Which Method Keeps Bob on the Log: Catching or Deflecting the Ball?

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Catching the ball allows Bob to slow its velocity and control the impact area, potentially reducing the force experienced. However, if Bob lets the ball bounce off, he not only absorbs its momentum but also adds to it by reversing the ball's direction, which could increase the force exerted on him. The discussion highlights that both methods have different implications for Bob's stability. Ultimately, letting the ball bounce off is more likely to topple Bob due to the additional momentum involved. This scenario illustrates key concepts of momentum and force in physics.
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This is a question for my physics class I am stuck with.

"Bob is standing on a log and his friend is trying to knock him off by throwing a ball at him. Bob can catch the ball or let it bounce off of him. Which of the two is more likely to topple Bob: catching the ball or letting it bounce off?"

If he catches the ball, he could possibly dampen the blow by slowing its velocity. He can also control its impact area that way and therefore lessen the force of the ball in terms of force per area. However, if he stands still then his intertia is greater. So I can see either answer working.

Am I thinking too hard on this or is it some sort of trick question? I'm a physics noob and I can't think stuff like this out very well. Thoughts and comments and flames welcome :o)
 
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It's not a trick question. If he bounces the ball, not only does he absorb the ball's momentum but he adds to it by reversing the direction of the ball.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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