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High School Physics : Momentum

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Example Question #31 : Momentum

A 175kg crate slides along the floor for before stopping. If it was initially moving with a velocity of $7.7\frac{m}{s}$ , what is the force of friction?

Possible Answers:

-336.88N

-612.88N

-332.81N

-181.44N

-346.12N

Correct answer:

-336.88N

Explanation:

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

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

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

$\frac{kg*m}{s^2}*s=N*s=Ft$

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.

$(175kg* 0\frac{m}{s})-(175kg* 7.7\frac{m}{s})=F(4s)$

$-(175kg* 7.7\frac{m}{s})=F(4s)$

$(-1347.5\frac{kg\cdot m}{s})=F(4s)$

$\frac{(-1347.5\frac{kg\cdot m}{s})}{4s}}=F$

-336.88N=F

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Example Question #32 : Momentum

A 3.2kg hammer moving with a velocity of $17\frac{m}{s}$ strikes a nail. The two are in contact for 0.02s , after which the hammer has a velocity of $0\frac{m}{s}$ . How much force went into the nail?

Possible Answers:

1123N

1510N

4781N

3220N

2720N

Correct answer:

2720N

Explanation:

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

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

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

$\frac{kg*m}{s^2}*s=N*s=Ft$

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

$(3.2kg* 0\frac{m}{s)}-(3.2kg* 17\frac{m}{s})=(F)(0.02s)$

$-(3.2kg* 17\frac{m}{s})=(F)(0.02s)$

$(-54.4\frac{kg \cdot m}{s})=(F)(0.02s)$

$\frac{-54.4\frac{kg \cdot m}{s}}{0.02s}=F$

-2720N=F

This equation solves for the force of the nail on the hammer, as we were looking purely at the momentum of the hammer; however, we need to find the force of the hammer on the nail. Newton's third law states that $F_{hammer}=-F_{nail}$ .

This means that if the nail exerts -2720N of force on the hammer, then the hammer must exert 2720N of force on the nail; therefore, our answer will be 2720N .

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Example Question #301 : Motion And Mechanics

A 117kg crate slides along a floor with a starting velocity of $21\frac{m}{s}$ . If the force due to friction is -330N , how long will it take before the crate slides to a stop?

Possible Answers:

6.6s

9.2s

11.2s

7.5s

14.5

Correct answer:

7.5s

Explanation:

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

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

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

$\frac{kg*m}{s^2}*s=N*s=Ft$

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.

$(117kg* 0\frac{m}{s})-(117kg* 21\frac{m}{s})=-330N* \Delta t$

$-(117kg* 21\frac{m}{s})=-330N* \Delta t$

$(-2457\frac{kg\cdot m}{s})=-330N* \Delta t$

$\frac{(-2457\frac{kg\cdot m}{s})}{-330N}= \Delta t$

$7.5s= \Delta t$

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Example Question #34 : Momentum

Courtney pushes a car 1200kg with 40N of force along a frictionless surface. If the car is initially at rest, how long does she need to push for the car to reach a velocity of $10\frac{m}{s}$ ?

Possible Answers:

150s

300s

500s

600s

200s

Correct answer:

300s

Explanation:

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

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

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

$\frac{kg*m}{s^2}*s=N*s=Ft$

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 car is not moving at the beginning, its initial velocity is zero. Plug in the given values and solve for the time.

$(1200kg* 10\frac{m}{s})-(1200kg* 0\frac{m}{s})=(40N)\Delta t$

$12000\frac{kg\cdot m}{s}=(40N) \Delta t$

$\frac{12000\frac{kg\cdot m}{s}}{40N}=\Delta t$

$300s=\Delta t$

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Example Question #35 : Momentum

A 0.05kg ball hits a brick wall with a velocity of $9.7\frac{m}{s}$ and bounces back at the same speed. If the ball is in contact with the wall for 0.2s , what is the value of the force exerted by the wall on the ball?

