Question

A 0.145-kg baseball pitched at 31.0 m/s is hit on a horizontal line drive straight back at the pitcher at 46.0 m/s. If the contact time between bat and ball is \({\bf{5}}{\bf{.00 \times 1}}{{\bf{0}}^{{\bf{ - 3}}}}\;{\bf{s}}\), calculate the force (assumed to be constant) between the ball and bat.

Short answer

The force between the ball and bat is 2,230 N.

Step-by-step solution

Newton’s second law

According to Newton’s second law, the rate of change of momentum of an object is equal to the net force applied on it, i.e.,

\(F = \frac{{\Delta p}}{{\Delta t}}\).

Here,\(\Delta p\)is the change in momentum of the object in\(\Delta t\)time.

In this problem, theforce between the ball and bat is equal to the rate of change of momentum of the baseball.

Given information

Mass of the baseball,\(m = 0.145\;{\rm{kg}}\).

Contact time between bat and ball,\(\Delta t = 5.00 \times {10^{ - 3}}\;{\rm{s}}\).

Ifthe direction of motion of the ball from bat to the pitch is considered the positive direction, then

The initial velocity of the baseball,\(u = - 31.0\;{\rm{m/s}}\).

The final velocity of the baseball,\(v = 46.0\;{\rm{m/s}}\).

Determination of force between the ball and bat

From Newton’s second law, the averageforce on the ball is as follows:

\(\begin{array}{c}\bar F = \frac{{\Delta p}}{{\Delta t}}\\ = \frac{{m\left( {v - u} \right)}}{{\Delta t}}\\ = \frac{{\left( {0.145\;{\rm{kg}}} \right)\left[ {46.0\;{\rm{m/s}} - \left( { - 31.0\;{\rm{m/s}}} \right)} \right]}}{{\left( {5.00 \times {{10}^{ - 3}}\;{\rm{s}}} \right)}}\\ = \;2.23 \times {10^3}\;{\rm{N}}\\ = 2230\;{\rm{N}}\end{array}\)

Since force is assumed to be constant, this average force of 2230 N is equal to the force between the ball and bat.