What is the speed of recoil of the chair and cat?

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Homework Help Overview

The problem involves a cat jumping onto a rolling desk chair, with the goal of determining the speed of recoil of both the chair and the cat after the jump. The scenario includes specific measurements for the cat's mass, the chair's mass, and the distance from the cat to the chair.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of kinematic equations to determine time and horizontal velocity components. There is an exploration of momentum conservation principles in the context of an inelastic collision.

Discussion Status

Participants have engaged in a back-and-forth regarding the calculations and the principles applied, with some confirming the correctness of equations used. There is a recognition of the conservation of momentum in the context of the problem, though no explicit consensus on the final outcome has been reached.

Contextual Notes

Participants are navigating through the implications of the problem setup, including the assumption that the chair was initially at rest and the nature of the collision as inelastic.

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Homework Statement


A cat crouches on the floor at a distance of 1.7 m from a desk chair of height 0.45 m. The cat jumps onto the chair , landing with zero vertical velocity. The desk chair has coasters and rolls soomthly away when the cat lands. The cats mass is 6.4 kg , the chairs mass is 12 kg. what is the speed of recoil of the chair and cat?





The Attempt at a Solution


1/2gt^2=0.45m
t=0.303s
1.7/0.303=5.61ms horizontal velocity component
5.61*6.4=12v
v=2.99m/s
 
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Sneakatone said:
1/2gt^2=0.45m
t=0.303s
1.7/0.303=5.61ms horizontal velocity component
So far, so good.

5.61*6.4=12v
After the "collision" the chair and cat move as one body.
 
so would it be 5.61*6.4=(12+6.4)v?
 
Sneakatone said:
so would it be 5.61*6.4=(12+6.4)v?
That's correct.
 
thank you!
was that center of mass equation used or was that m1v11=m2v1?
 
Sneakatone said:
thank you!
was that center of mass equation used or was that m1v11=m2v1?
That was conservation of momentum applied to an inelastic collision.

m1v1 + m2v2 = (m1 + m2)vf

(Here, v2 = 0, since the chair was initially at rest.)
 
I see now,
thank you very much!
 

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