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 $\vec{p}$ of the particle’s mass mand velocity $\vec{v}$ .

$\vec{p} = m \vec{v}$

Newton’s second law

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

$\overset{}{\sum F} = \frac{d p}{d t}$

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’s second law

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

Substituting the values into the formula.

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

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

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Short Answer

Step by step solution

The formula used in this exercise for finding average force.

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

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