【CH09】Center of Mass and Momentum Homework of ... - 物理學系

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Fundamentals of Physics Halliday & Resnic −3 質量中心在 t = 300× 10 s時的速度為 −3 −3 −3 mv 1 1+ mv 2 2 m1(9.8)(300× 10 ) + 2 m1(9.8)(300× 10 − 100× 10 ) vcm = = m1+ m2 m1+ 2m1 = 2.3 m/ s 東海大學物理系【Problem 9-13】 Figure 9-44 shows an arrangement with an air track, in which a cart is connected by a cord to a hanging block. The cart has mass m1= 0.6kg, and its center is initially at xy coordinates (-0.5m, 0 m); the block has mass m2= 0.4kg, and its center is initially at xy coordinates (0, -0.1m). The mass of the cord and pulley are negligible. The cart is released from rest, and both cart and block move until the cart hits the pulley. The friction between the cart and the air track and between the pulley and its axle is negligible. (a) In unit-vector notation, what is the acceleration of the center of mass of the cart–block system? (b) What is the velocity of the com as a function of time t? (c) Sketch the path taken by the com. (d) If the path is curved, determine whether it bulges upward to the right or downward to the left, and if it is straight, find the angle between it and the x axis. 圖 9-44,滑車以一條繩子和積木連接,滑車質量 m1= 0.6kg,一開始滑車中心位置為(-0.5m, 0 m),積木質量 m2= 0.4kg,一開始積木中心位置為(0, -0.1m)。假設繩子和滑輪無質量。滑車從靜止釋放,滑車和積木滑動,直到滑車碰到滑輪,滑車和氣軌無摩擦,(a)滑車、積木系統的質量中心,(b)以時間 t 為函數寫出質量中心的速度,(c)畫出質量中心的運動軌跡, (d)如果軌跡是一條曲線,那曲線是向上突起或向下凹?如果軌跡是一條直線,那它和 x 軸的夾角為何? (圖 9-44) mg ( m m) a mg 2 (0.4)(9.8) 2 a = = = 3.92 m/ s m1+ m2 0.6 + 0.4 ma 1 1+ ma 2 2 (0.6)(3.92 i) + (0.4)( −3.92 j) a 2.35 1.57 cm = = = i− j m1+ m2 0.6 + 0.4 (b) v (2.35 1.57 cm = acmt = i− j) t (c) The last sentence of our answer for part (b) implies that the path of the center-of-mass is a straight line. :(a) 2 = 1+ 2 Fundamentals of Physics Halliday & Resnic (d) −1 −1.57 0 tan =− 33.4 2.35 【Problem 9-15】 A shell is shot with an initial velocity 0 v of 20 m/s, at an angle of 東海大學物理系 0 θ 0 = 60 with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass (Fig. 9-46). One fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragment land, assuming that the terrain is level and that air drag is negligible? 0 一顆子彈,以 v0= 20 m/ s,水平夾角 θ 0 = 60 的初速射出,在射程頂端,子彈分成 2 塊相同質量的碎片,如圖 9-46,爆破後,其中一塊碎片速度立刻變 0,直接往下垂直掉落。問另一片碎片離發射的槍多遠?假設地形是水平,空氣阻力忽略不計。 v = v − gt = v sinθ − gt : y 0y 0 0 v0sinθ 0 v y = 0 時, t = g (圖 9-46) v 20 x = v t = v t = = = m 2 2 0x( 0cos θ0) 0 sinθ0cosθ0 g 0 0 sin 60 cos 60 2 9.8 17.7 0 y 1 2 2 2 v0 sin θ0 2 2 0 (20 )(sin 60 ) 15.3 y = v t− gt 2 = 2 g = (2)(9.8) = m m mv cosθ = V 2 0 V0 = 2v0cosθ0 = 2(20) cos 60 = 20 m/ s 在最高點時, x = 17.7m , y = 15.3m。 1 2 y = y0− gt 2 y = 0時, t = 2y0 g x = x0 + V0t = x0 + V0 2y0 = 17.7 + (20) g 2(15.3) = 53m 9.8 後半段分裂成兩段,因為動量守恆,所以 0 0 0 【Problem 9-16】 Ricardo, of mass 80 kg, and Carmelita, who is lighter, are enjoying Lake Merced at dusk in a 30 kg canoe. When the canoe is at rest in the placid water, they exchange seats, which are 3 m apart and symmetrically located with respect to the canoe’s center. If the canoe moves 40 cm horizontally
