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HW6-phy2048-Spring-2014-solution

# 010 s thus the average force is found to be 100 n c

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Unformatted text preview: The time of landing is t = 2 y0 / g and the x coordinate of the landing point is 2 (15.3 m ) 2 y0 x = x0 + V0t = x0 + V0 = 17.7 m + ( 20 m/s ) = 53 m. g 9.8 m/s 2 9P22. (a) Since the force of impact on the ball is in the y direction, px is conserved: pxi = pxf ⇒ mvi sin θ1 = mvi sin θ 2 . With θ1 = 30.0°, we find θ2 = 30.0°. (b) The momentum change is r Δp = mvi cos θ 2 −ˆ − mvi cos θ 2 +ˆ = −2 ( 0.165 kg ) ( 2.00 m/s ) ( cos 30° ) ˆ j j j () () ˆ = (−0.572 kg ⋅ m/s)j. 9P30. (a) By Eq. 9- 30, impulse can be determined from the “area” under the F(t) curve. Keeping in mind that the area of a triangle is 1 (base)(height)...
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