Kinetic Energy: .00020kg Bullet vs 6.4 x 10^7 kg Ocean Liner

[Jimsac]

Not understanding a formula for this problem

Which has more kinectic energy a .00020-kg bullet traveling at 400m/s or a 6.4 10 to 7th power -kg ocean liner traveling at 10m/s (20 knots)?

I know that w=fd=mad=ma(1/2)at2) =1/2m(at)2 +1/2mv2 Is this the right formula?

 

[Nylex]

KE = (1/2)mv^2. Use that to work out the KE for the bullet and the ocean liner, then compare them.

 

[Andrew Mason]

Not understanding a formula for this problem

Which has more kinectic energy a .00020-kg bullet traveling at 400m/s or a 6.4 10 to 7th power -kg ocean liner traveling at 10m/s (20 knots)?

I know that w=fd=mad=ma(1/2)at2) =1/2m(at)2 +1/2mv2 Is this the right formula?

You can just look at this and see that the ocean liner has much more kinetic energy. It is comparing the energy of .bb gun to the Titanic at full speed! So it isn't a great question. How about a highpowered rifle bullet and a more modest ship moving at a more modest speed :

a 20 gram (.02 kg) bullet at 1000 m/sec, and a 1,500 Tonne ship at 10 cm/sec

AM