500 kg cannon and is at rest on a horizontal, frictionless surface on a clear and sunny day. When a crazy cool 100 kg clown decided to show how the cannon works they decided to get fired horizontally from the barrel of the cannon and the cannon recoils with a speed of 5 m/s. A very inquisitive Physics student figured out how fast the clown was traveling when leaving the cannon. What value did they come up with?

Explanation:

Given that,

Mass of cannon

M1= 500kg and initially at rest

U1 = 0m/s

Mass of clown

M2 = 100kg

so it was initial at rest before this time, therefore, U2 = 0

Recoils speed of cannon V1 =5m/s, the recoils speed is after the cannon has left the barrel.

Using construction of linear momentum

Momentum before collision is equal to momentum after collision

The initial momentum is zero since the two bodies are until at rest

And the final momentum is

M1•V1 + M2•V2

Then,

P(initial) = P(final)

0 = M1•V1 + M2•V2

0 = 500 × 5 + 100 × V2

0 = 2500 +100•V2

100•V2 = -2500

V2 = -2500/100

V2 = —25m/s

So, the final velocity of the clown is -25m/s, opposite direction of the cannon

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more

0.11 m/s and 0.28 m/s should be correct