{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

section%202%20soln

# section%202%20soln - is mv 2 2 2 Mv 2 c 2 At the maximum...

This preview shows pages 1–5. Sign up to view the full content.

Friday, November 07, 2008 10:25 AM Midterm 2 Prob 1 Page 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
1 Solution for Problem 2 1) Before the collision the total momentum is mv 0 + Mv 0 . The total mo- mentum should be conserved, since there are no external forces in x-direction. Also since there is no force in x-direction acting on the block m (no friction) during the collision, the momentum of the block won’t change (just after the collision). Then we can find the velocities of the block and carts: the velocity of the block v 0 the velocity of two carts after the collision Mv 0 = 2 Mv c and therefore v c = v 0 2 2) To find the height we can use the conservation of energy. The energy just after the collision (we can’t use the initial energy, since the collision is inelastic)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: is mv 2 2 + 2 Mv 2 c 2 . At the maximum height the relative velocity of the block (respect to carts) is zero, so the block and the carts move as a whole system. To ﬁnd this velocity we again can use the conservation of momentum (this velocity is not equal to v c ). ( m + M ) v = ( m + 2 M ) v f v f = m + M m + 2 M v The conservation of energy gives us: mv 2 2 + 2 Mv 2 c 2 = mgh + ( m + 2 M ) v 2 f 2 mv 2 + 2 M ‡ v 2 · 2 = 2 mgh + ( m + M ) 2 (2 M + m ) v 2 h = Mv 2 4 g (2 M + m )...
View Full Document

{[ snackBarMessage ]}