A baseball of mass #0.141# #kg# is thrown at a speed of #70# #ms^-1# from a height of #2.4# #m#. How much energy was transferred between the bat and the ball?

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Answer 1 · 2017-07-15T04:04:54

If the assumptions I have stated are accurate, the energy transferred to the bat is #347.1-15.86=331.24# #J#

We are not told how high the bat is. I will assume it is #1.2# #m# high. The explanation can be adapted if there is a different answer.

We also have to assume a final velocity for the ball. I am going to assume that it leaves the bat at #15# #ms^-1#: call it a bunt.

The energy of the ball before the collision with the bat is made up of kinetic and potential energy:

#E = E_p+E_k=mg\Deltah+1/2mv^2#
#=0.141xx9.8xx(2.4-1.2)+1/2xx0.141xx70^2#
#=1.66+345.45=347.1# #J#

The energy of the ball immediately after the collision is all kinetic energy"

#E_k=1/2mv^2=1/2xx0.141xx15^2=15.86# #J#

The difference - the energy transferred to the bat - is #347.1-15.86=331.24# #J#