Understanding Conservation of Momentum - AP Physics C Electricity & Magnetism

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Question

In an inelastic collision, which combination of quantities is conserved?

Answer

In a perfectly inelastic collision, the two colliding objects stick together; the two colliding objects deform, but mass is still conserved. Momentum is conserved during collisions of any sort, including inelastic collisions.

Kinetic energy is reduced during an inelastic collision, and is only conserved in elastic collisions. During inelastic collisions, some kinetic energy is lost to the environment in the form of heat or sound.

The problem does not give any information regarding position, and thus we cannot comment on any changes or lack of changes in potential energy.

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Question

In an elastic collision, what combination of quantities is conserved?

Answer

The primary difference between elastic and inelastic collisions is the conservation of kinetic energy. Kinetic energy is conserved in elastic collisions, but is not conserved in inelastic collisions. Momentum is always conserved, regardless of collision type. Mass is conserved regardless of collision type as well, but the mass may be deformed by an inelastic collision, resulting in the two original masses being stuck together.

There is no description in the problem reagrding position, so we cannot comment on potential energy.

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Question

A 200kg car is traveling west at $30\frac{m}{s}$ . It collides with a 150kg car that was at rest. Following the collision, the second car moves with a velocity of $10\frac{m}{s}$ west. Assuming that the collision is elastic, what is the velocity of the first car after the collision?

Answer

The collision is assumed to be elastic, so both momentum and kinetic energy are conserved. Use the law of conservation of momentum:

$\sum p_i=\sum p_f$

Momentum is the product of velocity and mass:

We can expand the summation for the initial and final conditions:

$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$

Use the given values to fill in the equation and solve for $v_{1f}$ :

$(200kg)(30\frac{m}{s})+(150kg)(0\frac{m}{s})=(200kg)v_{1f}+(150kg)(10\frac{m}{s})$

$6000\frac{m\cdot kg}{s}=(200kg)v_{1f}+1500\frac{kg\cdot m}{s}$

$v_{1f}=\frac{6000\frac{m\cdot kg}{s}-1500\frac{kg\cdot m}{s}}{200kg}=22.5\frac{m}{s}$

Since the final velocity is positive, we know that the car is still traveling toward the west.

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Question

Two train cars, each with a mass of 2400 kg, are traveling along the same track. One car is traveling with a velocity of $50\frac{m}{s}$ east, while the other travels with a velocity of $45\frac{m}{s}$ west. The two cars collide and stick together as one mass. What is the magnitude and direction of the resulting velocity?

Answer

Use the law of conservation of momentum:

$\sum p_i=\sum p_f$

Momentum is the product of velocity and mass:

We can expand the summation for the initial and final conditions:

$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$

Note that we are working with an inelastic collision, meaning that the two masses stick together after the collision. Because of this, they will have the same final velocity:

$v_{1f}=v_{2f}$

$m_1v_{1i}+m_2v_{2i}=v_f(m_1+m_2)$

Use the given values to fill in the equation and solve for $v_{f}$ . Keep in mind that we must designate a positive direction and a negative direction. We will use east as positive and west as negative.

$(2400kg)(50\frac{m}{s})+(2400kg)(-45\frac{m}{s})=v_f(2400kg+2400kg)$

$12000\frac{kg\cdot m}{s}=v_f(4800kg)$

$v_f=\frac{1200\frac{kg\cdot m}{s}}{4800kg}=2.5\frac{m}{s}$

Since the final velocity is positive, we can determine that they train cars are traveling toward the east.

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Question

A bullet is fired at at a block of wood that is moving in the opposite direction at a speed of . The bullet passes through the block and emerges with the speed of , while the block ends up at rest.

What is the mass of the block?

Answer

This problem is a conservation of momentum problem. When doing these types of problems, the equation to jump to is:

It is given to us that is or , $v_{1i}$ is , is unknown, $v_{2i}$ is .

$v_{1f}$ is and $v_{2f}$ is .

With all this information given, the only unknown is .

Plugging everything in, we get:

$\ 0.02\cdot300 + m_2\cdot15 = 0.02\cdot130 + 0\ m_2 = 0.227kg$

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Question

In an inelastic collision, which combination of quantities is conserved?

