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A 0.145 kg ball moving in the +x direction at 14 m/s is hit by a bat. The ball's final velocity is 20.0 m/s in the -x direction. The bat acts on the ball for a duration of 0.010 s.

Find the average force exerted on the ball by the bat. Include the sign of the force to indicate the direction of the force.

User Danboh
by
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1 Answer

2 votes

Answer:

Approximately
(-4.9* 10^(2)\; {\rm N}) (rounded to two significant figures.)

Step-by-step explanation:

The impulse
\mathbf{J} on an object is equal to the change in the momentum
\mathbf{p} of this object. Dividing impulse by duration
\Delta t of the contact would give the average external force that was exerted on this object.

If an object of mass
m is moving at a velocity of
\mathbf{v}, the momentum of that object would be
\mathbf{p} = m\, \mathbf{v}.

Let
m denote the speed of this ball. Let
\mathbf{v}_(0) denote the velocity of this ball before the contact, and let
\mathbf{v}_(1) denote the velocity of the ball after the contact.

Momentum
\mathbf{p}_(0) of this ball before the contact:
\mathbf{p}_(0) = m\, \mathbf{v}_(0).

Momentum
\mathbf{p}_(1) of this ball after the contact:
\mathbf{p}_(1) = m\, \mathbf{v}_(1).

Impulse, which is equal to the change in momentum:


\mathbf{J} = \mathbf{p}_(1) - \mathbf{p}_(0) = m\, \mathbf{v}_(1) - m\, \mathbf{v}_(0).

Since the velocity of the ball after the contact is in the
(-x) direction,
\mathbf{v}_(1) = -20.0\; {\rm m\cdot s^(-1)}. The average external force on this ball would be:


\begin{aligned} \mathbf{F} &= \frac{\mathbf{J}}{\Delta t} \\ &= \frac{m\, \mathbf{v}_(1) - m\, \mathbf{v}_(0)}{\Delta t} \\ &= \frac{m\, (\mathbf{v}_(1) - \mathbf{v}_(0))}{\Delta t} \\ &= \frac{0.145\; {\rm kg} * ((-20.0\; {\rm m\cdot s^(-1)}) - 14\; {\rm m\cdot s^(-1)})}{0.010\; {\rm s}} \\ &\approx -4.9 * 10^(2)\; {\rm N}\end{aligned}.

(Rounded to two significant figures.)