Question

a 250.0 gram baseball is thrown horizontally to a batter at 38 m/sec. the batter hits the ball and sends a line drive horizontally at -52 m/sec. if the bat delivered a force of -201 newtons, how long were the bat and ball in contact?

Answer 1

Approximately
`0.11\; {\rm s}`, assuming that the force delivered on the baseball is at a constant value, and that all other forces on the baseball are negligible.

Step-by-step explanation:

The impulse
`J` on the ball is equal to:

• the product of mass
  `m` and the change in velocity
  `\Delta v`, as well as
• the product of net force
  `F` and the duration
  `\Delta t` during which the force is applied.

In other words:

`m\, \Delta v = J = F\, \Delta t`.

In this question, the mass
`m` of the baseball, change in the velocity of the baseball
`\Delta v`, and the net force
`F` on the baseball are given. To find the duration of contact, rearrange the equation for impulse and solve for
`\Delta t`:

`\begin{aligned}\Delta t &= (J)/(F) = (m\, \Delta v)/(F)\end{aligned}`.

Apply unit conversion and ensure that all quantities are measured in standard units before substituting in the values:

• Mass of the ball:
  `m = 250.0\; {\rm g} = 0.2500\; {\rm kg}`.
• Change in velocity of the ball:
  `\Delta v = ((-52) - 38)\; {\rm m\cdot s^(-1)}` (subtract the initial value from the new value.)
• Net force on the baseball, which under the assumptions is approximately equal to the force from the bat:
  `F = (-201)\; {\rm N}`.

`\begin{aligned}\Delta t & = (m\, \Delta v)/(F) \\ &= ((0.2500)\, ((-52) - 38))/((-201))\; {\rm s} \\ &\approx 0.11\; {\rm s}\end{aligned}`.

In other words, the bat would be in contact with the baseball for approximately
`0.11\; {\rm s}` under the assumptions.