Conservation of Momentum in a Two-Sled System

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In a two-sled system, a 4.72 kg cat jumps between two 18.0 kg sleds on frictionless ice, prompting discussions on conservation of momentum. Initial calculations for the final speed of sled 2 yielded incorrect results, with participants correcting the momentum equations used. The correct approach involves equating the momentum of the cat and sleds before and after the jumps, leading to a final speed of sled 2 being recalculated. Misunderstandings about which sled's speed to solve for were clarified, emphasizing the importance of accurate momentum conservation application. The conversation concluded with participants successfully resolving the calculations and confirming the correct equations.
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Two 18.0 kg ice sleds are placed a short distance apart, one directly behind the other, as shown in the figure. A 4.72 kg cat, initially standing on sled 1, jumps across to sled 2 and then jumps back to sled 1. Both jumps are made at a horizontal speed of 3.79 m/s relative to the ice. What is the final speed of sled 2? (Assume the ice is frictionless.)

I tried using conservation of momentum.
(18.0 + 4.72)(3.79)= (18.0)v
solving for v gave me 4.78 m/s which wasn't right.
Can someone help?
thanks
 
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You didn't write momentum conservation correctly.

If v is the speed of the cat (relative to the ground) and m it's mass then the cat's momentum upon leaving sled 1 is mv. After landing on the sled (mass = M) the momentum is (m + M) V1 so mv = (m + M)V1. When the cat jumps off that sled, momentum is again conserved so (m+M)V1 = MV2 - mv. (Notice that the first sled doesn't really enter into the picture for this particular problem.)

You should be able to take it from there! :)
 
Thank you. I got that part, the answer was 1.99 m/s.
The second part asks for the final speed of sled 1.
I tried using what you did before, only this time the initial sled was sled 2 and final was sled 1.
MV2-mv= (m+M)V1
(18)(1.99)-(4.72)(3.79)= (18 + 4.72)V1
Solving for v gave me .789 m/s, which wasn't right.
Can someone help?
 
You're solving for V2, not V1. V1 = 1.99 m/s, not V2.
 
Ok I used (m+M)V1= MV2-mv
(4.72+18)(1.99)= (18)V2- (4.72)(3.79)
Solving for V2 gave me 3.51 m/s.. which wasn't right.
Is my equation even right?
 
Punch,

From what I wrote previously you can take a shortcut:

mv = (M+m)V_2 - mv

and the solution should jump right out! :)
 
Punchlinegirl said:
Ok I used (m+M)V1= MV2-mv
(4.72+18)(1.99)= (18)V2- (4.72)(3.79)
Solving for V2 gave me 3.51 m/s.. which wasn't right.
Is my equation even right?
Your equation is correct, but the speed of sled #2 after the first jump (V1) is not 1.99 m/s! (Sorry for not checking before.) You can redo your first calculation or...

Even better is to use Tide's last suggestion, which recognizes that the total momentum of "cat + sled #2" after the first jump must equal the momentum of the jumping cat: (m + M)V1 = mv. Thus your equation:
(m + M) V_1 = M V_2 - mv
becomes mv = M V_2 - mv
 
Ok. I got it. Thanks a bunch!
 
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