How to Calculate Impulse and Average Net Force in a Baseball Collision

[SohailS]

Hello, I am taking physics 12U through correspondence. I am having issues with this problem. I don't know why but I feel like I made a mistake in this. Can you guys check this question and verify my results.

Thanks, very much.

Homework Statement

A baseball with a mass of 0.152 kg is moving horizontally at 32.0 m/s [E], when it is struck by a bat for 0.00200 seconds. The velocity of the ball just after the collision is 52.0 m/s [W 20° N].

a) Find the impulse experienced by the ball. (6 marks)​

b) Find the average net force of the ball. (2 marks)​

Given:

m_{} = 0.152 kg​

v_{1} = 32.0 m/s [E]​

t{} = 0.00200 s​

v_{2} = 52 m/s [W 20° N]​

Homework Equations

\vec{ΔP}=\vec{F}_{NET}Δt​

F_{NET}Δt=m(v_{2}-v_{1})​

The Attempt at a Solution

a)​

\vec{ΔP}=?​

F_{NET}Δt=m(v_{2}-v_{1})​

Break down vectors into components

solve for components

\vec{P}_{x}=0.152(-45.033-32)=-11.7 Ns​

\vec{P}_{y}=0.152(26-0)=39.52 Ns​

P=\sqrt{(-11.7)^{2}+(39.52)^{2}}=41.2 Ns​

θ=tan^{-1}\frac{39.52}{11.7}=73°​

∴\vec{ΔP}=41.2 Ns [W 73° N]​

b)​

F_{NET}Δt=41.2 Ns​

\vec{F}_{NET}=\frac{41.2}{0.002}=20,609 N [W 73° N]​

 

[Doc Al]

\vec{P}_{y}=0.152(26-0)=39.52 Ns​

Check your arithmetic here.

 

[SohailS]

\vec{P}_{y}=0.152(26-0)=39.52 Ns​

Check your arithmetic here.

Thanks for pointing that out. I have corrected that error.
Is there anything else wrong with the solution?

solve for components

\vec{P}_{x}=0.152(-45.033-32)=-11.7 Ns​

\vec{P}_{y}=0.152(26-0)=3.952 Ns​

P=\sqrt{(-11.7)^{2}+(3.952)^{2}}=12.4 Ns​

θ=tan^{-1}\frac{3.952}{11.7}=19°​

∴\vec{ΔP}=12.4 Ns [W 19° N]​

b)​

F_{NET}Δt=12.4 Ns​

\vec{F}_{NET}=\frac{12.4}{0.002}=6179 N [W 19° N]​

 

[Doc Al]

Break down vectors into components

Check your computation of those components of the final velocity.

 

[SohailS]

Break down vectors into components

Check your computation of those components of the final velocity.

Wow, I can't believe I did that. I used 30 degrees instead of 20 there. hah!
I need to pay attention when I input the values into the calculator.

I have made the changes. Please let me know if you see anything else wrong with it. particularly part b) I don't know if that is how it is supposed to be calculated.

solve for components

\vec{P}_{x}=0.152(-48.86-32)=-12.29 Ns​

\vec{P}_{y}=0.152(17.79-0)=2.703 Ns​

P=\sqrt{(-12.29)^{2}+(2.703)^{2}}=12.58 Ns​

θ=tan^{-1}\frac{12.29}{2.703}=78°​

∴\vec{ΔP}=12.6 Ns [W 78° N]​

b)​

F_{NET}Δt=12.6 Ns​

\vec{F}_{NET}=\frac{12.6}{0.002}=6300 N [W 78° N]​

 

[Doc Al]

θ=tan^{-1}\frac{12.29}{2.703}=78°​

I think you mixed up your components when calculating the angle.

Other than that, your solution looks good. (Since the force of the bat is so much greater than the weight of the ball, it's OK to ignore the weight of ball in calculating the net force from the bat.)

 

[SohailS]

Thanks, I am going to get some sleep and then do the rest of the questions. I really botched that lol.
I really appreciate your help.