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AP Physics C: Mechanics : Momentum

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Example Questions

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Example Question #1 : Momentum

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

Possible Answers:

Momentum and kinetic energy

Mass, kinetic energy, and momentum

Mass and potential energy

Kinetic energy and potential energy

Mass and momentum

Correct answer:

Mass and momentum

Explanation:

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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Example Question #1 : Momentum

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

Possible Answers:

Mass and momentum

Momentum, kinetic energy, and potential energy

Potential energy and kinetic energy

Mass, momentum, and kinetic energy

Momentum, kinetic energy, and potential energy

Correct answer:

Mass, momentum, and kinetic energy

Explanation:

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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Example Question #1 : Momentum

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?

Possible Answers:

$22.5\frac{m}{s}\ \text{to the east}$

$0\frac{m}{s}$

$42.5\frac{m}{s}\ \text{to the west}$

$22.5\frac{m}{s}\ \text{to the west}$

$42.5\frac{m}{s}\ \text{to the east}$

Correct answer:

$22.5\frac{m}{s}\ \text{to the west}$

Explanation:

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:

p=mv

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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Example Question #141 : Mechanics Exam

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?

Possible Answers:

$2.5\frac{m}{s}\ \text{to the east}$

$47.5\frac{m}{s}\ \text{to the west}$

$2.5\frac{m}{s}\ \text{to the west}$

$47.5\frac{m}{s}\ \text{to the east}$

$0\frac{m}{s}$

Correct answer:

$2.5\frac{m}{s}\ \text{to the east}$

Explanation:

Use the law of conservation of momentum:

$\sum p_i=\sum p_f$

Momentum is the product of velocity and mass:

p=mv

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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Example Question #1 : Momentum

A $20\: g$ bullet is fired at $300\: m/s$ at a block of wood that is moving in the opposite direction at a speed of $15\: m/s$ . The bullet passes through the block and emerges with the speed of $130\: m/s$ , while the block ends up at rest.

What is the mass of the block?

Possible Answers:

$0.24\:kg$

$0.227\:kg$

$0.193\:kg$

$0.253\:kg$

$0.28\:kg$

Correct answer:

$0.227\:kg$

Explanation:

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

m_1v_1_i + m_2v_2_i = m_1v_1_f + m_2v_2_f

It is given to us that m_1 is $20\: g$ or $0.02\:kg$ , $v_{1i}$ is $300\: m/s$ , m_2 is unknown, $v_{2i}$ is $-15\:m/s$ .

$v_{1f}$ is $130\:m/s$ and $v_{2f}$ is .

With all this information given, the only unknown is m_2 .

Plugging everything in, we get:

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

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Example Question #1 : Understanding Elastic And Inelastic Collisions

We have two balls. The first ball has mass 0.54kg and is traveling 7.1m/s to the right. It collides head-on elastically with a second ball of mass 0.95kg traveling 2.8m/s to the left. After the collision, what is the speed and direction of each ball?

1-d_collision

$v_1^{'}=final\ velocity\ of\ ball\ 1$

$v_2^{'}=final\ velocity\ of\ ball\ 2$

Possible Answers:

$v_1^{'}=-2.3\frac{m}{s}$ to the left, $v_2^{'}=0.2\frac{m}{s}$ to the right

$v_1^{'}=-2.3\frac{m}{s}$ to the left, $v_2^{'}=-6.4\frac{m}{s}$ to the left

$v_1^{'}=-5.5\frac{m}{s}$ to the left, $v_2^{'}=4.4\frac{m}{s}$ to the right

$v_1^{'}=5.5\frac{m}{s}$ to the right, $v_2^{'}=4.4\frac{m}{s}$ to the right

$v_1^{'}=-2.3\frac{m}{s}$ to the left, $v_2^{'}=6.4\frac{m}{s}$ to the right

Correct answer:

$v_1^{'}=-5.5\frac{m}{s}$ to the left, $v_2^{'}=4.4\frac{m}{s}$ to the right

Explanation:

We must use conservation of momentum to tackle this problem.

m_1v_1+m_2v_2=m_1v_1'+m_2v_2'

We are to find and , the velocities of the balls after the collision. We know the following for the first ball:

$m_1=0.54\ \text{kg}$

$v_1=7.1\ \frac{m}{s}$

and we know the following for the second ball:

$m_2=0.95\ \text{kg}$

$v_2=-2.8\ \frac{m}{s}$ .

