Question

A 200-gram baseball traveling at 48 m/s is hit by a bat and rebounds in the opposite direction at 52 m/s. Find the average force of the bat on the ball if the time of contact is 2.0x10-3 s.

Answer 1

The question involves calculating the average force exerted on a baseball by a bat during a collision, employing the impulse-momentum theorem in physics.

Step-by-step explanation:

The student is asking about the calculation of the average force exerted by a bat on a baseball during impact using the concept of impulse and momentum. To find the average force, we can use the formula F = Δp / Δt, where Δp is the change in momentum and Δt is the time of contact. In this case, the baseball's initial momentum is pi = -200 g × 48 m/s (taking the direction towards the bat as negative) and its final momentum is pf = 200 g × 52 m/s.

The change in momentum, therefore, is pf - pi, which equals (200 g × 52 m/s) - (-200 g × 48 m/s), or 20000 g·m/s (remember to convert grams to kilograms to keep the units consistent). Given the time of contact Δt = 2.0 x 10-3 s, the average force exerted by the bat on the baseball can be calculated.

Answer 2

400N

Step-by-step explanation:

Given the following parameters

Mass m = 200g = 0.2kg

initial velocity u = 48m/s

Final velocity v = 52m/s

Time t = 2.0x10-3 s.

Impulse is expressed as I = Ft = m(v-u)

Ft = m(v-u)

F = m(v-u)/t

Substitute the given values into the formula

F = 0.2(52-48)/0.002

F = 0.2(4)/0.002

F = 0.8/0.002

F = 400N

Hence the average force of the bat on the ball is 400N