Acceleration Due to Gravity #3 - Questions and Answers

Question 1 A rocket is fired in deep space, where gravity is negligible. In the first second, it ejects 1 160 160 1 ​ of its mass as exhaust gas and has an acceleration of 16.0   m/s 2 16.0m/s 2 . What is the speed of the exhaust gas relative to the rocket? Answer From the thrust equation: 𝑀 𝑎 = 𝑣 𝑑 𝑀 𝑑 𝑡 Ma=v dt dM ​ where 𝑀 M is the rocket mass, 𝑎 a is acceleration, and 𝑣 v is the exhaust gas speed relative to the rocket: 𝑣 = 𝑀 𝑎 𝑑 𝑀 𝑑 𝑡 v= dt dM ​ Ma ​ Given: 𝑑 𝑀 𝑑 𝑡 = 𝑀 160 ,   𝑎 = 16.0   m/s 2 dt dM ​ = 160 M ​ ,a=16.0m/s 2 𝑣 = 𝑀 ⋅ 16.0 𝑀 160 = 2560   m/s . v= 160 M ​ M⋅16.0 ​ =2560m/s. Question 2 A golfer hits a golf ball at an angle of 2 5 ∘ 25 ∘ to the ground. If the golf ball covers a horizontal distance of 301.5   m 301.5m, what is the ball’s maximum height? Answer 1 The horizontal displacement: Δ 𝑥 = 𝑣 𝑥 𝑡 Δx=v x ​ t The maximum height is reached when 𝑣 𝑦 = 0 v y ​ =0: ℎ = 𝑣 0 𝑦 𝑡 − 1 2 𝑔 𝑡 2 h=v 0y ​ t− 2 1 ​ gt 2 From the horizontal motion, 𝑣 𝑥 = 𝑣 0 cos ⁡ 𝜃 , 𝑡 = Δ 𝑥 𝑣 𝑥 v x ​ =v 0 ​ cosθ,t= v x ​ Δx ​ 𝑡 = 301.5 𝑣 0 cos ⁡ ( 2 5 ∘ ) . t= v 0 ​ cos(25 ∘ ) 301.5 ​ . From vertical motion, 𝑣 𝑦 = 𝑣 0 sin ⁡ ( 2 5 ∘ ) − 𝑔 𝑡 ⇒ 0 = 𝑣 0 sin ⁡ ( 2 5 ∘ ) − 𝑔 𝑡 v y ​ =v 0 ​ sin(25 ∘ )−gt⇒0=v 0 ​ sin(25 ∘ )−gt 𝑡 = 𝑣 0 sin ⁡ ( 2 5 ∘ ) 𝑔 . t= g v 0 ​ sin(25 ∘ ) ​ . Equating 𝑡 t from both motions: 301.5 𝑣 0 cos ⁡ ( 2 5 ∘ ) = 𝑣 0 sin ⁡ ( 2 5 ∘ ) 𝑔 . v 0 ​ cos(25 ∘ ) 301.5 ​ = g v 0 ​ sin(25 ∘ ) ​ . Solve for 𝑣 0 v 0 ​ : 𝑣 0 2 = 301.5 ⋅ 𝑔 cos ⁡ ( 2 5 ∘ ) sin ⁡ ( 2 5 ∘ ) . v 0 2 ​ = cos(25 ∘ )sin(25 ∘ ) 301.5⋅g ​ . Substitute 𝑣 0 v 0 ​ into the height equation: ℎ = ( 𝑣 0 sin ⁡ ( 2 5 ∘ ) ) 2 2 𝑔 . h= 2g (v 0 ​ sin(25 ∘ )) 2 ​ . Answer 3 Step 1: Given data. Angle 𝜃 = 2 5 ∘ θ=25 ∘ , horizontal distance Δ 𝑥 = 301.5   m Δx=301.5m, and 𝑔 = 9.8   m/s 2 g=9.8m/s 2 . Step 2: Find 𝑣 0 v 0 ​ . From: Δ 𝑥 = 𝑣 0 2 sin ⁡ ( 2 𝜃 ) 𝑔 Δx= g v 0 2 ​ sin(2θ) ​ 𝑣 0 2 = Δ 𝑥 ⋅ 𝑔 sin ⁡ ( 2 𝜃 ) . v 0 2 ​ = sin(2θ) Δx⋅g ​ . Step 3: