Momentum - Physics

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Question

A tennis ball strikes a racket, moving at . After striking the racket, it bounces back at a speed of . What is the change in momentum?

Answer

The change in momentum is the final momentum minus the initial momentum, or $\Delta p=p_{final}-p_{initial}$ .

Notice that the problem gives us the final SPEED of the ball but not the final VELOCITY. Since the ball "bounced back," it begins to move in the opposite direction, so its velocity at this point will be negative.

Plug in our values to solve:

$\Delta p=p_{final}-p_{initial}$

$\Delta p=(0.12kg-45m/s)-(0.12kg35m/s)$

$\Delta p=-5.4kgm/s-4.2kgm/s$

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Question

A crate slides along the floor for before stopping. If it was initially moving with a velocity of , what is the force of friction?

Answer

The fastest way to solve a problem like this is with momentum.

Remember that momentum is equal to mass times velocity: . We can rewrite this equation in terms of force.

Using this transformation, we can see that momentum is also equal to force times time.

$\Delta p=F\Delta t$ can also be thought of as $(p_{final}-p_{initial} =F\Delta t)$ .

Expand this equation to include our given values.

$(mv_{final}-mv_{initial})=F\Delta t$

Since the object is not moving at the end, its final velocity is zero. Plug in the given values and solve for the force.

$(-600kg \cdot m/s)=F(10s)$

$\frac{(-600kg\cdot m/s)}{10s}=F$

We would expect the answer to be negative because the force of friction acts in the direction opposite to the initial velocity.

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Question

A crate slides along a floor with a starting velocity of . If the force due to friction is , how long will it take for the box to come to rest?

Answer

The fastest way to solve a problem like this is with momentum.

Remember that momentum is equal to mass times velocity: . We can rewrite this equation in terms of force.

Using this transformation, we can see that momentum is also equal to force times time.

$\Delta p=F\Delta t$ can also be thought of as $(p_{final}-p_{initial})=F\Delta t$ .

Expand this equation to include our given values.

$(mv_{final}-mv_{initial})=F\Delta t$

Since the box is not moving at the end, its final velocity is zero. Plug in the given values and solve for the time.

$(100kg0m/s)-(100kg21m/s)=(-8N)\Delta t$

$(-100kg*21m/s)=(-8N)\Delta t$

$(-2100kg \cdot m/s)=(-8N)\Delta t$

$(-2100kg \cdot m/s)-8N=\Delta t$

$262.5s= \Delta t$

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Question

The area under the curve on a Force versus time (F versus t) graph represents

Answer

If we were to examine the area under the curve of a constant force applied over a certain amount of time we would have a graph with a straight horizontal line. To find the area of that rectangle we would multiply the base times the height. The base would be the time (number of seconds the force was applied). The height would be the amount of force applied during this time. Force*time is equal to the impulse acting on the object which is equal to the change in momentum of the object.

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Question

A man with a mass of m is painting a house. He stands on a tall ladder of height h. He leans over and falls straight down off the ladder. If he is in the air for s seconds, what will be his momentum right before he hits the ground?

Answer

The problem tells us he falls vertically off the ladder (straight down), so we don't need to worry about motion in the horizontal direction.

The equation for momentum is:

We can assume he falls from rest, which allows us to find the initial momentum.

.

From here, we can use the formula for impulse:

$\Delta p=F\Delta t$

$(p_f-p_i)=F\Delta t$

We know his initial momentum is zero, so we can remove this variable from the equation.

$p_f=F\Delta t$

The problem tells us that his change in time is s seconds, so we can insert this in place of the time.

The only force acting upon man is the force due to gravity, which will always be given by the equation .

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Question

If an egg is dropped on concrete, it usually breaks. If an egg is dropped on grass, it may not break. What conclusion explains this result?

Answer

In both cases the egg starts with the same velocity, so it has the same initial momentum. In both cases the egg stops moving at the end of its fall, so it has the same final velocity. The only thing that changes is the time and force of the impact. The force is produced by the deceleration resulting from the time that the egg is in contact with its point of impact.

