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HW6: Momentum and Energy Due 11/2/2021 1) Describe how Conservation of Momentum leads relates to Newton’s 3rd law of motion. 2) Two objects are at rest on a frictionless surface. The first object has mass m1 and the second object has mass m2. The two objects interact with one another for a brief moment and then move in opposite directions. a) Compare the velocity of the two objects after they interact with one another when: i) m1 = m2 ii) m1 > m2 b) Compare the momentum of the system before and after the objects interact when: i) m1 = m2 ii) m1 > m2 3) Use the Conservation of Momentum to answer the following questions: a) A 4 kg object traveling with a velocity of 5 m/s strikes a 6 kg object at rest. The 6 kg object then begins to move with a velocity of 3 m/s. What is the velocity of the 4 kg object after the collision? b) A 7 kg object travels to the right with a speed of 4 m/s. Another 2 kg object travels to the left with a speed of 14 m/s. Find the velocities of both objects after they collide. c) A 4 kg object traveling towards the right at 3 m/s comes into contact with an object at rest. The objects stick together and travel towards the right at 1 m/s. What was the mass of the object at rest? 4) A 3kg object moving towards the right at 2 m/s strikes a 6 kg object at rest. The 6 kg object begins to move to the right at 8 m/s and down at 2 m/s. a) What is the net momentum in the left-right direction before the collision? b) What is the net momentum in the up-down direction before the collision? c) What is momentum of the 3kg object in the left-right direction after the collision? Explain. d) What is momentum of the 3kg object in the up-down direction after the collision? Explain. 5) Using Newton’s 3rd Law and the Conservation of Momentum, explain how a rocket ship can travel through space. 6) An object travels with a speed of 4 m/s toward a frictionless incline. a) How high up the incline will the object travel? b) If the angle of the incline is doubled (so the incline is steeper), how will this affect how high up the incline the object travels? Explain. 