ch09-p136 - 136 Let M = 22.7 kg and m = 3.63 be the mass of...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
On the other hand, as the cat lands on the second sled, it sticks to it and the system (sled plus cat) moves forward with a speed 2 0.4205 m/s. i f mv v Mm == + When the cat makes the second jump back to the first sled with a speed v i , momentum conservation implies 22 () 2 f fi i i Mv mv M m v mv mv mv =++ =+= which yields 2 2 0.975 m/s. i ff mv v M After the cat lands on the first sled, the entire system (cat and the sled) again moves together. By momentum conservation, we have 11 2 f i i M m v mv Mv mv mv mv += + = + = or 1 2 0.841 m/s. i ff mv v + (a) From the above, we conclude that the first sled moves with a speed
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

Ask a homework question - tutors are online