Answer: See table above.

Before Collision: 1 kg) * (+2 m/s) + (1 kg) * (1 m/s ) = 3 kg m/s

After Collision: (1 kg) * (1 m/s) + (1 kg) * (2 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.

Answer: See table above.

Before Collision: (2 kg) * (3 m/s) + (4 kg) * (1 m/s ) = 10 kg m/s

After Collision: (2 kg) * (1 m/s) + (4 kg) * (2 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.

General Note about Questions #45-#49:

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

After Collision: (6 kg) * v + (4 kg) * (10 m/s ) = (6 kg) * v + 40 kg m/s

(In the following steps, the units are dropped so that the algebra can be more easily followed.)

Answer: 7 m/s

Before Collision: (4 kg) * (8 m/s) + (2 kg) * (5 m/s ) =

After Collision: (4 kg) * v + (2 kg) * v

(In the following steps, the units are dropped so that the algebra can be more easily followed.)

Answer: 9 m/s

Before Collision: (3 kg) * (8 m/s) + (2 kg) * (-6 m/s ) = 12 kg m/s

After Collision: (3 kg) * (-2 m/s) + (2 kg) * v = -6 kg m/s + (2 kg) * v

(In the following steps, the units are dropped so that the algebra can be more easily followed.)

Answer: 4 m/s

Before Collision: (4 kg) * (10 m/s) + (3 kg) * (-12 m/s ) = 4 kg m/s

After Collision: (4 kg) * (-2 m/s) + (3 kg) * v = -8 kg m/s+ (3 kg) * v

(In the following steps, the units are dropped so that the algebra can be more easily followed.)

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

b. New value = 32 kg m/s • 2 • 0.5 = 32 kg•m/s

c. New value = 32 kg m/s • 0.5 • 2 = 32 kg•m/s

d. New value = 32 kg m/s • 0.5 • 0.5 = 8 kg•m/s

e. New value = 32 kg m/s • 4 • 1 = 128 kg•m/s

f. New value = 32 kg m/s • 4 • 0.5 = 64 kg•m/s

g. New value = 32 kg m/s • 0.5 • 4 = 64 kg•m/s

h. New value = 32 kg m/s • 3 • 2 = 192 kg•m/s

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 one-half the momentum would require one-half 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 one-half 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.

b. If the same object with the same momentum (M•v) were stopped by twice the original force, then only one-half the original time would be required to stop it. Since the momentum change has not been altered in this case (an object with a momentum of M•v is still being stopped), the impulse is not altered either.

c. If the same object with the same momentum (M•v) were stopped by ten times the original force, then only one-tenth the original time would be required to stop it. Since the momentum change has not been altered in this case (an object with a momentum of M•v is still being stopped), the impulse is not altered either.

d. If the same object with the same momentum (M•) were stopped by one-fifth the original force, then five times the original time would be required to stop it. Since the momentum change has not been altered in this case (an object with a momentum of M•v is still being stopped), the impulse is not altered either.

e. If an object with twice the mass is being stopped, then twice the momentum change would occur. Two times the momentum change means that an impulse which is two times the original impulse would be required. To achieve two times the impulse, the product of force and time must be doubled. Since the force is four times bigger, the time needs to be one-half the original time.

f. If an object with twice the mass and twice the velocity is being stopped, then four times the momentum change would occur. Four times the momentum change means that an impulse which is four times the original impulse would be required. To achieve four times the impulse, the product of force and time must be quadrupled. Since the force is four times bigger, the time needs to be the same as the original time.

g. If an object with one-half the mass and four times the velocity is being stopped, then twice the momentum change would occur. Two times the momentum change means that an impulse which is two times the original impulse would be required. To achieve two times the impulse, the product of force and time must be doubled. Since the force is two times bigger, the time needs to be the same as the original time.

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 post-explosion velocity of Cart B is 40 cm/s.

b. If Cart B has twice the mass as Cart A, then it must be propelled backwards with one-half the velocity. The post-explosion velocity of Cart B is 20 cm/s.

c. If Cart B has one-half the mass As Cart A, then it must be propelled backwards with two times the velocity. The post-explosion velocity of Cart B is 80 cm/s.

d. Changing the mass of either one of the carts does not alter the impulse or the momentum change which it experiences. An alteration in the mass merely alters the velocity with which the cart is propelled. A twice-as-massive cart acquires one-half the original velocity but the momentum change is the same as the original momentum change.

e. Changing the mass of either one of the carts does not alter the impulse or the momentum change which it experiences. An alteration in the mass merely alters the velocity with which the cart is propelled. A twice-as-massive cart acquires one-half the original velocity but the impulse is the same as the original impulse.

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 one-third the original value. The cart and brick move forward with a velocity of 20 cm/s.

b. Adding a brick with a mass of 3M will increase the total mass in motion from M (the cart's mass) to 4M. This fourfold increase in mass is accompanied by a fourfold decrease in velocity. The new velocity is one-fourth the original value. The cart and brick move forward with a velocity of 15 cm/s.

c. Adding a brick with a mass of 4M will increase the total mass in motion from M (the cart's mass) to 5M. This fivefold increase in mass is accompanied by a fivefold decrease in velocity. The new velocity is one-fifth the original value. The cart and brick move forward with a velocity of 12 cm/s.

d. Adding a brick with a mass of 5M will increase the total mass in motion from M (the cart's mass) to 6M. This six-fold increase in mass is accompanied by a six-fold decrease in velocity. The new velocity is one-sixth the original value. The cart and brick move forward with a velocity of 10 cm/s.

e. Adding a brick with a mass of 0.5M will increase the total mass in motion from M (the cart's mass) to 1.5M. This increases the mass by a factor of 1.5 and is accompanied by a decrease in velocity by a factor of 1.5. The new velocity is two-thirds the original value. The cart and brick move forward with a velocity of 40 cm/s.

f. Adding a brick with a mass of 0.25M will increase the total mass in motion from M (the cart's mass) to 1.25M. This increases the mass by a factor of 1.25 and is accompanied by a decrease in velocity by a factor of 1.25. The new velocity is four-fifths the original value. The cart and brick move forward with a velocity of 48 cm/s.