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

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The discussion centers on calculating the recoil speed of a chair when a cat jumps onto it. The cat, weighing 6.4 kg, jumps from a distance of 1.7 m and lands with zero vertical velocity, causing the chair, which weighs 12 kg, to roll away. The calculations involve determining the horizontal velocity of the cat and applying the conservation of momentum for the inelastic collision between the cat and chair. The final speed of the chair and cat moving together is found to be approximately 2.99 m/s. The conversation confirms the use of momentum conservation rather than the center of mass equation for this scenario.
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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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