Question

A cannon of mass m1=10000 kgs located on a smooth horizontal platform fires a shell of mass m2=200 kg in horizontal direction with a velocity v2=300m/s. Find the velocity of the cannon after it is shot

Answer 1

There is no external force to the system. Therefore we can use linear momentum conservation theorem.

momentum of the system before collision =momentum of the system before collision

m1V= m2V2 + m1V1 ------ (1)

Where, m1 = mass of the cannon = 10000 kg

            V = initial velocity of the cannon = 0

            m2 = mass of shell = 200 kg

            V2 = velocity of shell = 300 m/s

            V1 = final velocity of cannon

Substituting above value for equation (1),

   $10000*0 = 200*300 + 10000*V1\\ 10000V1 = -60000\\ V1 = -6 m/s$

Answer: velocity of the cannon = -6 m/s (----> direction)

            velocity of the cannon = 6 m/s (<---- direction)

Answer 2

Answer:

6m

Explanation:

There is no external force to the system. Therefore we can use linear momentum conservation theorem.

momentum of the system before collision =momentum of the system before collision

m1V= m2V2 + m1V1 ------ (1)

Where, m1 = mass of the cannon = 10000 kg

           V = initial velocity of the cannon = 0

           m2 = mass of shell = 200 kg

           V2 = velocity of shell = 300 m/s

           V1 = final velocity of cannon

Substituting above value for equation (1),

Answer: velocity of the cannon = -6 m/s

           velocity of the cannon = 6 m/s