A 145-g baseball approaches the bat at 31.4m/s. The batter imparts a 11.5kg-m/s

Question

A 145-g baseball approaches the bat at 31.4m/s. The batter imparts a 11.5kg-m/s impulse in the opposite direction. Part A: What is the ball's new speed ? Part B: If contact time was 1.10ms, what was the average force applied by the bat? Express your answer with the appropriate units.

Explanation / Answer

2) (i) By Impulse(I) = ?P =>-11.5 = m x ?v =>-11.5 = 145 x 10^-3 x [v(final) - v(initial)] =>-11.5 = 145 x 10^-3 x [v(f) - 31.4] =>v(f) = - 47.91 m/s[-ve indicating the direction of the velocity of ball is reversed] (ii) By F = ?P/?t =>F = -11.5/(1.10 x 10^-3) = -10454.5454 N [-ve indicate the direction of Force is opposite to the direction of ball]

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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