What is the impulse is delivered to the block?

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To determine the impulse delivered to the block by the arrow, one must first calculate the initial momentum of the arrow, which is 14.28 N. The final momentum depends on whether the arrow penetrates the block or sticks to it, with the latter leading to a single object post-collision. If the arrow sticks, conservation of momentum can be used to find the new velocity of the combined mass. The discussion highlights the ambiguity in the term "penetrates," impacting the ability to solve the problem definitively. Understanding these concepts is crucial for solving impulse-related questions in physics.
brad sue
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Hi,
I would like to have some help the following problem
If a 32 g arrow moving at 160km/h penetrates a block of wood suspended by a rope, what is the impulse is delivered to the block?
The impulse= momentumfinal-momentuminitial.
We know that momentuminitial= 0.32 kg* 44.44m/s=14,28 N.
But what is the momentumfinal??
I do not understand this concept well even after read the texbook. Can someone explain me what is going on ?
Thank you
B
 
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I think the question is poorly written, as whether it can be solved or not depends on the definition of the word "penetrates". To me, that means it passes through the block and continues on - in which case, there isn't enough information to solve the problem uniquely.

So we'll make the other assumption - that the arrow simply sticks in the block. In this case, there is only one object after the collision occurs. Do you see a way to determine the velocity of that object? Think conservation.
 
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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