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For questions #3740: Consider the before and aftercollision momentum vectors in the diagram below. Determine the magnitude and direction of the system momentum before and after the collision and identify whether or not momentum is conserved. Finally, determine the magnitude and direction of the net external impulse encountered by the system during the collision.


System Momentum Before Collision: 

System Momentum After Collision: 

Is momentum conserved? 

Net External Impulse During Collision: 

Answer: See table above. Before the collision, the two momentum values can be added: 30 units + 0 = 30 units. (An object at rest has zero momentum.) After the collision, the momentum vectors can be added: 30 units + 0 = 30 units. In each case, the total momentum is in the direction of the arrows. Since the total system momentum before the collision is the same as it is after the collision, the total momentum of the system can be considered to be conserved. As such, there is no net external impulse upon the system. It is considered isolated. 


System Momentum Before Collision: 

System Momentum After Collision: 

Is momentum conserved? 

Net External Impulse During Collision: 

Answer: See table above. Here, the dark ball has a leftward momentum, so its momentum must be assigned a  sign to distinguish its momentum from the rightward momentum of the lightcolored ball. So before the collision, the system has +25 units + 10 units of momentum  a total of +15 units. After the collision, the total momentum of the system is + 15 units (the "+" is the designated sign for a rightward momentum). Since the system momentum is the same before as after the collision, momentum is conserved and the system is considered isolated from net external impulses. 


System Momentum Before Collision: 

System Momentum After Collision: 

Is momentum conserved? 

Net External Impulse During Collision: 

Answer: See table above. Since the lightcolored ball has a leftward momentum, the leftward direction will be designated as the negative direction. Thus, the before and aftercollision calculations are: Before: +30 units + 0 units = +30 units The total momentum of the system has changed by 40 units. Thus, momentum is not conserved and the system experiences a net external impulse of 40 units or 40 N s, leftward. 


System Momentum Before Collision: 

System Momentum After Collision: 

Is momentum conserved? 

Net External Impulse During Collision: 

Answer: See table above. The summations of system momentum before and after the collision are: Before: +30 units + +20 units = +50 units. The system loses 5 units of momentum. Thus, momentum is not conserved. There must be a net external impulse of 5 units upon the system. That impulse is 5 N s, left. (The left is equivalent to saying negative.) 
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For questions #4144: Repeat the procedure performed in questions #3740. Note that these diagrams give velocity and mass values before and after the collision.


System Momentum Before Collision: 

System Momentum After Collision: 

Is momentum conserved? 

Net External Impulse During Collision: 

Answer: See table above. Before Collision: 1 kg) * (+2 m/s) + (1 kg) * (1 m/s ) = 3 kg m/s The total system momentum is the same before and after the collision. Thus, momentum is conserved and there is no net external impulse on the system. 


System Momentum Before Collision: 

System Momentum After Collision: 

Is momentum conserved? 

Net External Impulse During Collision: 

Answer: See table above. Before Collision: (2 kg) * (2 m/s) + (4 kg) * (1 m/s ) = 8 kg m/s The total system momentum is NOT the same before and after the collision. Thus, momentum is NOT conserved. There is a net external impulse on the system which is equal to the momentum change of the system. The net external impulse is +2 kg m/s. 


System Momentum Before Collision: 

System Momentum After Collision: 

Is momentum conserved? 

Net External Impulse During Collision: 

Answer: See table above. Before Collision: (2 kg) * (3 m/s) + (4 kg) * (1 m/s ) = 10 kg m/s The total system momentum is the same before and after the collision. Thus, momentum is conserved and there is no net external impulse on the system. 


System Momentum Before Collision: 

System Momentum After Collision: 

Is momentum conserved? 

Net External Impulse During Collision: 

Answer: See table above. Before Collision: (5 kg) * (2 m/s) + (2 kg) * (5 m/s ) = 0 kg m/s The total system momentum is the same before and after the collision. Thus, momentum is conserved and there is no net external impulse on the system. 
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For questions #4549, determine the unknown velocity value. Assume that the collisions occur in an isolated system.
General Note about Q#4549: In each of these problems, an expression for the total momentum before the collision is written by summing the product of mass and velocity for each object. The same is done with the total momentum after the collision. One must be careful to assign a negative momentum to any object which is moving leftward. When the velocity of an object is not known, the variable v should be placed in the expression in place of the actual velocity value. The two expressions for total momentum before and after the collision are set equal to each other and algebraic operations are used to determine the unknown velocities. This method is shown below. 

Answer: 6 m/s Before Collision: (6 kg) * (10 m/s) + (4 kg) * (4 m/s ) = 76 kg m/s (In the following steps, the units are dropped so that the algebra can be more easily followed.) 36 = 6v v = 6 m/s 

Answer: 7 m/s Before Collision: (4 kg) * (8 m/s) + (2 kg) * (5 m/s ) = (In the following steps, the units are dropped so that the algebra can be more easily followed.) v = 7 m/s 

Answer: 9 m/s Before Collision: (3 kg) * (8 m/s) + (2 kg) * (6 m/s ) = 12 kg m/s (In the following steps, the units are dropped so that the algebra can be more easily followed.) 18 = 2v v = 9 m/s 

Answer: 4 m/s Before Collision: (4 kg) * (10 m/s) + (3 kg) * (12 m/s ) = 4 kg m/s (In the following steps, the units are dropped so that the algebra can be more easily followed.) 12 = 3v v = 4 m/s 

Answer: 11.3 m/s Before Collision: (4 kg) * (12 m/s) + (3 kg) * (10 m/s ) = 18 kg m/s (In the following steps, the units are dropped so that the algebra can be more easily followed.) 34.0 = 3v v = 11.3 m/s 
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For questions #5052, determine the total kinetic energy of the system before and after the collision and identify the collision as being either perfectly elastic, partially inelastic/elastic or perfectly inelastic.


