# Homework Help: Jumping cat & momentum

1. Oct 27, 2005

### Punchlinegirl

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

2. Oct 28, 2005

### Tide

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! :)

3. Oct 28, 2005

### Punchlinegirl

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?

4. Oct 28, 2005

### Staff: Mentor

You're solving for V2, not V1. V1 = 1.99 m/s, not V2.

5. Oct 28, 2005

### Punchlinegirl

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?

6. Oct 28, 2005

### Tide

Punch,

From what I wrote previously you can take a shortcut:

$$mv = (M+m)V_2 - mv$$

and the solution should jump right out! :)

7. Oct 29, 2005

### Staff: Mentor

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 $$mv = M V_2 - mv$$

8. Oct 29, 2005

### Punchlinegirl

Ok. I got it. Thanks a bunch!