Possible Answers:

-19.40N

-4.85N

-4.35N

-9.70N

-5.98N

Correct answer:

-4.85N

Explanation:

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

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

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

$\frac{kg*m}{s^2}*s=N*s=Ft$

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 what we know from the question, we can solve for the force.

$(0.05kg*-9.7\frac{m}{s})-(0.05kg* 9.7\frac{m}{s})=F(0.2s)$

$(-0.485\frac{kg\cdot m}{s})-(0.485\frac{kg\cdot m}{s})=F(0.2s)$

$(-0.97\frac{kg\cdot m}{s})=F(0.2s)$

$\frac{(-0.97\frac{kg\cdot m}{s})}{0.2s}=F$

-4.85N=F

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

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Example Question #36 : Momentum

A 0.75kg ball strikes a 2.5kg ball at rest. After the collision the 0.75kg ball is moving with a velocity of $20\frac{m}{s}$ and the second ball is moving with a velocity of $22\frac{m}{s}$ . What is the initial velocity of the first ball?

Possible Answers:

$93.33\frac{m}{s}$

$101.33\frac{m}{s}$

$83.67\frac{m}{s}$

$55.17\frac{m}{s}$

$99.25\frac{m}{s}$

Correct answer:

$93.33\frac{m}{s}$

Explanation:

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.

$m_1v_{1initial}+m_2v_{2initial}=m_1v_{1final}+m_2v_{2final}$

Plug in the given values and solve for $\small v_{1initial}$ .

$(0.75kg*v_{1initial})+(2.5kg* 0\frac{m}{s})=(0.75kg*20\frac{m}{s})+(2.5kg* 22\frac{m}{s})$

$(0.75kg* v_{1initial})+(0)=(15\frac{kg\cdot m}{s} )+(55\frac{kg\cdot m}{s})$

$(0.75kg* v_{1initial})=(70\frac{kg\cdot m}{s} )$

$v_{1initial}=\frac{70\frac{kg\cdot m}{s}}{0.75kg}$

$v_{1initial}=93.33\frac{m}{s}$

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Example Question #31 : Momentum

A 0.05kg ball strikes a piece of paper moving at $40\frac{m}{s}$ . It breaks through the paper and continues in the same direction. If the paper exerted a force of -18N on the ball and the two are in contact for 0.01s , what is the final momentum of the ball?

Possible Answers:

$0.91\frac{kg\cdot m}{s}$

$1.82\frac{kg\cdot m}{s}$

$1.13\frac{kg\cdot m}{s}$

$3.64\frac{kg\cdot m}{s}$

$2.42\frac{kg\cdot m}{s}$

Correct answer:

$1.82\frac{kg\cdot m}{s}$

Explanation:

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

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

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

$\frac{kg*m}{s^2}*s=N*s=Ft$

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

Since we are solving for $p_{final}$ , we can leave that part of the equation alone and expand the rest to include our given values.

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

From here, plug in the given values and solve for the final momentum.

$p_{final}-(0.05kg*40\frac{m}{s})=(-18N)(0.01s)$

$p_{final}-(2\frac{kg\cdot m}{s})=(-0.18N\cdot s)$

At this point, remember that $1N=1\frac{kg\cdot m}{s^2}$ to see that both sides are now working in the same units.

$p_{final}=-0.18\frac{kg\cdot m}{s}+2\frac{kg\cdot m}{s}$

$p_{final}=1.82\frac{kg\cdot m}{s}$

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Example Question #11 : Calculating Momentum

A 12kg ball moving at $76.4\frac{m}{s}$ strikes a second ball at rest. After the collision the 12kg ball is moving with a velocity of $22.2\frac{m}{s}$ and the second ball is moving with a velocity of $13.7\frac{m}{s}$ . What is the mass of the second ball?

Possible Answers:

45.8kg

92.3kg

31.2kg

11.7kg

29.3kg

Correct answer:

29.3kg

Explanation:

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.

$m_1v_{1initial}+m_2v_{2initial}=m_1v_{1final}+m_2v_{2final}$

Plug in the given values and solve for $\small m_2$ .