After plugging these values into our conservation of momentum equation, it is clear that we can't use this equation alone to find and ; however, since we are dealing with an elastic collision, we can use the relation below.

v_1-v_2=v_2'-v_1'

This relation can be derived using conservation of momentum and conservation of kinetic energy equations. Remember that kinetic energy is only conserved if the collision is elastic. We can use this relation to eliminate either or in our conservation of momentum equation.

v_1'=v_2'-v_1+v_2

Plug this into our conservation of momentum equation.

m_1v_1+m_2v_2=m_1(v_2'-v_1+v_2)+m_2v_2'

So now is eliminated and we can solve for .

m_1v_1+m_2v_2=m_1v_2'-m_1v_1+m_1v_2+m_2v_2'

2m_1v_1+(m_2-m_1)v_2=(m_1+m_2)v_2'

$v_2'=\frac{2m_1v_1+(m_2-m_1)v_2}{m_1+m_2}$

$v_2'=\frac{2(0.54\ \text{kg})(7.1\ \frac{m}{s})+(0.95\ \text{kg}-0.54\ \text{kg})(-2.8\ \frac{m}{s})}{0.54\ \text{kg}+0.95\ \text{kg}}$

$v_2'=4.4\ \frac{m}{s}$

This is the speed of the second ball, and it is traveling to the right because of the positive value. Use this positive value to find by using the relation we found earlier.

v_1'=v_2'-v_1+v_2

$v_1'=4.4\ \frac{m}{s}-7.1\ \frac{m}{s}-2.8\ \frac{m}{s}$

$v_1'=-5.5\ \frac{m}{s}$

This is the speed of the first ball and it is traveling to the left due to the negative sign.

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Example Question #141 : Ap Physics C

A 30kg cart travels at 9m/s and it hits another cart of mass 46kg traveling at 4m/s in the opposite direction. After the collision, they stick together to form one cart. Find the speed of this cart

Possible Answers:

$v=0.04\frac{m}{s}$

$v=1.13\frac{m}{s}$

$v=5.12\frac{m}{s}$

$v=1.57\frac{m}{s}$

$v=2.32\frac{m}{s}$

Correct answer:

$v=1.13\frac{m}{s}$

Explanation:

For the 30 kg cart, we know

$m_1=30\ \text{kg}$

$v_1=9\ \frac{m}{s}$

and for the 46 g cart, we know

$m_2=46\ \text{kg}$

$v_2=-4\ \frac{m}{s}$ .

After the collision, we have v_1'=v_2'=v .

Use conservation of momentum to solve this problem.

m_1v_1+m_2v_2=m_1v_1'+m_2v_2'

m_1v_1+m_2v_2=m_1v+m_2v

m_1v_1+m_2v_2=(m_1+m_2)v

$v=\frac{m_1v_1+m_2v_2}{m_1+m_2}$

$v=\frac{(30\ \text{kg})(9\ \frac{m}{s})+(46\ \text{kg})(-4\ \frac{m}{s})}{30\ \text{kg}+46\ \text{kg}}$

$v=1.13\ \frac{m}{s}$

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Example Question #1 : Momentum

There are two skaters. The male skater with mass 68kg travels 15m/s North. He approaches a 60kg female skater who is travel 12m/s East; they approach each other at right angles. When they meet, they hold on to each other. At what direction and speed do they move after they meet?

Possible Answers:

$v=9.8\frac{m}{s}$

$\theta=54.8^o$ from the female skater's direction

None of these

$v=8.7\frac{m}{s}$

$\theta=31.4^o$ from the female skater's direction

$v=13.4\frac{m}{s}$

$\theta=54.8^o$ from the female skater's direction

$v=10.0\frac{m}{s}$

$\theta=54.8^o$ from the female skater's direction

Correct answer:

$v=9.8\frac{m}{s}$

$\theta=54.8^o$ from the female skater's direction

Explanation:

This is a two-dimensional inelastic collision problem and we use conservation of momentum to solve. We know the following.

Male skater: $m_1=68\ \text{kg}$ , $v_{1y}=15\ \frac{m}{s}$

Female skater: $m_2=60\ \text{kg}$ , $v_{2x}=12\ \frac{m}{s}$

First, write down two equations representing conservation of momentum. One equation represents momentum in the x-direction (East-West direction), which is $m_2v_{2x}=(m_1+m_2)v\cos\theta$ ,and the other gives the momentum in the y-direction (North-South direction), $m_1v_{1y}=(m_1+m_2)v\sin\theta$ .

Take the y-direction of momentum and divide it by the x-direction momentum.

$\frac{m_1v_{1y}}{m_2v_{2x}}=\frac{(m_1+m_2)v\sin\theta}{(m_1+m_2)v\cos\theta}$

Simplify.

