- #1

kahunah

- 10

- 0

## Homework Statement

A baseball with a mass of 0.125 kg is moving horizontally at 32.0 m/s [E], when it is stuck by a bat for 0.00200 seconds. The velocity of the ball just after the collision is 52.0 m/s [W 20 N].

## Homework Equations

p = m Δv

Fnet = ∆p/∆t

there are also these equations

Fnetx ∆t = m(v2x – v1x)

Fnety ∆t = m(v2y – v1y)

## The Attempt at a Solution

i've gone over this a few times but nothing looks right to me...

Given

m for mass of baseball = 0.125 kg

v1 for the velocity of the baseball before the collision = 32.0 m/s [E]

v2 for the velocity of the baseball after the collision = 52.0 m/s [W 20 N]

Δt for duration of time the baseball struck the bat = 0.00200 s

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

Let [E] be positive

To solve for the horizontal x-component of v2

v2x = 52.0 m/s * cos (20)

v2x = 48.86 m/s

To solve for the vertical y-component of v2

v2y = 52.0 m/s * sin (20)

v2y = 17.79 m/s Since v2x is in the [W] direction, let v2x be negative

To calculate the impulse, if

p = m Δv

Then,

Δv = √((v2x – v1x)2 – (v2y – v1y)2)

Δv = √((-48.86 m/s – 32.0 m/s)2 – (17.79 m/s – 0 m/s)2)

Δv = √((-80.86 m/s)2 – (17.79 m/s)2)

Δv = √((6538.34 m/s) – (316.48 m/s))

Δv = √(6221.87 m/s)

Δv = 78.88 m/s

So,

p = m Δv

p = (0.125 kg)(78.88 m/s)

p = 8.86 kgm/s

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

Fnet = ∆p/∆t

Fnet = (8.86 kgm/s)/(0.00200 s)

Fnet = 4430 N <--- that's WAY too much force

I also know there's an angle in there somewhere that i SHOULD be using force to calculate for the direction of the impluse

tan θ = Fnety/Fnetx?

i don't know honestly I'm so confused at this point...PLEASE help!