7) At the top of a 3m high frictionless ramp, an object of 3kg is moving at a speed of 2 m/s. a) What is the gravitational potential energy of the object? b) How fast will it be traveling when it reaches the bottom of the ramp? c) How fast will the object be traveling when it is halfway down the ramp (i.e., at a height of 1.5 m above the ground)? d) Did the mass of the object matter? Explain why or why not. HW6: Momentum and Energy 1) When two objects collide, their total momentum is conserved. This leads to one object gaining momentum (speeding up), and the other losing momentum (slowing down). This is due to Newton’s Third Law of Motion, which states that for each action, there is an equal and opposite reaction. In our case, when the two objects collide, each of them exerts a force on the other. Those forces are of equal magnitude and of opposite direction, according to Newton’s Third Law. It’s those forces that cause the colliding objects to either slow down or speed up, or even change directions. The system that is made up of the two objects has a conserved momentum, because no external forces are acting upon it. 2) a. i. If 𝑚1 = 𝑚2, then, after they interact with each other, they will swap velocities. This is because: ⃗⃗⃗⃗ 𝑃0 = 𝑃⃗ ⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ 𝑚1 ∗ 𝑉 𝑖1 + 𝑚2 ∗ 𝑉𝑖2 = 𝑚1 ∗ 𝑉𝑓1 + 𝑚2 ∗ 𝑉𝑓2 Since the masses m1 and m2 are equal, and since the objects are moving in 1 dimension: 𝑉𝑖1 + 𝑉𝑖2 = 𝑉𝑓1 + 𝑉𝑓2 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1: 𝑉𝑖1 − 𝑉𝑓1 = 𝑉𝑓2 − 𝑉𝑖2 Also, since the collision is elastic (they don’t stick together), we have that: 1 1 1 1 2 2 𝑚1 ∗ 𝑉𝑖12 + 𝑚2 ∗ 𝑉𝑖22 = 𝑚1 ∗ 𝑉𝑓1 + 𝑚2 ∗ 𝑉𝑓2 2 2 2 2 2 2 ⇒ 𝑉𝑖12 + 𝑉𝑖22 = 𝑉𝑓1 + 𝑉𝑓2 2 2 ⇒ 𝑉𝑖12 − 𝑉𝑓1 = 𝑉𝑓2 − 𝑉𝑖22 ⇒ (𝑉𝑖1 − 𝑉𝑓1 )(𝑉𝑖1 + 𝑉𝑓1 ) = (𝑉𝑓2 − 𝑉𝑖2 )(𝑉𝑓2 + 𝑉𝑖2 ) According to Equation 1, we can simplify 𝑉𝑖1 − 𝑉𝑓1and 𝑉𝑓2 − 𝑉𝑖2on each side, since they are equal and not 0. 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2: (𝑉𝑖1 + 𝑉𝑓1 ) = (𝑉𝑓2 + 𝑉𝑖2 ) Adding Equation 1 and Equation 2, we obtain: 2 ∗ 𝑉𝑖1 = 2 ∗ 𝑉𝑓2 ⇒ 𝑉𝑖1 = 𝑉𝑓2 Subtracting Equation 1 from Equation 2, we obtain: 2 ∗ 𝑉𝑓1 = 2 ∗ 𝑉𝑖2 ⇒ 𝑉𝑓1 = 𝑉𝑖2 This goes to show that when two objects of equal masses collide without sticking together, they swap masses. ii. m1 > m2 Conservation of Momentum: ⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ 𝑚1 ∗ 𝑉 𝑖1 + 𝑚2 ∗ 𝑉𝑖2 = 𝑚1 ∗ 𝑉𝑓1 + 𝑚2 ∗ 𝑉𝑓2 𝑚1 (𝑉𝑖1 − 𝑉𝑓1 ) = 𝑚2 (𝑉𝑓2 − 𝑉𝑖2 ) 𝑚1 (𝑉𝑓2 − 𝑉𝑖2 ) = 𝑚2 (𝑉𝑖1 − 𝑉𝑓1 ) (𝑉 −𝑉 ) Since m1>m2, (𝑉 𝑓2− 𝑉 𝑖2 ) > 0 𝑖1 𝑓1 (𝑉𝑓2 − 𝑉𝑖2 ) > (𝑉𝑖1 − 𝑉𝑓1 ) This means that the change in absolute velocity of the object of mass m2 will be greater than the change of the absolute velocity of the