Conservation of energy minimum speed

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Homework Help Overview

The problem involves a scenario where a person named Gus, who is at the end of a rope, needs to reach a roof after being struck by a cannonball. The context includes concepts from conservation of energy and momentum, with specific parameters such as the height of the rope and the masses of Gus and the cannonball.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of conservation of energy and momentum to determine the minimum speed of the cannonball. There are attempts to derive equations relating the variables involved, with some questioning the correctness of their calculations and assumptions.

Discussion Status

Several participants have provided equations and reasoning related to the problem, with some expressing agreement on the approaches taken. The discussion includes verification of the methods used, particularly regarding the treatment of the collision as inelastic.

Contextual Notes

Participants are working within the constraints of the problem as stated, including specific values for mass and height, and are exploring the implications of these values on the calculations. There is an emphasis on ensuring the correct application of physical principles without reaching a definitive conclusion.

firezap
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Homework Statement


Gus is at end of rope. He needs to reach roof. His friend fires cannon ball horizontally hitting Gus and get stuck in belt. Find minimum speed of cannon ball for Gus to reach roof. 50m is rope. 99kg is Gus. 1kg is ball. 45° angle in diagram


Homework Equations


conservation of energy
Et1 = Et2
1/2mv^2 = mgh
momentum
p = mv


The Attempt at a Solution


1/2(1)v^2 = 99(9.8)(50xsin45°)
1/2v^2 = 34301.75
v = 260 m/s
is this right?
 

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conservation of energy:
1/2 (m+M) V² = (m+M)gh
V² = 2gh
conservation of momentum:
mv = (m+M)V
v = (m+M) sqrt(2gh) / m
is this correct?
 
Yes.
 
firezap said:
conservation of energy:
1/2 (m+M) V² = (m+M)gh
V² = 2gh
conservation of momentum:
mv = (m+M)V
v = (m+M) sqrt(2gh) / m
is this correct?
Looks good. The collision of cannonball and person is treated as an inelastic collision.
 

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