HW6-phy2048-Spring-2014-solution

# 00m1 where we have used the fact found most easily

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Unformatted text preview: quations from Ch. 2 (since we are ignoring air friction) or simple energy conservation from Ch. 8. Choosing the latter approach, we have 1 ( 5.0 kg )(1.2 m s )2 = ( 5.0 kg ) 9.8 m s2 h 2 which gives the result h = 0.073 m. ( ) 3 PHY2048 Spring 2014 HW6 9P58. We think of this as having two parts: the first is the collision itself, where the blocks “join” so quickly that the 1.0- kg block has not had time to move through any distance yet, and then the subsequent motion of the 3.0 kg system as it compresses the spring to the maximum amount xm. The first part involves momentum conservation (with +x rightward): ! m1v1 = (m1+m2)v ⇒ (2.0 kg)(4.0 m s) = (3.0 kg)v !...
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## This document was uploaded on 03/24/2014.

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