object of mass m1, in all cases, if m1>m2. b. i. The momentum of the system is always conserved, which means that it is the same before and after the collision of the objects. The two objects ‘swap’ the absolute value of their momentums before and after collision, since they have the same mass. The momentum of the system is therefore conserved. ii. The proportion of change of the velocity of the object of smaller mass to the change of velocity to the object of larger mass is equal to the ratio of both masses (proven in part a. ii.), which means that the final momentum is still equal to the initial momentum of the system. The momentum of the system is still conserved, even when one object has a larger mass than the other. 3) a. Let m1 = 4kg, Vi1 = 5m/s, m2 = 6kg, Vi2 = 0m/s, Vf2 = 3m/s Conservation of Momentum: ⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ 𝑚1 ∗ 𝑉 𝑖1 + 𝑚2 ∗ 𝑉𝑖2 = 𝑚1 ∗ 𝑉𝑓1 + 𝑚2 ∗ 𝑉𝑓2 4 ∗ 5𝑖⃗ + 𝑚2 ∗ ⃗⃗0⃗ = 4 ∗ 𝑉𝑓1⃗𝑖 + 6 ∗ 3 ∗ 𝑖⃗ 20 = 4 ∗ 𝑉𝑓1 + 18 𝑉𝑓1 = 20 − 18 1 = 𝑚/𝑠 4 2 b. Let m1 = 7kg, Vi1 = 4m/s, m2= 2kg, Vi2 = 14m/s. Conservation of Momentum: ⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ 𝑚1 ∗ 𝑉 𝑖1 + 𝑚2 ∗ 𝑉𝑖2 = 𝑚1 ∗ 𝑉𝑓1 + 𝑚2 ∗ 𝑉𝑓2 𝑖 7 ∗ 4𝑖⃗ − 2 ∗ 14𝑖⃗ = 7 ∗ 𝑉𝑓1 ∗ 𝑖⃗ + 2 ∗ 𝑉𝑓2 𝑖⃗ 2 7 ∗ 𝑉𝑓1 = − 2 ∗ 𝑉𝑓2 ⇒ 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1: 𝑉𝑓1 = − 7 𝑉𝑓2 Conservation of Kinetic Energy: 1 1 1 1 2 2 𝑚1 ∗ 𝑉𝑖12 + 𝑚2 ∗ 𝑉𝑖22 = 𝑚1 ∗ 𝑉𝑓1 + 𝑚2 ∗ 𝑉𝑓2 2 2 2 2 2 2 7 ∗ 42 + 2 ∗ 142 = 7 ∗ 𝑉𝑓1 + 2 ∗ 𝑉𝑓2 Replacing Vf1 in the above equation with its value found in Equation 1, we get: 2 2 2 504 = 7 ∗ (− 𝑉𝑓2 ) + 2 ∗ 𝑉𝑓2 7 18 2 𝑉 = 504 7 𝑓2 2 𝑉𝑓2 = 196 𝑉𝑓2 = √196 = +14, since the 2kg object moves to the right after collision 2 2 From Equation 1, we can find Vf1: 𝑉𝑓1 = − 7 𝑉𝑓2 = − 7 ∗ 14 = −4 𝑚/𝑠 This means that the 7 kg object is moving at a velocity of 4m/s to the left after collision. c. Let m1 = 4kg, Vi1=3m/s, and Vf = 1m/s. Conservation of Momentum: ⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗ 𝑚1 ∗ 𝑉 𝑖1 + 𝑚2 ∗ 𝑉𝑖2 = (𝑚1 + 𝑚2 ) ∗ 𝑉𝑓 4 ∗ 3 𝑖 + 𝑚2 ∗ ⃗⃗0⃗ = (4 + 𝑚2 ) ∗ 1 ∗ 𝑖⃗ 12 = 4 + 𝑚2 𝑗 𝑚2 = 8𝑘𝑔 4) Let m1=3kg, Vi1 = 2m/s, m2=6kg, Vf2x = 8m/s, Vf2y = 2m/s. a. Before collision, in the left-right direction: ⃗⃗⃗⃗⃗⃗⃗⃗ ⃗⃗⃗ 𝑝0𝑥 = 𝑚1 ∗ 𝑉 𝑖1𝑥 + 𝑚2 ∗ 0 = 3 ∗ 2𝑖 = 6𝑖 b. Before collision, in the up-down direction: ⃗⃗⃗ = 0𝑗 𝑝0𝑦 = 𝑚1 ∗ ⃗⃗⃗⃗⃗⃗⃗⃗ 𝑉𝑖1𝑦 + 𝑚2 ∗ 0 c. After collision, in the left-right direction: ⃗⃗⃗⃗⃗⃗⃗⃗ 𝑝𝑥 = 𝑚1 ∗ ⃗⃗⃗⃗⃗⃗⃗⃗⃗ 