Answer

In a perfectly inelastic collision, the two colliding objects stick together; the two colliding objects deform, but mass is still conserved. Momentum is conserved during collisions of any sort, including inelastic collisions.

Kinetic energy is reduced during an inelastic collision, and is only conserved in elastic collisions. During inelastic collisions, some kinetic energy is lost to the environment in the form of heat or sound.

The problem does not give any information regarding position, and thus we cannot comment on any changes or lack of changes in potential energy.

Compare your answer with the correct one above

Question

In an elastic collision, what combination of quantities is conserved?

Answer

The primary difference between elastic and inelastic collisions is the conservation of kinetic energy. Kinetic energy is conserved in elastic collisions, but is not conserved in inelastic collisions. Momentum is always conserved, regardless of collision type. Mass is conserved regardless of collision type as well, but the mass may be deformed by an inelastic collision, resulting in the two original masses being stuck together.

There is no description in the problem reagrding position, so we cannot comment on potential energy.

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Question

A 200kg car is traveling west at $30\frac{m}{s}$ . It collides with a 150kg car that was at rest. Following the collision, the second car moves with a velocity of $10\frac{m}{s}$ west. Assuming that the collision is elastic, what is the velocity of the first car after the collision?

Answer

The collision is assumed to be elastic, so both momentum and kinetic energy are conserved. Use the law of conservation of momentum:

$\sum p_i=\sum p_f$

Momentum is the product of velocity and mass:

We can expand the summation for the initial and final conditions:

$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$

Use the given values to fill in the equation and solve for $v_{1f}$ :

$(200kg)(30\frac{m}{s})+(150kg)(0\frac{m}{s})=(200kg)v_{1f}+(150kg)(10\frac{m}{s})$

$6000\frac{m\cdot kg}{s}=(200kg)v_{1f}+1500\frac{kg\cdot m}{s}$

$v_{1f}=\frac{6000\frac{m\cdot kg}{s}-1500\frac{kg\cdot m}{s}}{200kg}=22.5\frac{m}{s}$

Since the final velocity is positive, we know that the car is still traveling toward the west.

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Question

Two train cars, each with a mass of 2400 kg, are traveling along the same track. One car is traveling with a velocity of $50\frac{m}{s}$ east, while the other travels with a velocity of $45\frac{m}{s}$ west. The two cars collide and stick together as one mass. What is the magnitude and direction of the resulting velocity?

Answer

Use the law of conservation of momentum:

$\sum p_i=\sum p_f$

Momentum is the product of velocity and mass:

We can expand the summation for the initial and final conditions:

$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$

Note that we are working with an inelastic collision, meaning that the two masses stick together after the collision. Because of this, they will have the same final velocity:

$v_{1f}=v_{2f}$

$m_1v_{1i}+m_2v_{2i}=v_f(m_1+m_2)$

Use the given values to fill in the equation and solve for $v_{f}$ . Keep in mind that we must designate a positive direction and a negative direction. We will use east as positive and west as negative.

$(2400kg)(50\frac{m}{s})+(2400kg)(-45\frac{m}{s})=v_f(2400kg+2400kg)$

$12000\frac{kg\cdot m}{s}=v_f(4800kg)$

$v_f=\frac{1200\frac{kg\cdot m}{s}}{4800kg}=2.5\frac{m}{s}$

Since the final velocity is positive, we can determine that they train cars are traveling toward the east.

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Question

A bullet is fired at at a block of wood that is moving in the opposite direction at a speed of . The bullet passes through the block and emerges with the speed of , while the block ends up at rest.

What is the mass of the block?

Answer

This problem is a conservation of momentum problem. When doing these types of problems, the equation to jump to is:

It is given to us that is or , $v_{1i}$ is , is unknown, $v_{2i}$ is .

$v_{1f}$ is and $v_{2f}$ is .

With all this information given, the only unknown is .

Plugging everything in, we get:

$\ 0.02\cdot300 + m_2\cdot15 = 0.02\cdot130 + 0\ m_2 = 0.227kg$

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