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AP Physics 1 : Impulse and Momentum

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Example Question #261 : Linear Motion And Momentum

A blue rubber ball weighing 20g is rolling with a velocity of $2\frac{m}{s}$ when it hits a still red rubber ball with a weight of 40g . After this elastic collision, what are the speeds and directions of the blue and red balls respectively?

Possible Answers:

$\frac{4}{3} \frac{m}{s}$ to the left, $\frac{2}{3} \frac{m}{s}$ to the right

At rest, $1 \frac{m}{s}$ to the right

$\frac{4}{3} \frac{m}{s}$ to the right, $\frac{2}{3} \frac{m}{s}$ to the left

$\frac{2}{3} \frac{m}{s}$ to the left, $\frac{4}{3} \frac{m}{s}$ to the right

At rest, $2 \frac{m}{s}$ to the right

Correct answer:

$\frac{2}{3} \frac{m}{s}$ to the left, $\frac{4}{3} \frac{m}{s}$ to the right

Explanation:

Because we are solving for two velocities (two unknowns), we need two equations. We can use the conservation of linear momentum:

$m_{1}v_{1}+m_{2}v_{2} = m_{1}v_{1}'+m_{2}v_{2}'$

Because we know that the collision is elastic, we know that kinetic energy is conserved:

$\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}=\frac{1}{2}m_{1}v_{1}'^{2}+\frac{1}{2}m_{2}v_{2}'^{2}$

The red ball starts at rest so

$v_{2}=0$

The above equations can then be simplified and one can solve for $v_{1}'$ and $v_{2}'$ .

$v_{1}'=\left ( \frac{m_{1}-m_{2}}{m_{1}+m_{2}} \right )v_{1}=\left ( \frac{20-40}{20+40} \right )2=-\frac{2}{3}$

Negative implies the ball is moving to the left.

$v_{2}'=\left ( \frac{2m_{1}}{m_{1}+m_{2}} \right )v_{1}=\left ( \frac{2(20)}{20+40} \right )2=\frac{4}{3}$

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Example Question #267 : Linear Motion And Momentum

Jennifer has a mass of 70kg . She is riding her skateboard east at $5\frac{m}{s}$ . She collides with Tommy, who is riding in the opposite direction at $4\frac{m}{s}$ . After they collide, they continue in Tommy's direction at $1\frac{m}{s}$ . Find the mass of Tommy.

Possible Answers:

100kg

70kg

120kg

140kg

Correct answer:

140kg

Explanation:

Using conservation of momentum

p_i=p_f

m_1v_1_i+m_2v_2_i=(m_1+m_2)v_f

Solving for m_2

$m_2=m_1\frac{(v_f-v_1_i)}{(v_2_i-v_f)}$

Plugging in values:

$m_2=70*\frac{(-1-5)}{(-4--1)}$

(A negative value is assumed for Tommy's direction for simplicity, a negative could have been used for Jennifer's instead, it will yield the same result.

m_2=140kg

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Example Question #268 : Linear Motion And Momentum

A 50 ton ship traveling at 30 knots collides with a 80 ton ship traveling at 25 knots . The two ships are $30 \degree$ apart prior to collision, and stick together afterward. What is the final speed of the combined ships?

Possible Answers:

23.4 knots

45.2knots

18.3knots

68.1knots

31.6knots

Correct answer:

23.4 knots

Explanation:

A coordinate system isn't immediately given, so the first step is to establish an x- and y-direction. Let the first ship travel in the positive y-direction with the second traveling $30 \degree$ from the y-axis.

The linear momentum for the first ship is given by $\vec{p_1} = m_1\vec{v_1}= (50ton)(30 knots)\hat{y}$

and for the second ship by $\vec{p_2} = m_2\vec{v_2}= (80ton)(25 knots)cos(30\degree)\hat{x} + (80ton)(25 knots)sin(30\degree)\hat{y}$

Conservation of linear momentum says that $\vec{p}_{final} = \vec{p_1} + \vec{p_2} = m_{final}\vec{v}_{final}$ . Thus the final velocity can be expressed as

$\vec{v}_{final} = \frac{1}{80ton + 50ton}([(80ton)(25knots)cos(30\degree)]\hat{x} + [(50ton)(30knots) + (80ton)(25 knots)sin(30\degree)]\hat{y})$

Further,

$\vec{v}_{final} = 13.32knots\hat{x}+19.23knots\hat{y}$ $\vec{v}_{final} = 13.32knots\; \hat{x}+19.23knots\;\hat{y}$

The final speed is then $v_{final} = \sqrt{13.32^2 + 19.23^2} = 23.4knots$

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Example Question #269 : Linear Motion And Momentum

A 3kg ball is launched across a flat, frictionless surface at a constant velocity of $+10.0 \frac{m}{s}$ in the rightward direction. While traveling at $+10.0 \frac{m}{s}$ , it collides in a perfectly inelastic manner with a stationary 7kg ball and the two balls continue traveling in the rightward direction, stuck to one another. What is the velocity of the 3-and-7 kg ball unit?

