HW6 Solutions

# 05 kg 25 kg f m v v m m 5 a if the collision is

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0.5 kg 2.5 kg f m v v m m 5. (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:

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1 2 m 2 v 2 2 = E th = f k d = 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). 6. The wheel has angular velocity 0 = +1.5 rad/s = +0.239 rev/s at t = 0, and has constant value of angular acceleration < 0, which indicates our choice for positive sense of rotation. At t 1 its angular displacement (relative to its orientation at t = 0) is 1 = +20 rev, and at t 2 its angular displacement is 2 = +40 rev and its angular velocity is 2 0 . (a) We obtain t 2 using Eq. 10-15: 2 0 2 2 2 1 2(40 rev) 335 s 2 0.239 rev/s t t which we round off to 2 2 3.4 10 s t . (b) Any equation in Table 10-1 involving can be used to find the angular acceleration; we select Eq. 10-16. 2 4 2 2 2 2 2 2 1 2(40 rev) 7.12 10 rev/s 2 (335 s) t t     which we convert to = 4.5 10 3 rad/s 2 . (c) Using 1 0 1 1 2 1 2 t t (Eq. 10-13) and the quadratic formula, we have 2 2 4 2 0 0 1 1 4 2 2 (0.239 rev/s) (0.239 rev/s) 2(20 rev)( 7.12 10 rev/s ) 7.12 10 rev/s t   which yields two positive roots: 98 s and 572 s. Since the question makes sense only if t 1 < t 2 we conclude the correct result is t 1 = 98 s. 7. If we make the units explicit, the function is 3 3 2 2 ) / 0 . 1 ( ) / 0 . 3 ( ) / 0 . 4 ( t s rad t s rad t s rad
but generally we will proceed as shown in the problem letting these units be understood. Also, in our manipulations we will generally not display the coefficients with their proper number of significant figures. (a) Equation 10-6 leads to . 3 6 4 ) 3 4 ( 2 3 2 t t t t t dt d Evaluating this at t = 2 s yields 2 = 4.0 rad/s. (b) Evaluating the expression in part (a) at t = 4 s gives 4 = 28 rad/s. (c) Consequently, Eq. 10-7 gives avg 2 rad / s 4 2 4 2 12 . (d) And Eq. 10-8 gives . 6 6 ) 3 6 4 ( 2 t t t dt d dt d Evaluating this at t = 2 s produces 2 = 6.0 rad/s 2 . (e) Evaluating the expression in part (d) at t = 4 s yields 4 = 18 rad/s 2 . We note that our answer for avg does turn out to be the arithmetic average of 2 and 4 but point out that this will not always be the case.

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