How Is Momentum Conserved in a Bullet and Ballistic Pendulum Collision?

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In a ballistic pendulum scenario, a 7g bullet is fired into a 1.5kg block, emerging at 200m/s while the block rises to a height of 12cm. The initial speed of the bullet is calculated to be 528m/s, with a kinetic energy loss of 835.7J. The conservation of momentum is applied using the equation mu + mu = mv + mv, leading to the determination of the pendulum's final velocity. The potential energy at the maximum height is equated to the kinetic energy at the moment of collision to find the necessary values. This discussion effectively illustrates the principles of momentum conservation in collisions.
stupif
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1. a 7g bullet is fired into a 1.5kg ballistic pendulum. the bullet emerges from the block with a speed of 200m/s, and the block rises to a maximun height of 12cm. find the initial speed of the bullet and the kinetic energy lost of the bullet.
answer: 528m/s, 835.7j
2. mu+ mu = mv + mv
0.007u + 1.5(0)= 0.007(200) + 1.5kg(v)
how to know the final velocity of ballistic pendulum?? help me...please...
thank you
 
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The pendulum's max height is 12cm, so its KE at the collision = PE at 12cm.

so 0.5mv2=mg(0.12).
 
i got it...thank you very much~
 
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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