A 7.0-g bullet is fired into a 1.5-kg ballistic pendulum. The bullet emerges fro

Question

A 7.0-g bullet is fired into a 1.5-kg ballistic pendulum. The bullet emerges from the block with a speed of 200m/s, and the block rises to a maximum height of 12cm. Find the initial speed of the bullet.

What I have trouble with is how to place the bullet or the clock into the conservation properties. I think of the problem as the bullet hitting the block and going through while applying a certain amount of energy to the block. So, I applied the conservation of energy property to the block with KE+PE (at bottom)=KE+PE (at top). I solved for the initial velocity of the block which I received as 1.5m/s~. I then took this answer and used it as the final velocity in a conservation of momentum property with the bullet and the block collision. The final velocity I recevied was around 580m/s I believe. I don't have my ntoes with me right now but once i have them in front maybe I can add more information.

Explanation / Answer

Given: the mass of the bullet is m = 7.0 g = 7*10-3kg the mass of the pendulum is M = 1.5kg the final velocity of the bullet is v = 200 m/s the height of the pendulum rised is h = 12 cm = 0.12m
According to the energy conservation theorem initial kinetic energy of the bullet is equal to the sum of the potential energy of the pendulum and final kinetic energy of the bullet
if we let the initial velocity be V, then we have                       
1/2mV2 = Mgh + 1/2mv2                            
V =[(2Mgh+mv2)/(m)]
plug in the values and calculate for V: V=[(2*1.5*9.8*0.12+7*10-3*2002)/(7*10-3)]=201.256 m/s Hope this helps. If you still have any questions, feel free to message me.
Given: the mass of the bullet is m = 7.0 g = 7*10-3kg the mass of the pendulum is M = 1.5kg the final velocity of the bullet is v = 200 m/s the height of the pendulum rised is h = 12 cm = 0.12m
According to the energy conservation theorem initial kinetic energy of the bullet is equal to the sum of the potential energy of the pendulum and final kinetic energy of the bullet
if we let the initial velocity be V, then we have                       
1/2mV2 = Mgh + 1/2mv2                            
V =[(2Mgh+mv2)/(m)]
plug in the values and calculate for V: V=[(2*1.5*9.8*0.12+7*10-3*2002)/(7*10-3)]=201.256 m/s Hope this helps. If you still have any questions, feel free to message me.
                           
V =[(2Mgh+mv2)/(m)]
plug in the values and calculate for V: V=[(2*1.5*9.8*0.12+7*10-3*2002)/(7*10-3)]=201.256 m/s Hope this helps. If you still have any questions, feel free to message me.

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