[tex]E_k= \frac{1}{2}mv^2 [/tex]
Therefore we can substitute values in:
The kinetic energy of the bowling ball is [tex] \frac{1}{2}*7.14*1.92^2=13.160448 [/tex]
We can now substitute this in for the ping pong ball (note the mass of the ping pong ball is 0.0023kg as Ek is kgm^2s-1):
[tex]13.160448= \frac{1}{2}*0.0023*v^2 \\ v^2= \frac{2*13.160448}{0.0023} v^2=11443.87=11400ms^{-1} [/tex]
v=sqrt(11443.87)=106.97m/s
=107m/s To 3sf as the values in the question are given to 3sf.
