A baseball player hits a baseball with a bat. Just before the player hits the ba

Question

A baseball player hits a baseball with a bat. Just before the player hits the ball, the incoming ball's speed is 40 m/s. The ball's mass is .7kg. The ball was in contact with the bat for 40 ms. There is a force-time graph recorded by the force probe that is shown. It looks like an upside down V, with the peak at about 4000 N at 20ms. They asked me to find the impulse. I assumed the impulse would be the velocity by the mass over time, which was like 487. 5. However, that's wrong. They also asked me to find the velocity of the ball right after it's been hit by the baseball. I got the answer for that wrong too. If someone can explain what exactly they are looking for that would be nice.

Explanation / Answer

impulse is given by   force* del(T)   or   I=p=mvf-mvi.
where Vf is the final velocity and Vi is the initial velocity

initial velocity = 40m/s

peak value of force = 4000 N

as we know that impulse is given by F*del(T) , the area under the curve will give the impulse.

as it is a triangle , base = 40 m s= 40x10-3

height = 4000N

so, impulse = area under the curve= 0.5 * 40*10-3 s * 4000 N = 80 N-s

now we need to find the velocity

again imulse is given by

I = del(P) =mvf-mvi.

so,

80 N= 0.7 ( Vf - 40 )

=> 114.28 = ( Vf - 40 )

=> Vf = 154.28 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.