Question

In t-ball, young players use a ball to hit a stationary ball off a stand. The $140g$ ball has about the same mass as a baseball, but it is large and softer. In one hit the ball leaves the bat at $12m/s$ after being in contact with the bat for $2.0ms$.Assume constant acceleration during the hit.

Answer 1

Hint: We will mark the initial and final velocity along with the time taken. And then using the equation $v = u + at$ we can find the acceleration of the body. Then by applying Newton's second law we will get the net force acting on the ball.

Formula used:
$v = u + at$
$F=ma$

Complete step by step solution:
Let us have a closer and detailed look at the given details.
It was said in the question that the mass of the ball is $140g$. It is not given in SI units so let us convert it.
$m = 140 \times {10^{ - 3}}kg$
As we know that the young players are practising by hitting at a stationary ball, so before launching the hit the velocity is zero.
i.e. $u = 0m/s$
after the hit, the leaves the ball with a velocity of $12m/s$
i.e.$v = 12m/s$
and the ball was in contact with the bat for, let us convert it to SI units.
$t = 2 \times {10^{ - 3}}s$
Now we can use the given data to find the acceleration experienced by the ball.
$\eqalign{
  & v = u + at \cr
  & \Rightarrow 12 = 0 + (a \times 2 \times {10^{ - 3}}) \cr
  & \Rightarrow a = \dfrac{{12}}{{2 \times {{10}^{ - 3}}}} = 6000m/{s^2} \cr} $
Thus, the acceleration experienced by the ball is $6000m/{s^2}$.
Also, from Newton's second law of motion we know that the net force acting on the ball can be expressed as the product of mass of the ball and the net acceleration experienced by the ball.
$\eqalign{
  & F = ma \cr
  & \Rightarrow F = 140 \times {10^{ - 3}} \times 6000 \cr
  & \therefore F = 840N \cr} $
Thus, the total force acting on the ball is $840N$.

So, we conclude by stating that the total acceleration of the ball is $6000m/{s^2}$
And the net force acting on the ball at that time interval is $840N$.

Note:
Remember to convert the units into SI units. Also, the force obtained here is the total force experienced. Instantaneous force on the body might be different and we can’t find it with the given values.