Total System Kinetic Energy Before Collision: 

Total System Kinetic Energy After Collision: 

Perfectly Elastic, Partially Inelastic/Elastic or Perfectly Inelastic? 

Answer: See table above. Before Collision: KE_{total} = 0.5•(4 kg)•(5 m/s)^{2} + 0.5•(4 kg)•(0 m/s)^{2} = 50 J The system kinetic energy is conserved in this collision. That is, the total kinetic energy is the same before the collision as after the collision. The critical requirement for a collision to be perfectly elastic is that the total system kinetic energy is conserved. Thus, this collision is a perfectly elastic collision. 


Total System Kinetic Energy Before Collision: 

Total System Kinetic Energy After Collision: 

Perfectly Elastic, Partially Inelastic/Elastic or Perfectly Inelastic? 

Answer: See table above. Before Collision: KE_{total} = 0.5•(4 kg)•(4 m/s)^{2} + 0.5•(2 kg)•(2 m/s)^{2} = 36 J The system kinetic energy is NOT conserved in this collision. That is, the total kinetic energy is NOT the same before the collision as after the collision. The critical requirement for a collision to be perfectly inelastic is that the colliding objects move as a single unit with the same velocity after the collision. These objects do "stick together" after the collision and move as a single unit at the same velocity. Thus, the collision is a perfectly inelastic collision. 


Total System Kinetic Energy Before Collision: 

Total System Kinetic Energy After Collision: 

Perfectly Elastic, Partially Inelastic/Elastic or Perfectly Inelastic? 

Answer: See table above. Before Collision: KE_{total} = 0.5•(10 kg)•(6 m/s)^{2} + 0.5•(6 kg)•(1 m/s)^{2} = 183 J The system kinetic energy is NOT conserved in this collision. That is, the total kinetic energy is NOT the same before the collision as after the collision. Thus, the collision does not meet the criterion for a perfectly elastic collision. The objets do NOT "stick together" after the collision; they move with different speeds. Thus, the collision does not meet the criterion for a perfectly inelastic collision. This collision is a partially inelastic/elastic collision. 
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53. An object with a mass M and a velocity v has a momentum of 32 kg•m/s. An object with a mass of ...
Answer: See Answers above. Momentum is the product of mass and velocity. As such, the momentum of an object is directly proportional to the mass and directly proportional to the velocity. If the mass of an object is altered by some factor, then the momentum of the object is altered by that same factor. So if the mass is doubled, the momentum is doubled; the new value would be two times the original value. And if the mass is tripled, then the momentum is tripled; the new value would be three times the original value. If the velocity of an object is altered by some factor, then the momentum of the object is altered by that same factor. So if the velocity is doubled, the momentum is doubled; the new value would be two times the original value. And if the velocity is tripled, then the momentum is tripled; the new value would be three times the original value. This is shown in the work below. a. New value = 32 kg m/s • 2 • 2 = 128 kg•m/s 
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54. An object with a mass M and a velocity v undergoes a collision and encounters a force of F for a time of t. The collision brings the object to a final rest position ...
Answer: See Answers above. In a collision, the impulse encountered by an object causes and is equal to the momentum change of the object. For objects being brought to rest, the change in momentum is simply equal to the original momentum. An object with twice the momentum would require twice the impulse to stop it. An object with onehalf the momentum would require onehalf the impulse to stop it. The momentum is the product of mass and velocity, so any alteration in the mass or the velocity or both would alter the required impulse by that same factor. The impulse results from a force acting over time and is the product of force and time. So twice the impulse can be achieved by the same force acting for twice the time or twice the force acting for the same time or even onehalf the force acting for four times the time. These principles are illustrated below. a. Two times the mass means two times the original momentum. Two times the impulse would be required to stop an object with two times the momentum. 
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55. Two carts are placed next to each other on a lowfriction track. The carts are equipped with a springloaded mechanism which allows them to impart an impulse to each other. Cart A has a mass of M and Cart B has a mass of M. The springloaded mechanism is engaged and then released. The impulse causes Cart A to be propelled forward with a velocity of 40 cm/s.
Answer: See Answers above. Cart A and Cart B begin at rest. The original momentum of the system is 0 units. If momentum is to be conserved, the total momentum of the system of two carts must also be 0 units. This means that the momentum of Cart A must have the same magnitude as the momentum of Cart B. That is, after the spring is released, the product of mass and velocity for Cart A must equal the product of mass and velocity for Cart B. This principle is illustrated in the reasoning below. a. Since the two carts have the same mass, they must also have the same velocity in order to have the same magnitude of momentum. The postexplosion velocity of Cart B is 40 cm/s. 
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56. A cart with a mass of M is moving along a lowfriction track with a speed of 60 cm/s. A brick is gently dropped from rest upon the cart. After the collision the cart and brick move together.
Answer: See Answers above. This is a special case of a perfectly inelastic collision. The total system momentum before the collision is possessed solely by the moving cart. After the collision, the total system momentum is the combined momentum of the brick and the cart. Since the brick and cart travel at the same velocity after the collision, the momentum is simply the sum of their masses multiplied by their velocity. In effect, the total mass which is in motion is increased by some factor as a result of the subsequent motion of the brick. For momentum to be conserved, the velocity of the cart (and the brick which is on top of it) must be decreased by that same factor. This principle is used to reason towards the answers to these questions. a. Adding a brick with a mass of 2M will increase the total mass in motion from M (the cart's mass) to 3M. This threefold increase in mass is accompanied by a threefold decrease in velocity. The new velocity is onethird the original value. The cart and brick move forward with a velocity of 20 cm/s. 
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Last update: 5/6/02