Calculate maximum height. ℎ = 𝑣 0 2 ( sin ⁡ 𝜃 ) 2 2 𝑔 . h= 2g v 0 2 ​ (sinθ) 2 ​ . Substitute values: ℎ = ( 301.5 ⋅ 9.8 ) ⋅ ( sin ⁡ 2 ( 2 5 ∘ ) ) 2 ⋅ 9.8 ⋅ cos ⁡ 2 ( 2 5 ∘ ) . h= 2⋅9.8⋅cos 2 (25 ∘ ) (301.5⋅9.8)⋅(sin 2 (25 ∘ )) ​ . ℎ = 29.82   m . h=29.82m. Question 3 The mass of Venus is 81.5 % 81.5% that of the Earth, and its radius is 94.9 % 94.9% that of the Earth. a) Compute the acceleration due to gravity on Venus. b) If a rock weighs 75.0   N 75.0N on Earth, what would it weigh on Venus? Answer a) Acceleration due to gravity on Venus: 𝑔 𝑉 = 𝐺 𝑀 𝑉 𝑅 𝑉 2 . g V ​ =G R V 2 ​ M V ​ ​ . Using: 𝑀 𝑉 = 0.815 𝑀 𝐸 ,   𝑅 𝑉 = 0.949 𝑅 𝐸 ,   𝑔 𝐸 = 𝐺 𝑀 𝐸 𝑅 𝐸 2 . M V ​ =0.815M E ​ ,R V ​ =0.949R E ​ ,g E ​ =G R E 2 ​ M E ​ ​ . 𝑔 𝑉 = 𝑔 𝐸 0.815 ( 0.949 ) 2 . g V ​ =g E ​ (0.949) 2 0.815 ​ . 𝑔 𝑉 = 9.8 0.815 0.9006 = 8.86   m/s 2 . g V ​ =9.8 0.9006 0.815 ​ =8.86m/s 2 . b) Weight on Venus: 𝑊 𝑉 = 𝑚 𝑔 𝑉 ,   𝑚 = 75.0 9.8 . W V ​ =mg V ​ ,m= 9.8 75.0 ​ . 𝑊 𝑉 = 75.0 9.8 ⋅ 8.86 = 67.8   N . W V ​ = 9.8 75.0 ​ ⋅8.86=67.8N. Question 4 A basketball player accelerates uniformly from rest to 6.0   m/s 6.0m/s in 1.5   s 1.5s. Find the distance. Answer Using: 𝑥 = 𝑣 0 𝑡 + 1 2 𝑎 𝑡 2 . x=v 0 ​ t+ 2 1 ​ at 2 . Find acceleration: 𝑎 = 𝑣 − 𝑣 0 𝑡 = 6.0 − 0 1.5 = 4.0   m/s 2 . a= t v−v 0 ​ ​ = 1.5 6.0−0 ​ =4.0m/s 2 . 𝑥 = 0 + 1 2 ( 4.0 ) ( 1.5 ) 2 = 4.5   m . x=0+ 2 1 ​ (4.0)(1.5) 2 =4.5m. Question 5 A stuntman jumps horizontally at 5.0   m/s 5.0m/s from a 1 5 ∘ 15 ∘ incline. Will he clear a gap 4.0   m 4.0m wide to a roof 2.5   m 2.5m shorter? Answer Horizontal motion: 𝑥 = 𝑣 𝑥 𝑡 ,   𝑣 𝑥 = 𝑣 0 cos ⁡ ( 1 5 ∘ ) . x=v x ​ t,v x ​ =v 0 ​ cos(15 ∘ ). Vertical motion: 𝑦 = 𝑣 𝑦 𝑡 − 1 2 𝑔 𝑡 2 ,   𝑣 𝑦 = 𝑣 0 sin ⁡ ( 1 5 ∘ ) . y=v y ​ t− 2 1 ​ gt 2 ,v y ​ =v 0 ​ sin(15 ∘ ). Solve 𝑡 t when 𝑦 = − 2.5   m y=−2.5m: − 2.5 = 5.0 sin ⁡ ( 1 5 ∘ ) 𝑡 − 4.9 𝑡 2 . −2.5=5.0sin(15 ∘ )t−4.9t 2 . Find 𝑡 t, then check 𝑥 = 5.0 cos ⁡ ( 1 5 ∘ ) 𝑡 x=5.0cos(15 ∘ )t: 𝑥 = 4.11   m > 4.0   m . x=4.11m>4.0m. He clears the gap safely.