𝑉𝑓1𝑥 + 𝑚2 ∗ 𝑉 𝑓2𝑥 = 6𝑖 3 ∗ 𝑉𝑓1𝑥 ∗ 𝑖⃗ + 6 ∗ 8 ∗ 𝑖 = 6𝑖 6− 6∗8 𝑉𝑓1𝑥 = 𝑖 = −14𝑚/𝑠 3 This means that the 3kg object starts moving to the left, at 14m/s after collision. ⃗⃗⃗⃗⃗⃗⃗⃗ Its momentum in the left-right direction is 𝑚1 ∗ 𝑉 𝑖1𝑥 = −14 ∗ 3𝑖 = −42𝑖 d. After collision, in the up-down direction: 𝑝0𝑦 = 𝑚1 ∗ ⃗⃗⃗⃗⃗⃗⃗⃗⃗ 𝑉𝑓1𝑦 + 𝑚2 ∗ ⃗⃗⃗⃗⃗⃗⃗⃗⃗ 𝑉𝑓2𝑦 3 ∗ 𝑉𝑓1𝑦 ∗ 𝑗⃗ + 6 ∗ (−2) ∗ 𝑗 = 0𝑗 𝑖 12 = 4𝑚/𝑠 3 This means that the 3kg object starts moving up at 4m/s after collision. Its momentum in the up-down direction is: 𝑚1 ∗ ⃗⃗⃗⃗⃗⃗⃗⃗ 𝑉𝑖1𝑦 = 3 ∗ 4𝑗 = 12𝑗 5) According to Newton’s Third Law, every action has an equal and opposite reaction. This means that in order for the rocket ship to travel through space, there should be something pushing it forward. The acceleration of the oxidized fuel molecules that emerge from the nozzle of the rocket are what push it forward to travel through space. 6) Since the object is traveling on a frictionless incline, the mechanical energy will be conserved. At its highest point on the incline, the object’s velocity will be zero. a. 𝐸0 = 𝐸 𝐾0 + 𝑈0 = 𝐾 + 𝑈 1 𝑚 ∗ 𝑣 2 + 0 = 0 + 𝑚𝑔ℎ 2 1 𝑔ℎ = 𝑣 2 2 1 2 ℎ= 𝑣 2𝑔 b. The angle of incline does not affect the height at which the object will travel, because the formula for the height that we found in part a. doesn’t take the angle of the incline into consideration. The object will still travel for a height of 1 2 ℎ= 𝑣 2𝑔 𝑉𝑓1𝑦 = However, the distance traveled by the object will be less than the distance traveled with half the angle. 7) a. 𝑈0 = 𝑚𝑔ℎ 𝑈0 = 3 ∗ 9.8 ∗ 3 𝑈0 = 88.2𝐽 b. Since the ramp is frictionless, the mechanical energy will be conserved. At the bottom of the ramp, the gravitational potential energy is zero. 𝐾0 + 𝑈0 = 𝐾 + 𝑈 1 𝐾 = 𝑚𝑣0 2 + 𝑈0 2 1 𝑚𝑣 2 = 0.5 ∗ 3 ∗ 22 + 88.2 2 0.5 ∗ 3 ∗ 𝑣 2 = 94.2 94.2 𝑣2 = = 62.8 1.5 𝑣 = √62.8 = 7.92𝑚/𝑠 c. We follow the same procedure as before. 𝐾0 + 𝑈0 = 𝐾 + 𝑈 1 1 𝑚𝑔ℎ 𝑚𝑣0 2 + 𝑚𝑔ℎ = 𝑚𝑣 2 + 2 2 2 1 2 1 2 𝑔ℎ 𝑣 + 𝑔ℎ = 𝑣 + 2 0 2 2 1 2 1 2 𝑔ℎ 1 2 𝑔ℎ 𝑣 = 𝑣0 + 𝑔ℎ − = 𝑣0 + 2 2 2 2 2 𝑣 2 = 𝑣0 2 + 𝑔ℎ 𝑣 2 = 22 + 9.8 ∗ 3 = 33.4 𝑣 = √33.4 = 5.78𝑚/𝑠 d. The mass of the object did not matter, because in the equation of conservation of mechanical energy, we were able to simplify the mass of the object in order to find its velocity at a given point. We were able to write the equation in this form, without using mass: 1 2 1 𝑔ℎ 𝑣0 + 𝑔ℎ = 𝑣 2 + 2 2 2
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