Unformatted text preview: 68. (a) If the collision is perfectly elastic, then Eq. 9-68 applies v2 =
2m1 2m1 2 v= 2gh = 3 2gh m1+ m2 1i m1+ (2.00)m1 where we have used the fact (found most easily from energy conservation) that the speed of block 1 at the bottom of the frictionless ramp is 2gh (where h = 2.50 m). Next, for block 2’s “rough slide” we use Eq. 8-37:
1 2 2 m2 v2 = ΔEth = fk d = μk m2 g d . where μk = 0.500. Solving for the sliding distance d, we find that m2 cancels out and we obtain d = 2.22 m. m1 (b) In a completely inelastic collision, we apply Eq. 9-53: v2 = v (where, as m1+ m2 1i above, v1i = 2gh ). Thus, in this case we have v2 = 2gh /3. Now, Eq. 8-37 (using the total mass since the blocks are now joined together) leads to a sliding distance of d = 0.556 m (one-fourth of the part (a) answer). ...
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