As the time of contact increases, acceleration decreases and force decreases.

As the time of contact decreases, acceleration increases and force increases.

$a=\Delta v\Delta t$

The grass will have less force on the egg, allowing for a lesser acceleration, due to a longer period of impact.

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Question

A ball moving at strikes a ball at rest. After the collision the ball is moving with a velocity of . What is the velocity of the second ball?

Answer

This is an example of an elastic collision. We start with two masses and end with two masses with no loss of energy.

We can use the law of conservation of momentum to equate the initial and final terms.

$m_1 v_{1initial} + m_2 v_{2initial} = m_1 v_{1final} + m_2 v_{2final}$

Plug in the given values and solve for $v_{2final}$ .

$(10kg)(13m/s)+(20kg)(0ms)=(10kg)(7m/s)+(12kg)(v_{2final})$

$30kgm/s + okgm/s = (70kgm/s) + (20kg)(v_{2final})$

$130kgm/s-70kgm/s=20kg v_{2final}$

$60kgm/s= 20kg v_{2final}$

$\frac{60kgm/s}{20kg} = v_{2final}$

$3m/s = v_{2final}$

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Question

A ball is thrown west at and collides with a ball while in the air. If the balls stick together in the crash and fall straight down to the ground, what was the velocity of the second ball?

Answer

We know that if the balls fell straight down after the crash, then the total momentum in the horizontal direction is zero. The only motion is due to gravity, rather than any remaining horizontal momentum. Based on conservation of momentum, the initial and final momentum values must be equal. If the final horizontal momentum is zero, then the initial horizontal momentum must also be zero.

In our situation, the final momentum is going to be zero.

Use the given values for the mass of each ball and initial velocity of the first ball to find the initial velocity of the second.

$v_2=\frac{-120N}{14kg} = -8.6m/s$

The negative sign tells us the second ball is traveling in the opposite direction as the first, meaning it must be moving east.

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Question

A object moves to the right at . It collides head on with a object moving to the left at . Which statement is correct?

Answer

The total momentum before the collision is equal to the momentum of each object added together.

Remember that moving to the left means that the object has a negative velocity

Total momentum =

According to the law of conservation of momentum, the total momentum at the end must equal the total momentum at the beginning. Since the momentum at the beginning was , the momentum at the end would also be .

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Question

Two equal mass balls (one green and the other yellow) are dropped from the same height and rebound off of the floor. The yellow ball rebounds to a higher position. Which ball is subjected to the greater magnitude of impulse during its collision with the floor?

Answer

The impulse is equal to the change in momentum. The change in momentum is equal to the mass times the change in velocity.

The yellow ball rebounds higher and therefore has a higher velocity after the rebound. Since it has a higher velocity after the collision, the overall change in momentum is greater. Therefore since the change in momentum is greater, the impulse is higher.

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Question

A tennis ball strikes a racket, moving at . After striking the racket, it bounces back at a speed of . What is the change in momentum?

Answer

The change in momentum is the final momentum minus the initial momentum, or $\Delta p=p_{final}-p_{initial}$ .

Notice that the problem gives us the final SPEED of the ball but not the final VELOCITY. Since the ball "bounced back," it begins to move in the opposite direction, so its velocity at this point will be negative.

Plug in our values to solve:

$\Delta p=p_{final}-p_{initial}$

$\Delta p=(0.12kg-45m/s)-(0.12kg35m/s)$

$\Delta p=-5.4kgm/s-4.2kgm/s$

Compare your answer with the correct one above

Question

A crate slides along the floor for before stopping. If it was initially moving with a velocity of , what is the force of friction?

Answer

The fastest way to solve a problem like this is with momentum.

Remember that momentum is equal to mass times velocity: . We can rewrite this equation in terms of force.

Using this transformation, we can see that momentum is also equal to force times time.

$\Delta p=F\Delta t$ can also be thought of as $(p_{final}-p_{initial} =F\Delta t)$ .

Expand this equation to include our given values.

$(mv_{final}-mv_{initial})=F\Delta t$

Since the object is not moving at the end, its final velocity is zero. Plug in the given values and solve for the force.

