ch09-p121 - v 1 f = V = m 1 m 1 + m 2 v 1 i = + 1.00 m/s ....

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

View Full Document Right Arrow Icon
v 1 f = m 1 m 2 m 1 + m 2 v 1 i = 200 g 600 g v 1 i = 1 3 (3.00 m/s) = 1.00 m/s . (a) The impulse is therefore J = m 1 v 1 f m 1 v 1 i = (0.200 kg)(–1.00 m/s) – (0.200 kg)(3.00 m/s) = – 0.800 N . s = – 0.800 kg . m/s, or | J | = –0.800 kg . m/s. (b) For the completely inelastic collision Eq. 9-75 applies
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: v 1 f = V = m 1 m 1 + m 2 v 1 i = + 1.00 m/s . Now the impulse is J = m 1 v 1 f m 1 v 1 i = (0.200 kg)(1.00 m/s ) (0.200 kg)(3.00 m/s) = 0.400 N . s = 0.400 kg . m/s. 121. Using Eq. 9-67, we have after the elastic collision...
View Full Document

Ask a homework question - tutors are online