Peter Nguyen Physics homework #4: Chapter 5 Think and Solve: 27, 28, 29, 30 27) π= Δπ£ Time Interval -> = 25π/π 0.05π = 500π/π 2 So, force= Mass x Acceleration = 0.003πππ₯500π/π 2 = 1.5π 28) π= ππππ π πππ΅ -> π = ππππ = π. π π/ππ 29) ππ πππ ππ πππ πππ πππππππ ππππππ ππππππ πππ ππππ ππ πππ πππππππππ πππππ ππππ = √(ππ )π + (ππ )π -> ππππ = √(π π΅)π + (π π΅)π = π π΅ ππππ ππ΅ ππππ = ππ → π = →π= = π. π π/ππ π πππ 30a) mass of two pucks =m ππ = π π ππ ππ, ππ = π ππ ππ π π1 = π2 (2π)π2π = (π)ππ Δv2m (2π)π2π = (π)ππ → (2π) ( Δt ) = (π)( Δvm Δt ) 1 Δv2m = 2 Δvm 30b) mass of the single puck is: π£π = 0.4 π/π 1 1 speed of the double mass puck is: Δv2m = 2 Δvm = 2 (0.4 π/π ) = 0.2 π/π Think and Rank: 31, 32, 33 31) In the case of all three (a, b and c), the van is moving with constant velocity. So, the acceleration of the van in all three cases is zero. According to Newton’s second law of motion, the force exerted by the Van on the trailer is directly proportional to the acceleration of the force. Therefore, the force exerted by the van on the trailer in all three cases is zero. A=B=C 32a) Mass of each block= π, πΉ = πππ΄ πΉ πΉ πΉ ππ΄ = π , ππ΅ = 2π , πΉ = (π + π + π) = ππΆ = 3πππΆ , ππΆ = 3π Acceleration highest to least: A> B> C 32b) πΉ πΉ πΉ 2 π π = πππ΅ → π = π 2π = π = πππ → π = π 3π = 3 Ranking of tension in the rope connected to single box= B, C 33a) πΉ= ππ ππ‘ βΆ π = π‘βπ ππππππ‘π’π πππ π‘ ππ π‘βπ π‘πππ Given that frictional force due to air and ice on the three identical pucks are negligible, there is no dragging force acting on the three pucks. The speed of the motion and direction of the pucks will not change until there is no external force acting on them. There is no external force acting on the pucks, so the three pucks moving with their speeds in the same direction. The forces acting on the three pucks is 0 N therefore the forces acting on three pucks are equal. 33b) Expression is momentum= π = ππ£ ππ‘ →πΉ= π(ππ£) ππ‘ The puck which has the greater speed will require more time to stop. In the diagram, puck C has greater speed than Pucks A, B and puck B has greater speed than puck A. Pucks by the force needed to stop them is C>B>A Think and Explain: 37, 38, 45, 47, 53, 58, 62, 63 37a) In figure 1, it’s shown that when we hold an apple over our head then gravitational pull acts on it. This action force is the form of the weight of the apple pushing our head in the downward direction (ππ)The reaction force is the push extended by our head on the apple in the upward direction=R. These two actions and reaction forces act at the point where the apple is in contact with our head and are equal in magnitude and opposite in direct so that the apple rests on our head. 37b) When the same apple falls through air, its weight force is then pulled in a downward direction. This is the action force – mg weight of the apple. The reaction force here is the pull of the apple on the earth. The earth will accelerate upward under this reaction but due to its large mass this acceleration is not noticeable in upward direction, The other force is air resistance which opposes the fall of the apple given by π π in reaction to the air resistance the ball pushes the air molecules aside. 38a) As a person steps off a curb, he pushes the curb in a downward direction. This downward push on the curb is action force. The curb pushes the person with equal force in upward direction. This upward push on the person is the reaction force. 38b) When a person pats the back of another person, their hand exerts some force on the back of the other person. This force exerted by the hand on the back is the action force. As a result, the second person’s back exerts an equal and opposite force on the first person’s hand. The force exerted by back on the hand is the reaction force. 38c) When a wave hits a rocky shore, the waves exert force on the shore in direction of motion of waves. The force exerted on the shore is action force. The rocky shore exerts an equal force in backwards direction. This force acting on the waves is the reaction force. 45) π€ = ππ m=mass of the person and g is the acceleration due to gravity Yes, the normal, the normal force on the person and the force of gravity on the person are equal in magnitude and opposite in direction. When the person is in mechanical equilibrium then the two forces are cancelling to each other. In the figure, n is the normal force acting on the person by the floor which is pointed along the upward direction. The force gravity acting on the person is directed aling the downward direction. Since the person has no acceleration in upward or downward direction, the value of net forces on the standstill person is zero. Second law of motion applied -> πΉ = π + ππ → 0 = π + ππ → π = −ππ Weight of the body and normal and normal reaction of the floor are equal and opposite. No, weight and normal reaction are not an action reaction pair. Because according to Newton’s third law of motion, the action and reaction force always act on different objects. Thus, it involves interaction of two objects. But in the given example, both weight and normal reaction force are acting on the same object. Therefore, they do not comprise action and reaction as action and reaction act on the different bodies. 47) πΉπ = ππ , The person pulls the rope which has free end below his ties to move in a upward direction. The person applies enough pull on the lower part of the rope, and this pull force is equal to the tension in the rope. The tension is the rope is balanced by the weight or force of the gravity of the person. Therefore, the rope climber pulls downward on the rope to move upward by balancing the tension force in the rope with his weight or force of gravity. 53) Push on the ground would make you win the game. If team B pushes the ground with their feet, then the ground in reaction pushes them back with an equal force. If team B applies their force on the ground in a direction slanted forward, then the reactional force would be in a slanted backward direction. This would further help team B pull harder on the rope and ultimately allow them to win. 58) The relative velocity of an object is the velocity of the object measured by an observer in another moving frame. Rain falling vertically makes slanted streaks on the sides of windows of a moving car because o the observed relative velocity of the rain from the car. Thus, the rain falling vertically makes slanted streaks because of relative speed between rain and car. πππ‘ π£ππππ ππ π ππππ ππ π‘βπ ππππ, πππ π£πππ π ππππ ππ π‘βπ πππ, π‘ππ0 = πππππ ππππ → π‘ππ45° = πππππ ππππ → ππΆππ = πππππ speed of car is equal to speed of rain 62a) Figure 1 shows a stone of weight W suspended by a string. The stone is under influence of two forces, weight of the stone W acting downwards and tension in the string T acting upwards. Weight of the stone is shown by vector W and tension in the string is shown by vector T. Vectors are drawn to their relative length and their respective direction. 62b) According to Newton’s first law of motion, an object is in mechanical equilibrium only if net force acting on it is zero. The stone is in mechanical equilibrium as it is not moving in upwards or downwards direction. It implies that net force on the stone is zero. πΉ = 0 → π + (−π) + 0 → π = π Here, weight W is taken negative to account for downwards direction. Tension in the string is equal in magnitude to weight of the stone but opposite in direction because that net force acting on the stone is zero. 63a) The forces acting on the stone are wright and tension. The tension is upward and weight is downward. Since the stone is acceleration upward, the resultant force is in an upward direction. This implies that tension is larger than weight of the stone. The diagram represents the force vectors and its magnitude. We can see that the upward arrow (force vector) is larger than the downward force weight. 63b) From the diagram we can see that tension force is greater than the weight of the stone because the resultant force is upward. So the resultant force is along the upward direction causing the stone to move in upward direction.