$(-600kg \cdot m/s)=F(10s)$

$\frac{(-600kg\cdot m/s)}{10s}=F$

We would expect the answer to be negative because the force of friction acts in the direction opposite to the initial velocity.

Compare your answer with the correct one above

Question

A crate slides along a floor with a starting velocity of . If the force due to friction is , how long will it take for the box to come to rest?

Answer

The fastest way to solve a problem like this is with momentum.

Remember that momentum is equal to mass times velocity: . We can rewrite this equation in terms of force.

Using this transformation, we can see that momentum is also equal to force times time.

$\Delta p=F\Delta t$ can also be thought of as $(p_{final}-p_{initial})=F\Delta t$ .

Expand this equation to include our given values.

$(mv_{final}-mv_{initial})=F\Delta t$

Since the box is not moving at the end, its final velocity is zero. Plug in the given values and solve for the time.

$(100kg0m/s)-(100kg21m/s)=(-8N)\Delta t$

$(-100kg*21m/s)=(-8N)\Delta t$

$(-2100kg \cdot m/s)=(-8N)\Delta t$

$(-2100kg \cdot m/s)-8N=\Delta t$

$262.5s= \Delta t$

Compare your answer with the correct one above

Question

The area under the curve on a Force versus time (F versus t) graph represents

Answer

If we were to examine the area under the curve of a constant force applied over a certain amount of time we would have a graph with a straight horizontal line. To find the area of that rectangle we would multiply the base times the height. The base would be the time (number of seconds the force was applied). The height would be the amount of force applied during this time. Force*time is equal to the impulse acting on the object which is equal to the change in momentum of the object.

Compare your answer with the correct one above

Question

A man with a mass of m is painting a house. He stands on a tall ladder of height h. He leans over and falls straight down off the ladder. If he is in the air for s seconds, what will be his momentum right before he hits the ground?

Answer

The problem tells us he falls vertically off the ladder (straight down), so we don't need to worry about motion in the horizontal direction.

The equation for momentum is:

We can assume he falls from rest, which allows us to find the initial momentum.

.

From here, we can use the formula for impulse:

$\Delta p=F\Delta t$

$(p_f-p_i)=F\Delta t$

We know his initial momentum is zero, so we can remove this variable from the equation.

$p_f=F\Delta t$

The problem tells us that his change in time is s seconds, so we can insert this in place of the time.

The only force acting upon man is the force due to gravity, which will always be given by the equation .

Compare your answer with the correct one above

Question

If an egg is dropped on concrete, it usually breaks. If an egg is dropped on grass, it may not break. What conclusion explains this result?

Answer

In both cases the egg starts with the same velocity, so it has the same initial momentum. In both cases the egg stops moving at the end of its fall, so it has the same final velocity. The only thing that changes is the time and force of the impact. The force is produced by the deceleration resulting from the time that the egg is in contact with its point of impact.

As the time of contact increases, acceleration decreases and force decreases.

As the time of contact decreases, acceleration increases and force increases.

$a=\Delta v\Delta t$

The grass will have less force on the egg, allowing for a lesser acceleration, due to a longer period of impact.

Compare your answer with the correct one above

Question

A crate slides along the floor for before stopping. If it was initially moving with a velocity of $3\frac{m}{s}$ , what is the force of friction?

Answer

The fastest way to solve a problem like this is with momentum.

Remember that momentum is equal to mass times velocity: . We can rewrite this equation in terms of force.

$mv=(kg)(\frac{m}{s})=(kg)(\frac{m}{s})(\frac{s}{s})$

$\frac{kgm}{s^2}s=Ns=Ft$

Using this transformation, we can see that momentum is also equal to force times time.

$\Delta p=F\Delta t$ can also be thought of as $(p_{final}-p_{initial})=F\Delta t$ .

Expand this equation to include our given values.

$(mv_{final}-mv_{initial})=F\Delta t$

Since the crate is not moving at the end, its final velocity is zero. Plug in the given values and solve for the force.

