Question

When the batter hits the ball, a net force of 1150 N, opposite to the direction of the ball's initial motion, acts on the ball for 9.0 x 10-3 s during the hit. What is the final velocity of the ball?

Answer 1

Answer:

This question is not complete as we cannot find the velocity without knowing the mass of the ball.

Explanation:

While the question may not be complete, I could guide on what to do.

From Newton's second law of motion,

An impressed force is proportional to the rate of change of momentum.

Mathematically, this is written as

$F = \frac{(mv - mu)}{t}$

Where

$F$ is the impressed force, $m$ is the mass of the object, $u$ and $v$ are the initial and final velocities respectively.

This equation can be written as

$Ft = mv - mu$

Substituting the values from the question,

$1150\times9.0\times10^{-3}=mv-mu$

Assuming the collision is perfectly elastic, then $v=-u$ (being perfectly elastic means it rebounds with the same initial velocity; the negative sign indicates that the rebound is opposite in direction). Thus, we have

$10.35=2mv$

Then the final velocity $v$ is

$v=\frac{10.35}{2m}$

If $m$ is known, then you have your answer.

Answer 2

The final velocity of the ball after it was hit is $\frac{10.35}{m}$ (m/s).

The given parameters:

• Net force on the ball, F = 1150 N
• Time of ball's action, t = $9\times 10^{-3} \ s$

Let the mass of the ball, = m

The final velocity of the ball is determined by applying Newton's second law of motion, as follows;

$F = ma = \frac{mv}{t} \\\\F = \frac{mv}{t} \\\\mv = Ft\\\\v = \frac{Ft}{m}$

$v = \frac{1150 \times 9\times 10^{-3}}{m} \\\\v = \frac{10.35}{m}$

Thus, the final velocity of the ball after it was hit is $\frac{10.35}{m}$ (m/s).

Learn more about Newton's second law here: https://brainly.com/question/3999427