Will someone check my work to see if Im on the right track?

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SUMMARY

The discussion centers on a momentum and collisions problem involving a 50g bullet and a 1.00 kg ballistic pendulum. The bullet strikes the pendulum at 1000 m/s and exits at 800 m/s, leading to an impulse calculation of 10 Ns imparted to the block. The participant correctly applies the kinematic equation v² = u² + 2as to determine the block's rise, concluding it reaches a height of 5.1 meters. The analysis confirms that energy conservation is not necessary for this inelastic collision scenario.

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camel-man
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This is collisions and momentum problem. for some reason it just seems like a short problem and I am use to doing long problems so I am here to check my work.

the questions is

A 50g bullet is fired vertically up at a 1.00 kg ballistic pendulum . The bullet strikes the block at
1000 m/s and emerges from it at 800 m/s. How far does the block rise? Assume that the
duration of the collision is instantaneous, in other words, during the collision no external forces
were acting on the system

my answer

Impulse given to block=0.050x1000-0.500x800=10=change in the block's momentum=1.00xv→v=10m/s. Using v²=u²+2as with v=0, u=10, a=-g=-9.8, we get 0=100-2x9.8s→s=5.1m.
[The collision isn't elastic but this doesn't matter since we aren't using energy conservation to solve the problem].

Am i missing something else?
 
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hi camel-man! :wink:
camel-man said:
Impulse given to block=0.050x1000-0.500x800=10=change in the block's momentum=1.00xv→v=10m/s. Using v²=u²+2as with v=0, u=10, a=-g=-9.8, we get 0=100-2x9.8s→s=5.1m.
[The collision isn't elastic but this doesn't matter since we aren't using energy conservation to solve the problem].

Am i missing something else?

looks ok to me :smile:

(though that's not how a ballistic pendulum works :confused:)
 

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