Question

A special 10.0 kg cannonball is build that has a fracture exactly through the center, with a small radio-controlled explosive in the center. When the explosive is triggered, the two cannonball halves, each of mass 5.0 kg, travel in opposite direction. This cannonball is dropped from a cliff that is 20 m high. After falling for 10 m, the explosive is triggered. Immediately after the explosion, the top half of the cannonball is instantaneously at rest before it starts to fall again, while the bottom half of the cannonball continues directly downward. Ignore air resistance for all questions, and assume that the explosive itself was massless. Immediately after the explosion, what is the downward velocity of the bottom half of the cannonball?

Answer 1

Let, V be the the velocity of the cannon ball after leaving the cliff. From energy conservation;

M g h1 = 1/2 M V

V = sqrt ( 2 g h1)

V = sqrt ( 2 x 9.8 x 20) = 19.8 m/s

Momentum of the cannon ball before explosion:

Pi = MV = 10 x 19.8 = 198 kg-m/s

After explosion, all the instantneous energy that the top portion has was PE and as it falls PE gets converted to KE, again from energy conservation:

m g h2 = 1/2 mv1^2

v1 = sqrt ( 2 g h2)

v1 = sqrt ( 2 x 9.8 x 10) = 14 m/s

its momentum is : P1 = m1v1 = 5 x 14 = 70 kg-m/s

From conservation of momentum

Pi = Pf

Pi = P1 + P2

198 = 70 + P2

P2 = 198- 70 = 128

m2v2 = 128 => v2 = 128/m2 = 128/5 = 25.6 m/s

Hence, v2 = 25.6 m/s