Question

A 0.145-kg baseball pitched horizontally at 27.0 m/s strikes a bat and pops straight up to a height of 31.5 m. If the contact time between bat and ball is 2.5 ms, calculate the average force between the ball and bat during contact.

Answer 1

Answer:

The average force between the ball and bat during contact is 3006.72 N.

Explanation:

Given that,

Mass of the baseball, m = 0.145 kg

Initial speed of the baseball, u = 27 m/s

The baseball pops straight up to a height of 31.5 m.

The time of contact, $t=2.5\ ms=2.5\times 10^{-3}\ s$

The rate of change of momentum is equal to the external force applied. It is given by :

$F=\dfrac{m(v-u)}{t}$....(1)

v can be given by using the conservation of energy:

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.8\times 31.5}$

v = -24.84 m/s (it will be negative as the ball pops straight up)

Force between the bat and the ball is given by :

$F=\dfrac{0.145(-24.84-27)}{2.5\times 10^{-3}}$

F = -3006.72 N

So, the average force between the ball and bat during contact is 3006.72 N. Hence, this is the required solution.

Answer 2

The average force between the ball and the bat during the contact is 1,566 N.

The given parameters;

• mass of the ball, m = 0.145 kg
• horizontal velocity of the ball, u = 27 m/s
• height of the ball's displacement, h = 31.5 m
• the contact time of the ball, t = 2.5 ms

The average force between the ball and the bat during the contact is determined by applying Newton's second law of motion.

F = ma

where;

• a is the acceleration of the ball
• m is the mass of the ball

The average force is calculated as follows;

$F = ma = m\frac{v}{t} \\\\F = \frac{mv}{t} \\\\F = \frac{0.145 \times 27}{2.5 \times 10^{-3}} \\\\F = 1,566 \ N$

Thus, the average force between the ball and the bat during the contact is 1,566 N.

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