Problem_Set_6_Solutions

# Problem_Set_6_Solutions - 56 The total momentum immediately...

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56. The total momentum immediately before the collision (with + x upward) is p i = (3.0 kg)(20 m/s) + (2.0 kg)( –12 m/s) = 36 kg·m/s. Their momentum immediately after, when they constitute a combined mass of M = 5.0 kg, is p f = (5.0 kg) v G . By conservation of momentum, then, we obtain v G = 7.2 m/s, which becomes their "initial" velocity for their subsequent free-fall motion. We can use Ch. 2 methods or energy methods to analyze this subsequent motion; we choose the latter. The level of their collision provides the reference ( y = 0) position for the gravitational potential energy, and we obtain K 0 + U 0 = K + U ¡ 1 2 Mv 2 0 + 0 = 0 + Mgy max . Thus, with v 0 = 7.2 m/s, we find y max = 2.6 m.

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1 2 m 2 v 2 2 = Δ E th = f k d = μ k m 2 g d . where k = 0.500. Solving for the sliding distance d , we find that m 2 cancels out and we obtain d = 2.22 m. (b) In a completely inelastic collision, we apply Eq. 9-53: v 2 = m 1 m 1 + m 2 v 1 i (where, as above, v 1 i = 2 gh ). Thus, in this case we have v 2 = 2 gh / 3. Now, Eq. 8-37 (using the total mass since the blocks are now joined together) leads to a sliding distance of 0.556 m d = (one-fourth of the part (a) answer). 68. (a) If the collision is perfectly elastic, then Eq. 9-68 applies v 2 = 2 m 1 m 1 + m 2 v 1 i = 2 m 1 m 1 + (2.00) m 1 2 gh = 2 3 2 gh 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 2 gh (where h = 2.50 m). Next, for block 2’s “rough slide” we use Eq. 8-37:
78. We use Eq. 9-88. Then rel 6090 kg ln 105 m/s (253 m/s) ln 108 m/s.

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Problem_Set_6_Solutions - 56 The total momentum immediately...

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