Question

A 7.14 kg bowling ball moves at 1.92 m/s. how fast must a 2.3 g ping pong ball move so that the two balls have the same kinetic energy?

Answer 1

$E_k= \frac{1}{2}mv^2$
Therefore we can substitute values in:
The kinetic energy of the bowling ball is $\frac{1}{2}*7.14*1.92^2=13.160448$

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):
$13.160448= \frac{1}{2}*0.0023*v^2 \\ v^2= \frac{2*13.160448}{0.0023} v^2=11443.87=11400ms^{-1}$
v=sqrt(11443.87)=106.97m/s
=107m/s To 3sf as the values in the question are given to 3sf.