Fundamentals of Physics Halliday & Resnic relative to a pier post, what is Carmelita’s mass? 東海大學物理系里卡多(質量 80 公斤)和卡梅莉塔(比較輕),都喜歡在黃昏默塞德湖在 30 公斤獨木舟。當獨木舟是靜止在平靜的水面上,他們交換座位,此時相隔 3 米,位於對稱方面的獨木舟的中心。如果獨木舟水平移動 40cm 相對於一個碼頭後,卡梅莉塔的質量為何? :Ricardo 質量: M R Carmelita 質量: M C L L M R( − x) = mx+ MC( + x) 2 2 40 x = = 20cm = 0.2m 2 L 3 MR( −x) −mx 80( −0.2) −(30)(0.2) M 2 2 C = = = 58kg L 3 + x + 0.2 2 2 【Problem 9-21】 0 A 0.3 kg softball has a velocity of 15 m/s at an angle of 35 below the horizontal just before making contact with the bat. What is the magnitude of the change in momentum of the ball while in contact with the bat if the ball leaves with a velocity of (a) 20m/s, vertical downward, and (b) 20m/s, horizontally back toward the pitcher? 一 0.3 公斤的壘球以速度 15m/s 水平向下 0 35 接觸球棒。問壘球碰觸到球棒時,壘球動量變化為何?如果球離開的速度(a)20m/s,垂直向下,和(b)20 米/秒,水平朝向投手? 0 0 0 :一開始動量 p 0 = (0.3 kg)(15 m / s)( ∠ 215 ) = 4.5cos 215 i + 4.5sin 215 j (a)20m/s垂直向下 0 0 0 動量: p = (0.3 kg)(20 m / s)( ∠− 90) = 6cos( − 90) i + 6sin( −90) j f −1 −3.42 0 p f − p0= 3.69 i−3.42 j( 動量變化: 5 kg ⋅ m / s (b) 20m/s 水平朝向投手 2 2 (3.69) + (3.42) = 5 , tan 3.69 = − 43 ) 動量: p 0 = (0.3 kg)(20 m / s)( ∠ 0 ) = 6 i + 0j f −1 −2.58 0 p f − p0= 9.68 i−2.58 j( 動量變化:10 kg ⋅ m / s 2 2 (9.68) + (2.58) = 10 , tan 9.68 = − 15 ) 【Problem 9-27】 A 1.2 kg ball drops vertically onto a floor, hitting with a speed of 25 m/s. It rebounds with an initial speed of 10 m/s. (a) What impulse acts on the ball during the contact? (b) If the ball is in contact with the floor for 0.02 s, what is the magnitude of the average force on the floor from the ball? Fundamentals of Physics Halliday & Resnic 東海大學物理系一顆球 1.2 公斤,垂直掉落到樓層板上,撞擊時速度為 25m/s,反彈後最初速度為 10m/s(a) 球接觸時,衝量為何?(b)如果球在接觸地面時間為 0.02 秒,球作用在樓層版上的平均力大小為何? :初速: v 25 / i =− m sj 末速: v 10 / f = m sj (a) J = mv f − mvi = (1.2)(10) −(1.2)( − 2.5) = 42 kg ⋅m/ s J 42 3 (b) Favg = = = 2.1× 10 N Δt 0.02 【Problem 9-31】 Figure 9-52 shows a 0.3kg baseball just before and just after it collides with a bat. Just before, the ball has velocity v1 0 of magnitude 12 m/s and angleθ 1 = 35 . Just after, it is traveling directly upward with velocity 2 v of magnitude 10.0 m/s. The duration of the collision is 2ms. What are the (a) magnitude and (b) direction (relative to the positive direction of the x axis) of the impulse on the ball from the bat? What are the (c) magnitude and (d) direction of the average force on the ball from the bat? 圖 9-52 顯示了一個 0.3 公斤的棒球被球棒打到的前後,一開始球速為 v1 ,大小為 12m/s,角 0 度 θ 1 = 35 。被球棒打到後,筆直以 v2 向上飛出, 大小為 10m/s。球棒接觸到球的時間為 2ms。問衝量的(a)大小(b)方向,還有平均施力的(c)大小(d)方向? (圖 9-52) 0 0 0 : p (0.3)(12)( 215 ) 3.6cos 215 3.6sin 215 i = ∠ = i+ j 0 0 0 p (0.3)(10)( 90 ) 3cos90 3sin 90 f = ∠ = i+ j J =Δ p= p 2.95 5.06 f − pi = 2 2 −15.06 0 i+ j( (2.95) + (5.06) = 5.85 , tan = 59.8 ) 2.95 (a) J = Δ p= 5.86 kg⋅ m/ s = 5.86N⋅ s (b) +x 軸夾 59.8 度 5.86N⋅s 3 (c) J = FavgΔ t = 5.86N⋅ s ⇒ Favg = ≈ 2.93× 10 N −3 2.00× 10 s (d) +x 軸夾 59.8 度
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【CH09】Center of Mass and Momentum Homework of ... - 物理學系

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