Question

A 150 g baseball pitched at a speed of 45 m/s is hit straight back to the pitcher at a speed of 60 m/s. What is the magnitude of the average force on the ball from the bat if the bat is in contact with the ball for 9.5 ms?

Answer 1

Answer:

The average force is 1578.94 N.

Explanation:

Given that,

Mass of baseball = 150 g

Speed = 45 m/s

Speed of pitcher = 60 m/s

Time = 9.5 ms

We need to calculate the average force

Using formula of impulse

$J=\Delta p$

$J=m\Delta v$...(I)

$J=F\Delta t$....(II)

From equation (I) and (II)

$F=\dfrac{m(v_{f}-v_{i})}{\Delta t}$

Where, m = mass of baseball

$v_{f}$ = final velocity

$v_{i}$ = Initial velocity

$\Delta t$ = time

Put the value into the formula

$F=\dfrac{150\times10^{-3}\times(60-(-40))}{9.5\times10^{-3}}$

$F=1578.94\ N$

Hence, The average force is 1578.94 N.

Answer 2

Answer:

F = 1657.89 N

Explanation:

given,

mass of the baseball, m = 150 g

initial speed, u = 45 m/s

final speed, v = 60 m/s

time of contact,t = 9.5 ms

we know,

impulse is equal to change in momentum

J = m (v - u)

J = 0.15 x (60-(-45))

J = 0.15 x 105

J = 15.75 Kg.m/s

We also know that impulse

J = F x Δ t

F x 9.5 x 10⁻³ = 15.75

F = 1657.89 N

The magnitude of the average force is equal to F = 1657.89 N