need help with physics

Question

In a slow-pitch softball game, a 0.200-kg softball crosses the plate at 20.0 m/s at an angle of 40.0° below the horizontal. The batter hits the ball toward center field, giving it a velocity of 44.0 m/s at 30.0° above the horizontal. (Take positive î to be the horizontal direction from the plate toward center field and take positive ĵ to be the upward vertical direction.)
(a) Determine the impulse delivered to the ball.

 

(b) If the force on the ball increases linearly for 4.00 ms, holds constant for 20.0 ms, and then decreases linearly to zero in another 4.00 ms, what is the maximum force on the ball?

Answer 1

Hi Porcia!

 

I think Part (a) breaks down into a few steps:

 

1.  Determine the horizontal (x) and vertical (y) components of the initial and final momenta

2.  Determine the change in in momentum in the x direction, and the change in momentum in the y direction.  These will be the components of the overall change in momentum.

3.  Combine those change components into the overall change in momentum, calculating its magnitude and direction.  By the impulse momentum theorem, the overall change in momentum equals the impulse delivered to the ball.

 

For the initial (incoming) case, you should draw a diagram indicating the direction of motion, to see how the velocity components (and thus the momentum components) are calculated from the given information.  What I find is that:

 

p_xo = mv_xo

v_xo = -(v_ocosθ) = -(20 m/s)cos(40^o)  (negative because the ball is presumably moving, horizontally, AWAY from center field)

p_xo = (0.200 kg)(-20 m/s)(cos(40^o)

 

In the same way, for the vertical direction:

 

p_yo = (0.200 kg)(-20 m/s)(sin(40^o))

 

You can do the same trigonometric calculation for the final case, when the ball is heading up and out toward center field (so both momentum components -- p_fx and p_fy -- will be positive).

 

Then, Δp_x = p_xf - p_xo  and Δp_y = p_yf - p_yo.  These are the x and y components of the overall change in momentum.

 

Then, to calculate the overall change in momentum Δp:

 

Magnitude:  calculate using Pythagorean theorem with both components

Direction:  calculate -- with respect to the x-axis -- using an inverse trigonometric function, such as θ = tan^-1(Δp_y/Δp_x)  

 

[NOTE:  Because of the ball's final direction of travel compared to its initial direction, you can be pretty sure that the change in momentum will point into the first quadrant]

 

The change in momentum equals the impulse delivered, by the impulse-momentum theorem, so this is the answer to (a).

 

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One way to define impulse is as the area under a plot of force versus time over the duration of the event that changes the object's momentum.  In calculus terms, by the way, this means impulse is the integral of force with respect to time.

 

In this case, the described plot has a shape whose area under the plot can be calculated in a relative few steps, geometrically.  The plot is a symmetric shape, and the area underneath can basically be broken down into three parts:  two symmetric (and mirror-image) right triangles connected to a central rectangle.  The triangles have a base (on the time axis) of 0.004 s, and a height of F_max (the maximum force achieved during this event).  The central rectangle has a base of 0.020 s and a height of F_max, as well.

 

So the area under this plot can be calculated by:

 

A = 2*(area of triangle) + area of rectangle

A = 2(1/2)(0.004 s)(F_max) + (0.020 s)(F_max)

A = 0.024F_max

 

This equals the impulse calculated in Part (a).  Just set this equal to impulse and solve for F_max.

 

I hope this helps!  If this did not answer your questions, or if you would like to check an answer, just let me know.  I will also check this again later to see if I made an error somewhere.