How Fast Does Howard Fly from the Cannon?

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The discussion centers on the physics of momentum conservation as it applies to Howard the Human Projectile being launched from a 300 kg circus cannon. Given Howard's mass of 75 kg and the cannon's recoil velocity of 2.5 m/s, the initial velocity at which Howard leaves the cannon is calculated to be 10 m/s. This conclusion is derived from the principle that the momentum of Howard moving forward equals the momentum of the cannon moving backward, ensuring that the total momentum remains zero before and after the launch.

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The amazing 'Howard the Human Projectile' is fired straight forward from a 300 kg circus cannon during a recent Ringling Bros. Barnum and Baily performance in La Plata. Howard has a mass of 75 kg and the cannon recoils with a velocity of 2.5 m/s, at what initial velocity does Howard leave the cannon?




a. 2.5 m/s

b. 10 m/s

c. 25 m/s

d. 187.5 m/s

I am completely stuck ):
 
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momentum is conserved
the momentum of howard going forward = momentum of cannon going backwards
 
Note that initially, the cannon+Howard have no velocity, and so zero momentum. So the sum of the momenta of the cannon and Howard afterward also has to equal zero, so that you conserve momentum (initial=final). Be sure to keep tabs on your directions (Howard moves opposite the direction of the cannon) because momentum is a vector (but you're dealing with one dimension, which simplifies it a bit).
 
If 300kg object moved backward at the rate of 2.5m/s, then how far did a 75kg object move back?

(2.5/75)*300
 

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