Where $m$ is the mass, $v_f$ is the final velocity, and $v_i$ is the initial velocity.

STEP 6

Combining these, we get: $J = m \cdot v_f - m \cdot v_i$

STEP 7

Impulse is the average force multiplied by the time the force acts: $J = F \cdot t$ Where $F$ is the average force and $t$ is the time.

STEP 8

We're given $F =$ 8900 N and $t =$ 1.3 × 10⁻³ s.
Let's plug those in: $J = 8900 \text{ N} \cdot 1.3 \times 10^{-3} \text{ s}$ $J = 11.57 \text{ kg} \cdot \text{m/s}$ So, our impulse is 11.57 kg⋅m/s.

STEP 9

Now, let's go back to our impulse-momentum equation: $J = m \cdot v_f - m \cdot v_i$

STEP 10

We know $J =$ 11.57 kg⋅m/s, $m =$ 0.145 kg, and $v_i =$ -44 m/s (negative because it's going towards the bat).
We want to find $v_f$ .

STEP 11

Let's plug in our values: $11.57 \text{ kg} \cdot \text{m/s} = 0.145 \text{ kg} \cdot v_f - 0.145 \text{ kg} \cdot (-44 \text{ m/s})$

STEP 12

STEP 13

The ball's speed as it leaves the bat is 35.79 m/s.

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Studdy Solution

STEP 1

STEP 2

1. Impulse-Momentum Theorem2. Calculate the Impulse3. Find the Final Velocity

STEP 3

STEP 4

Mathematically, the impulse-momentum theorem is written as: J=Δp J = \Delta p J=Δp Where JJJ is the **impulse** and Δp\Delta pΔp is the **change in momentum**.

STEP 5

Momentum is simply mass times velocity.So, the change in momentum is: Δp=m⋅vf−m⋅vi \Delta p = m \cdot v_f - m \cdot v_i Δp=m⋅vf​−m⋅vi​ Where mmm is the **mass**, vfv_fvf​ is the **final velocity**, and viv_ivi​ is the **initial velocity**.

STEP 6

Combining these, we get: J=m⋅vf−m⋅vi J = m \cdot v_f - m \cdot v_i J=m⋅vf​−m⋅vi​

STEP 7

Impulse is the **average force** multiplied by the **time** the force acts: J=F⋅t J = F \cdot t J=F⋅t Where FFF is the **average force** and ttt is the **time**.

STEP 8

STEP 9

Now, let's go back to our impulse-momentum equation: J=m⋅vf−m⋅vi J = m \cdot v_f - m \cdot v_i J=m⋅vf​−m⋅vi​

STEP 10

We know J=J = J= **11.57 kg⋅m/s**, m=m = m= **0.145 kg**, and vi=v_i = vi​= **-44 m/s** (negative because it's going *towards* the bat).We want to find vfv_fvf​.

STEP 11

Let's plug in our values: 11.57 kg⋅m/s=0.145 kg⋅vf−0.145 kg⋅(−44 m/s) 11.57 \text{ kg} \cdot \text{m/s} = 0.145 \text{ kg} \cdot v_f - 0.145 \text{ kg} \cdot (-44 \text{ m/s}) 11.57 kg⋅m/s=0.145 kg⋅vf​−0.145 kg⋅(−44 m/s)

STEP 12

STEP 13

The ball's speed as it leaves the bat is **35.79 m/s**.

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Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. Impulse-Momentum Theorem
2. Calculate the Impulse
3. Find the Final Velocity

Mathematically, the impulse-momentum theorem is written as: $J = \Delta p$ Where $J$ is the impulse and $\Delta p$ is the change in momentum.

Momentum is simply mass times velocity.
So, the change in momentum is: $\Delta p = m \cdot v_f - m \cdot v_i$ Where $m$ is the mass, $v_f$ is the final velocity, and $v_i$ is the initial velocity.

Combining these, we get: $J = m \cdot v_f - m \cdot v_i$

Impulse is the average force multiplied by the time the force acts: $J = F \cdot t$ Where $F$ is the average force and $t$ is the time.

Now, let's go back to our impulse-momentum equation: $J = m \cdot v_f - m \cdot v_i$

We know $J =$ 11.57 kg⋅m/s, $m =$ 0.145 kg, and $v_i =$ -44 m/s (negative because it's going towards the bat).
We want to find $v_f$ .

Let's plug in our values: $11.57 \text{ kg} \cdot \text{m/s} = 0.145 \text{ kg} \cdot v_f - 0.145 \text{ kg} \cdot (-44 \text{ m/s})$

The ball's speed as it leaves the bat is 35.79 m/s.