$(100kg* 0\frac{m}s{})-(100kg3\frac{m}{s})=F(3s)$

$(-100kg3\frac{m}{s})=F(3s)$

$(-300\frac{kgm}{s})=F(3s)$

$\frac{(-300\frac{kgm}{s})}{3s}}=F$

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Question

A crate slides along the floor for before stopping. If it was initially moving with a velocity of $12\frac{m}{s}$ , what is the force of friction?

Answer

The fastest way to solve a problem like this is with momentum.

Remember that momentum is equal to mass times velocity: . We can rewrite this equation in terms of force.

$mv=(kg)(\frac{m}{s})=(kg)(\frac{m}{s})(\frac{s}{s})$

$\frac{kgm}{s^2}s=Ns=Ft$

Using this transformation, we can see that momentum is also equal to force times time.

$\Delta p=F\Delta t$ can also be thought of as $(p_{final}-p_{initial})=F\Delta t$ .

Expand this equation to include our given values.

$(mv_{final}-mv_{initial})=F\Delta t$

Since the object is not moving at the end, its final velocity is zero. Plug in the given values and solve for the force.

$(50kg* 0\frac{m}{s})-(50kg* 12\frac{m}{s})=F(10s)$

$(-50kg* 12\frac{m}{s})=F(10s)$

$(-600\frac{kg\cdot m}{s})=F(10s)$

$\frac{(-600\frac{kg\cdot m}{s})}{10s}}=F$

We would expect the answer to be negative because the force of friction acts in the direction opposite to the initial velocity.

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Question

A crate slides along a floor with a starting velocity of $11\frac{m}{s}$ . If the force due to friction is , how long will it take before stopping?

Answer

The fastest way to solve a problem like this is with momentum.

Remember that momentum is equal to mass times velocity: . We can rewrite this equation in terms of force.

$mv=(kg)(\frac{m}{s})=(kg)(\frac{m}{s})(\frac{s}{s})$

$\frac{kgm}{s^2}s=Ns=Ft$

Using this transformation, we can see that momentum is also equal to force times time.

$\Delta p=F\Delta t$ can also be thought of as $(p_{final}-p_{initial})=F\Delta t$ .

Expand this equation to include our given values.

$(mv_{final}-mv_{initial})=F\Delta t$

Since the crate is not moving at the end, its final velocity is zero. Plug in the given values and solve for the time that it is in motion.

$(300kg0\frac{m}{s})-(300kg11\frac{m}{s})=-35N\Delta t$

$(-300kg11\frac{m}{s})=-35N\Delta t$

$(-3300\frac{kgm}{s})=-35N\Delta t$

$\frac{(-3300\frac{kgm}{s})}{-35N}= \Delta t$

$94.29s= \Delta t$

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Question

A ball hits a brick wall with a velocity of $10\frac{m}{s}$ and bounces back at the same speed. If the ball is in contact with the wall for , what is the force exerted by the wall on the ball?

Answer

The fastest way to solve a problem like this is with momentum.

Remember that momentum is equal to mass times velocity: . We can rewrite this equation in terms of force.

$mv=(kg)(\frac{m}{s})=(kg)(\frac{m}{s})(\frac{s}{s})$

$\frac{kgm}{s^2}s=Ns=Ft$

Using this transformation, we can see that momentum is also equal to force times time.

$\Delta p=F\Delta t$ can also be thought of as $(p_{final}-p_{initial})=F\Delta t$ .

Expand this equation to include our given values.

$(mv_{final}-mv_{initial})=F\Delta t$

Even though the ball is bouncing back at the same "speed", its velocity will now be negative as it is moving in the opposite direction. Using these given values, we can solve for the force that acts on the ball.

$(0.5kg-10\frac{m}{s})-(0.5kg10\frac{m}{s})=F(0.1s)$

$(-5\frac{kgm}{s})-(5\frac{kgm}{s})=F(0.1s)$

$(-10\frac{kgm}{s})=F(0.1s)$

$\frac{(-10\frac{kgm}{s})}{0.1s}=F$

Our answer is negative because the force is moving the ball in the OPPOSITE direction from the way it was originally heading.

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