Possible Answers:

$-10 \frac{m}{s}$

$+10 \frac{m}{s}$

$-3 \frac{m}{s}$

$+30 \frac{m}{s}$

$+3 \frac{m}{s}$

Correct answer:

$+3 \frac{m}{s}$

Explanation:

This question tests your understanding of the concept of conservation of momentum. More specifically, this tests your understanding of perfectly inelastic collisions, or collisions in which the two colliding particles stick together and move together as a system.

By conservation of momentum, the initial total momentum of a system is equal to the final total momentum of the system. The formula for the momentum of an object is equal to the mass of the object multiplied by the velocity of the object. When applying the conservation of momentum formula to perfectly inelastic collisions, you consider the two objects as a single mass after they collide, since they stick together and travel together. Thus, this problem can be solved as follows:

$\textbf{\textup{Total Momentum (initial) = Total Momentum (final)}}$

$(mass_{ball1})(velocity_{ball1, initial}) + (mass_{ball2})(velocity_{ball2, initial}) = (mass_{ball1+2})(velocity_{ball1+2, final})$

$(3.0 kg) (+10.0 \frac{m}{s}) + (7.0 kg) (0 \frac{m}{s}) = (3.0 + 7.0 kg) (velocity_{ball1+2,final})$

$30 kg * \frac{m}{s} + 0 = 10 kg (velocity_{ball1+2,final})$

$30 kg * \frac{m}{s} = 10 kg (velocity_{ball1+2,final})$

$Velocity_{ball1+2,final}= +3 \frac{m}{s}$

The final velocity of the 3-and-7kg ball unit is therefore $+3 \frac{m}{s}$ (sign is positive because initially the 3kg ball was traveling to the right, and its velocity was positive, and when the ball unit formed and moved, it continued moving in the rightward, positive direction.

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Example Question #81 : Impulse And Momentum

A 5.0kg block is launched in the positive direction across a frictionless surface at a constant velocity of $+7.0\frac{m}{s}$ . It collides elastically with an 8.0kg block that then travels in the positive direction at a velocity of $+2.0\frac{m}{s}$ . What is the sign and magnitude of the velocity of the 5.0kg block after the collision? Assume that no energy is lost to friction during the collision or while the blocks travel.

Possible Answers:

$+7.6\frac{m}{s}$

$+19.0\frac{m}{s}$

$-7.6\frac{m}{s}$

$-3.8\frac{m}{s}$

$+3.8\frac{m}{s}$

Correct answer:

$+3.8\frac{m}{s}$

Explanation:

This question tests your understanding of the conservation of momentum, which states that momentum (mass multiplied by velocity) can neither be created nor destroyed in a system. In this elastic collision, we are presented with two blocks (one initially moving, and one initially stationary), that collide such that each block is moving separately afterwards and neither loses energy to friction or other forces. You can set up the equation as follows:

$\textup{Total Momentum(initial) = Total Momentum(final)}$

$(mass_{1} * velocity_{1})_{initial} + (mass_{2} * velocity_{2})_{initial} = (mass_{1} * velocity_{1})_{final} + (mass_{2} * velocity_{2})_{final}$

$(5.0 kg)(+7.0 \frac{m}{s}) + (8.0 kg)(0 \frac{m}{s}) = (5.0 kg)(velocity_{1,final}) + (8.0 kg)(+2.0 \frac{m}{s})$

$35.0 kg *\frac{m}{s} = (5.0 kg)(velocity_{1,final}) + 16.0 kg * \frac{m}{s}$

$(5.0 kg)(velocity_{1,final}) = 19.0 kg * \frac{m}{s}$

$velocity_{1,final} = +3.8 \frac{m}{s}$

Therefore, the final velocity of the 5.0kg block is $3.8 \frac{m}{s}$ in the positive direction.

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AP Physics 1 : Impulse and Momentum

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #261 : Linear Motion And Momentum

Example Question #267 : Linear Motion And Momentum

Example Question #268 : Linear Motion And Momentum

Example Question #269 : Linear Motion And Momentum

Example Question #81 : Impulse And Momentum

All AP Physics 1 Resources

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