$(12kg*76.4\frac{m}{s})+(m_2*0\frac{m}{s})=(12kg* 22.2\frac{m}{s})+(m_2* 13.7\frac{m}{s})$

$(916.8\frac{kg\cdot m}{s})+(0)=(266.4\frac{kg\cdot m}{s})+m_2(22.2\frac{m}{s})$

$(916.8\frac{kg\cdot m}{s})-(266.4\frac{kg\cdot m}{s})=m_2(22.2\frac{m}{s})$

$(650.4\frac{kg\cdot m}{s})=m_2( 22.2\frac{m}{s})$

$\frac{650.4\frac{kg\cdot m}{s}}{22.2\frac{m}{s}}=m_2$

29.3kg=m_2

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Example Question #39 : Momentum

A 3400kg car travelling at $28.7\frac{m}{s}$ rear ends another 3400kg car at rest. The two bumpers lock and the cars move forward together. What is their final velocity?

Possible Answers:

$28.7\frac{m}{s}$

$19.5\frac{m}{s}$

$15.1\frac{m}{s}$

$14.4\frac{m}{s}$

$13.3\frac{m}{s}$

Correct answer:

$14.4\frac{m}{s}$

Explanation:

This is an example of an inelastic collision, as the two cars stick together after colliding. We can assume momentum is conserved.

To make the equation easier, let's call the first car "1" and the second car "2."

Using conservation of momentum and the equation for momentum, p=mv , we can set up the following equation.

$m_1v_{1initial}+m_2v_{2initial}=(m_1+m_2)v_{final}$

Since the cars stick together, they will have the same final velocity. We know the second car starts at rest, and the velocity of the first car is given. Plug in these values and solve for the final velocity.

$(3400kg* 28.7\frac{m}{s})+(3400kg*0\frac{m}{s})=(3400kg+3400kg)v_{final}$

$97580\frac{kg\cdot m}{s}=(6800kg)v_{final}$

$\frac{97580\frac{kg\cdot m}{s}}{6800kg}=v_{final}$

$14.4\frac{m}{s}=v_{final}$

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Example Question #31 : Momentum

A 950kg strikes a 2200kg car at rest from behind. The bumpers lock and they move forward together. If the final velocity of the joined cars is equal to $15\frac{m}{s}$ , what was the initial speed of the first car?

Possible Answers:

$71.2\frac{m}{s}$

$50.2\frac{m}{s}$

$46.2\frac{m}{s}$

$33.2\frac{m}{s}$

$49.7\frac{m}{s}$

Correct answer:

$49.7\frac{m}{s}$

Explanation:

This is an example of an inelastic collision, as the two cars stick together after colliding. We can assume momentum is conserved.

To make the equation easier, let's call the first car "1" and the second car "2."

Using conservation of momentum and the equation for momentum, p=mv , we can set up the following equation.

$m_1v_{1initial}+m_2v_{2initial}=(m_1+m_2)v_{final}$

Since the cars stick together, they will have the same final velocity. We know the second car starts at rest, and the final velocity of the joined cars is given. Plug in these values and solve for the initial velocity of the first car.

$(950kg*v_{1initial})+(2200kg* 0\frac{m}{s})=(950kg+2200kg)* 15\frac{m}{s}$

$950kg* v_{1initial}=(3150kg)*15\frac{m}{s}$

$950kg* v_{1initial}=47250\frac{kg\cdot m}{s}$

$v_{1initial}=\frac{47250\frac{kg\cdot m}{s}}{950kg}$

$v_{1initial}=49.7\frac{m}{s}$

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High School Physics : Momentum

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #31 : Momentum

Example Question #32 : Momentum

Example Question #301 : Motion And Mechanics

Example Question #34 : Momentum

Example Question #35 : Momentum

Example Question #36 : Momentum

Example Question #31 : Momentum

Example Question #11 : Calculating Momentum

Example Question #39 : Momentum

Example Question #31 : Momentum

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