$\frac{m_1v_{1y}}{m_2v_{2x}}=\frac{\sin\theta}{\cos\theta}$

$\frac{m_1v_{1y}}{m_2v_{2x}}=\tan\theta$

Plug in the numbers to find the angle

$\frac{(68\ \text{kg})(15\ \frac{m}{s})}{(60\ \text{kg})(12\ \frac{m}{s})}=\tan\theta$

$\theta=54.8^o$

To find the speed at which they move at this angle we can use one of the momentum equations.

$m_2v_{2x}=(m_1+m_2)v\cos\theta$

Solve for v.

$v=\frac{m_2v_{2x}}{(m_1+m_2)\cos\theta}$

$v=\frac{(60\ \text{kg})(12\ \frac{m}{s})}{(68\ \text{kg}+60\ \text{kg})\cos(54.8)}$

$v=9.8\ \frac{m}{s}$

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Example Question #1 : Understanding Elastic And Inelastic Collisions

Object A has mass $\small 1.0kg$ and initially moves to the right at $\small 6.0\frac{m}{s}$ . It then collides with object B, which has mass $\small 5.0kg$ and was initially moving to the right at $\small 4.0\frac{m}{s}$ . If the two objects stick together after the collision, what percentage of the initial kinetic energy has been dissipated?

Possible Answers:

$4.7\%$

$2.9\%$

$6.4\%$

$3.1\%$

$1.5\%$

Correct answer:

$2.9\%$

Explanation:

Relevant equations:

$K = \frac{1}{2}mv^2$

$\Delta K = K_f - K_i$

p_i = p_f

p = mv

Use conservation of momentum to determine the final velocity of the objects.

p_i = p_f

$m_Av_{A,i}+m_Bv_{B,i}=m_Av_{A,f}+m_Bv_{B,f}$

$m_Av_{A,i}+m_Bv_{B,i}=(m_A + m_B)v_f$

$(1kg)(6\frac{m}{s})+(5kg)(4\frac{m}{s})=(1kg+5kg)v_f$

$v_f=4.33\frac{m}{s}$

Calculate the total initial kinetic energy of the two objects.

$K_{A,i} = \frac{1}{2}(1kg)(6\frac{m}{s})^2=18J$

$K_{B,i}=\frac{1}{2}(5kg)(4\frac{m}{s})^2=40J$

$K_i = K_{A,i}+K_{B,i}=18J+40J=58J$

Calculate the total final kinetic energy of the two objects.

$K_{A,f}=\frac{1}{2}(1kg)(4.33\frac{m}{s})^2=9.39J$

$K_{B,f}=\frac{1}{2}(5kg)(4.33\frac{m}{s})^2=46.94J$

$K_f = K_{A,f}+K_{B,f}=9.39J+46.94J = 56.33J$

Find the difference in kinetic energy.

$\Delta K = K_f-K_i = 56.33J - 58J = -1.67J$

This is the amount of kinetic energy lost during the inelastic collision. Express this amount as a percentage of the initial kinetic energy.

$\frac{|\Delta K|}{K_i}*100\% = \frac{1.67}{58}*100\% \approx 2.9\%$

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Example Question #1 : Momentum

A baseball player hits a $\small 150g$ baseball initially moving at $\small 35\frac{m}{s}$ , returning it at a speed of $\small 45\frac{m}{s}$ along the same path. If the ball was in contact with the bat for $\small 1ms$ , what magnitude of force did the ball experience during the moment of contact?

Possible Answers:

530N

80kN

1.5kN

42kN

12kN

Correct answer:

12kN

Explanation:

Relevant equations:

$J = m\Delta v$

$J = F_{avg}\Delta t$

Evaluate the impulse based on the mass and change of velocity.

$J = m\Delta v = (0.150kg)(80\frac{m}{s})=12 \frac{kg*m}{s}$

Use the total impulse and time in the second equation.

$m\Delta v = F_{avg}\Delta t$

$12\frac{kg*m}{s}=F_{avg}(0.001s)$

Solve for the average force.

$F_{avg}=\frac{12\frac{kg*m}{s}}{0.001s}=12,000N = 12kN$

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All AP Physics C: Mechanics Resources

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AP Physics C: Mechanics : Momentum

Study concepts, example questions & explanations for AP Physics C: Mechanics

All AP Physics C: Mechanics Resources

Example Questions

Example Question #1 : Momentum

Example Question #1 : Momentum

Example Question #1 : Momentum

Example Question #141 : Mechanics Exam

Example Question #1 : Momentum

Example Question #1 : Understanding Elastic And Inelastic Collisions

Example Question #141 : Ap Physics C

Example Question #1 : Momentum

Example Question #1 : Understanding Elastic And Inelastic Collisions

Example Question #1 : Momentum

All AP Physics C: Mechanics Resources

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