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High School Physics : Momentum

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Example Question #11 : Calculating Momentum

An 11.2kg ball moving at $39\frac{m}{s}$ strikes a 15kg ball at rest. After the collision the 11.2kg ball is moving with a velocity of $30\frac{m}{s}$ . What is the velocity of the second ball?

Possible Answers:

$7.6\frac{m}{s}$

$67.1\frac{m}{s}$

$5.9\frac{m}{s}$

$6.7\frac{m}{s}$

$3.4\frac{m}{s}$

Correct answer:

$6.7\frac{m}{s}$

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 $\small v_{2final}$ .

$(11.2kg*39\frac{m}{s})+(15kg*0\frac{m}{s})=(11.2kg*30\frac{m}{s})+(15kg*v_{2final})$

$436.8\frac{kg\cdot m}{s}+0=(336\frac{kg\cdot m}{s})+15kg( v_{2final})$

${436.8\frac{kg\cdot m}{s}}-{336\frac{kg\cdot m}{s}}=15kg( v_{2final})$

${100.8\frac{kg\cdot m}{s}}=15kg( v_{2final})$

$\frac{100.8\frac{kg\cdot m}{s}}{15kg}=v_{2final}$

$6.7\frac{m}{s}=v_{2final}$

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Example Question #41 : Momentum

A cat is playing with a 0.2kg toy mouse. He bats it with of force and is in contact with the toy for 0.1s . What is the new velocity of the mouse?

Possible Answers:

$0.2\frac{m}{s}$

$4\frac{m}{s}$

$2\frac{m}{s}$

$1\frac{m}{s}$

$0.5\frac{m}{s}$

Correct answer:

$1\frac{m}{s}$

Explanation:

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 $(p_{2}-p_{1})=F\Delta t$ .

Expand this equation to include our given values.

$(mv_{2}-mv_{1})=F\Delta t$

Since the toy starts at rest, we know the initial velocity.

$v_1=0\frac{m}{s}$

We are also given the mass, force, and time. Using these given values allows us to solve for the final velocity.

$0.2kg(v_2)-0.2kg(0\frac{m}{s})=2N(0.1s)$

$0.2kg(v_2)=2\frac{kg\cdot m}{s^2}( 0.1s)$

$0.2kg(v_2)=0.2\frac{kg\cdot m}{s}$

$v_2=\frac{0.2\frac{kg\cdot m}{s}}{0.2kg}$

$v_2=1\frac{m}{s}$

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Example Question #21 : Calculating Momentum

A 0.8kg hammer moving with a velocity of $12\frac{m}{s}$ strikes a nail, after which the hammer has a velocity of $0\frac{m}{s}$ . If the hammer strikes the nail with 190N of force, how long were the two in contact?

Possible Answers:

0.1s

0.05s

0.01s

1824s

Correct answer:

0.05s

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. We can plug in the given values on the left side of the equation.

$(0.8kg* 0\frac{m}{s})-(0.8kg*12\frac{m}{s})=F \Delta t$

The problem gave us the force of the hammer on the nail, but not the force of the nail on the hammer, which is what we need as we are looking purely at the momentum of the hammer.

Fortunately, Newton's third law can help us. It states that $F_{hammer}=-F_{nail}$ . This means that if the hammer exerts 190N of force on the nail, then the nail must exert -190N of force on the hammer.

We can plug that value in for the force, and solve for the time.

$(0)-(0.8kg* 12\frac{m}{s})=(-190N)(\Delta t)$

$(-0.8kg* 12\frac{m}{s})=(-190N)(\Delta t)$

$-9.6\frac{kg\cdot m}{s}=(-190N) \Delta t$

$\frac{-9.6\frac{kg\cdot m}{s}}{-190N}=\Delta t$

$0.05s=\Delta t$

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Example Question #41 : