Question

In t-ball, young players use a bat to hit a stationary ball off a stand. The 140 g ball has about the same mass as a baseball, but it is larger and softer. In one hit, the ball leaves the bat at 12 m/s after being in contact with the bat for 2.0 ms . Assume constant acceleration during the hit.a.What is the acceleration of the ball? Express your answer to two significant figures and include the appropriate units. a =
b. What is the net force on the ball during the hit?

Express your answer to two significant figures and include the appropriate units.

F =

Answer 1

Answer:

The acceleration is $6.0*10^3 \ m/s^2$

The net force on the ball during the hit is  $840N$

Explanation:

In order to gain a good understanding of the solution above we need to understated that the main concept used to solve this problem is Newton’s First and Second law .

Initially, calculate the change in momentum of the ball. Later, compare the expression for force to find the acceleration of ball. Finally, change in momentum is equivalent to the force experienced by the ball.

Now looking at the fundamentals

  Newton’s second law of motion in terms of momentum is as follows,

   $\frac{\Delta p}{\Delta t} = ma$

Here, $\Delta p$ is change in momentum, $\Delta t$ is change in time, m is mass and a is acceleration.

Momentum is given by the following formula,

   $p=mv$

Here, p is momentum, mm is mass and v is velocity.

The net force on the ball during the hit is given by

   $F=ma$

Here F is the force, m is the mass and a is the acceleration

Step One

Calculate change in momentum.

$\Delta p = m(v_{f}-v_{i} )$

Is final velocity of ball ${v_f}$ and ${v_i}$ is initial velocity of ball.

Substitute, 12 ${\rm{ m/s}}$ for ${v_f}$ 0 ${\rm{ m/s}}$ for ${v_i}$ and 140${\rm{ g}}$ for m

$\Delta p = (140g)(12\ m/s - 0\ m/s)$

     $=(140g)(\frac{1kg}{1000g} )(12\ m/s)$

    $= 1.680kg. m/s$

Now consider the following expression,

$\frac{\Delta p}{\Delta t} = ma$

Rearrange the above expression for acceleration.

$a = \frac{\Delta p}{\Delta tm}$

Substitute, $1.680{\rm{ kg}} \cdot {\rm{m/s}}$ for $\Delta p$ ,$2.0{\rm{ ms}}$ for $\Delta t$ and $140{\rm{ g}}$ for m .

$a = \frac{1,680kg . m/s}{(2.0ms)(\frac{1kg}{1000ms} )(140g)(\frac{1kg}{1000g} )}$

  $=6.0*10^3 m/s^2$

The acceleration of the ball$6.0*10^3\ m/s$

Note :

Initially ball was in rest so, initial velocity is $0{\rm{ m/s}}$ . Om the last line of calculation the term$\left( {\frac{{1{\rm{ s}}}}{{1000{\rm{ ms}}}}} \right)$ and $\left( {\frac{{1{\rm{ kg}}}}{{1000{\rm{ g}}}}} \right)$ in the denominator is multiplied to convert $2.0{\rm{ ms}}$ and $140{\rm{ g}}$ respectively into SI units.

Step Two

Rate of change in momentum is as follows,

$F = \frac{\Delta p}{\Delta t}$

Substitute,$1.680{\rm{ kg}} \cdot {\rm{m/s}}$ for $\Delta p$ and $2.0{\rm{ ms}}$ for $\Delta t$ .

$F = \frac{1.68-kg \cdot m/s}{2.0ms}$

   $= \frac{1.680kg\cdot m/s}{(2.0ms)(\frac{1m}{1000ms} )}$

    $= 840N$

Hence the net force on the charge is  840 N

Note :

From second law of motion force is equivalent to the rate of change of momentum.