Bullet imbedding in a block on a frictionless surface

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SUMMARY

The discussion focuses on the physics of a 10g bullet striking a 100g block at rest on a frictionless surface, resulting in an inelastic collision. The conservation of momentum is applied to determine the final combined velocity, calculated as v_f = 1000/11 m/s. The participants clarify that conservation of energy cannot be used in this scenario due to the inelastic nature of the collision, where kinetic energy is not conserved. Thus, only conservation of momentum is applicable for this problem.

PREREQUISITES
  • Understanding of inelastic collisions
  • Knowledge of conservation of momentum
  • Familiarity with kinetic energy concepts
  • Basic principles of Newton's laws of motion
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  • Study the principles of elastic vs. inelastic collisions
  • Learn about conservation of energy in different types of collisions
  • Explore detailed examples of momentum calculations in collisions
  • Investigate the implications of frictionless surfaces in physics problems
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Physics students, educators, and anyone interested in understanding collision dynamics and momentum conservation principles.

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A 10g bullet with v = 1000 m/s strikes a 100g block which is at rest. What is their combined velocity? Can this be done with conservation of energy? momemtum?

Here we have inelastic collision so by conservation of momentum
\[
10\cdot 1000 + 0 = 110\cdot v_f\Rightarrow v_f = \frac{1000}{11}m/s.
\]
Can this be done with CoE? If so, how?
 
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Re: bullet imbedding in a block on a frictionless surface

dwsmith said:
A 10g bullet with v = 1000 m/s strikes a 100g block which is at rest. What is their combined velocity? Can this be done with conservation of energy? momemtum?

Here we have inelastic collision so by conservation of momentum
\[
10\cdot 1000 + 0 = 110\cdot v_f\Rightarrow v_f = \frac{1000}{11}m/s.
\]
Can this be done with CoE? If so, how?

No, it cannot, because the collision is inelastic. An elastic collision, by definition, is a collision in which kinetic energy is conserved. With an inelastic collision, it is not conserved. Moreover, you're given no information about the force during the collision, so you can't use Newton's Second Law, or a more general conservation of energy approach. Conservation of Momentum is all that's available to you.
 

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