Help Solve Impulse Problem w/ Pendulum Bob

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The discussion revolves around calculating the impulse delivered to a pendulum bob by a bullet that passes through it. The initial momentum of the bullet is m*v, and after passing through, it exits with a speed of v/2, while the bob, initially at rest, gains some velocity v'. The impulse is expressed as the change in momentum, calculated as (m(v/2) + Mv') - mv. There is uncertainty about whether this approach leads to the correct answer, and the poster seeks confirmation or further assistance. The conversation highlights the complexities involved in momentum conservation during the interaction between the bullet and the pendulum bob.
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HELP!Impulse problem

Homework Statement


A bullet of mass m and speed v passes through a pendulum bob of large mass M and emerges with speed v/2. the bob is at the end of a sting of length l.What impulse is delivered to the bob by the bullet in this case.

The Attempt at a Solution


impulse=momentum change=P_f-P_i
therefore,
impulse=(m(v/2)+Mv')-mv

However, I don't think it is the final answer for this question, but I don't really know how to continue... can anyone help me with this question? thank you very much.
 
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I see nothing is wrong. The bullet comes in with v while the bob has v=0 then the bullet exits with v/2 and then the bob has a velocity of v' (or some velocity v)
 


djeitnstine said:
I see nothing is wrong. The bullet comes in with v while the bob has v=0 then the bullet exits with v/2 and then the bob has a velocity of v' (or some velocity v)

thank you very much. I am just making sure.
 
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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