Question

A 2kg shot is fired from a canon of mass 198 kg with a velocity 50 m/s with respect to gun. The K.E of recoil of gun is

Answer 1

ke = 1/2 mass velocity to the power sq.
=1/2 * 2 * (50) 2
=2500 joule

Answer 2

Answer:

25.25 J

Explanation:

m1=2kg

m2=198kg

v2=50m/s

v1=?

since m1*v1=m2*v2

        198*v2=2*50

        198*v2=100

               v2=100/198m/s

KE=1/2*mass of canon*(velocity of canon)^2

   =1/2*198*100/198*100/198

   =25.25252525J

which is approximately 25.25J