Question

Two 24.0 kg ice sleds are placed a short distance apart, one directly behind the other, as shown in the figure. A 3.25 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 2.93 m/s relative to the ice.

What is the final speed of sled 2?
What is the final speed of sled 1?

Answer 1

can you attach the figure?

Answer 2

Two 17.0 kg ice sleds are placed a short distance apart, one directly behind the other, as shown in the figure. A 4.27 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 3.26 m/s relative to the ice. What is the final speed of sled 2? (Assume the ice is frictionless.) My data values are slightly different, then the original post.

Answer 3

ohhhhh, the picture helps a ton!

Answer 4

have you gone over 'momentum' in your class yet?

Answer 5

Yes, m1v1=m2v2
I have gotten the first part to the question, its just when he returns to the 1st sled that I am not able to obtain

Answer 6

great ^_^
sooo
for the first part, does it look like this? am i doing this right? its been a while.
4.27*3.26 + 17*0 = v*(4.27+17)
v = 0.655

Answer 7

.655*4.27 + 17*.655 = -3.26*4.27 + v*17
v = 1.63 for velocity of sled 2

Answer 8

-3.26 because the cat is jumping to the left now, not to the right.

Answer 9

I got the final answer on sled 2 to be 1.31m/s. I used 4 equations to break up each component, as leaving, landing, leaving and landing on the sled. But thanks for your help, if anyone needs it I can write out the lengthy equation, but let me know!