How Does Hitting a Corked, Water-Filled Soda Bottle with a Fist Break It?

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A glass soda bottle filled with water and sealed with a cork breaks when a student strikes the cork with her fist. The fist, weighing 0.480 kg and moving at 5.00 m/s, creates a pressure that exceeds the bottle's maximum tolerance of 70.0 MPa. The collision between the fist and cork is elastic, lasting 1.20×10^-4 seconds. To find the force exerted by the fist, one can apply Newton's second law, focusing on the change in momentum over time. Understanding these dynamics is crucial for calculating the breaking force on the bottle.
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A glass soda bottle is emptied of soda and filled to the very top with water. A cork is carefully fitted into the top of the bottle, leaving no air between the cork and the water. (Intro 1 figure) The top of the bottle has a diameter of D_top = 2.00 cm and the bottom of the bottle has a diameter of D_bot = 6.50 cm. The glass breaks when it is exposed to p_max = 70.0 MPa of pressure.
A student hits the cork sharply with her fist and the bottom of the bottle breaks. The student's fist has a mass of m = 0.480 kg and moves downward at a speed of v_i = 5.00 m/s. It collides elastically with the cork and rebounds with the same speed. The collision lasts for t = 1.20×10−4 s. In this problem, the positive direction is upward.

Part A
What is the force that the fist exerts on the top of the bottle?
Express your answer in Newtons.

Where do I even begin with this?
 
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doggieslover said:
A glass soda bottle is emptied of soda and filled to the very top with water. A cork is carefully fitted into the top of the bottle, leaving no air between the cork and the water. (Intro 1 figure) The top of the bottle has a diameter of D_top = 2.00 cm and the bottom of the bottle has a diameter of D_bot = 6.50 cm. The glass breaks when it is exposed to p_max = 70.0 MPa of pressure.
A student hits the cork sharply with her fist and the bottom of the bottle breaks. The student's fist has a mass of m = 0.480 kg and moves downward at a speed of v_i = 5.00 m/s. It collides elastically with the cork and rebounds with the same speed. The collision lasts for t = 1.20×10−4 s. In this problem, the positive direction is upward.
Part A
What is the force that the fist exerts on the top of the bottle?
Express your answer in Newtons.

Hi doggieslover! Welcome to PF! :smile:

Hint: use good ol' Newton's second law … force = rate of change of momentum …

how much momentum is changed? and over what time? :smile:
 
Thanks
 
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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