Q10E A bat strikes a 0.145-kg basebal... [FREE SOLUTION]

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University Physics with Modern Physics

Hugh D. Young, Roger A. Freedman

Physics

14th edition Edition 路 1596 pages

ISBN: 9780321973610

Chapter 1: Q10E (page 264)

A bat strikes a 0.145-kg baseball. Just before impact, the ball is traveling horizontally to the right at $40 m / s$ ; when it leaves the bat, the ball is traveling to the left at an angle $30^{o}$ of above horizontal with a speed of 52 m/s. If the ball and bat are in contact for 1.75 ms, find the horizontal and vertical components of the average force on the ball.

Short Answer

Expert verified

The horizontal component of the average force on the ball is 7045 N towards the left and the vertical component of the average force is 2154 N.

Step by step solution

The formula used in this exercise for finding average force.

The momentum of a particle: The momentum $\overset{鈫�}{p}$ of the particle鈥檚 mass mand velocity $\overset{鈫�}{v}$ .

$\overset{鈫�}{p} = m \overset{鈫�}{v}$

Newton鈥檚 second law

The net force on a particle is equal to the rate of change of the particle鈥檚 momentum.

$\overset{}{鈭� F} = \frac{d p}{d t}$

Finding horizontal and vertical components of the average force on the ball.

Given in the question,

m =0.145 kg

Considering the right direction and above horizontal as the positive directions.

$V_{i} = 40 i m / s$

$v_{f} = - 52 \cos 30 i - 52 \sin 30 j m / s$

According to the Newton鈥檚 second law

$\begin{array}{l} \underset{}{鈭�} F \frac{d p}{d t} \\ \underset{}{鈭�} F = \frac{p_{2} - p_{1}}{鈭� t} \\ \underset{}{鈭�} F = \frac{m v_{f} - m v_{i}}{鈭� t} \end{array}$

Substituting the values into the formula.

$\overset{}{鈭� F} = \frac{(0.145) [- 52 \cos 30 i - 52 \sin 30 j] - (0.145) (40 i)}{1.75 脳 10^{- 3}} \overset{}{鈭� F} = \frac{- 12.32 i + 3.77 j}{1.75 脳 10^{- 3}} \overset{}{鈭� F} = - 7045 i + 2154 j N$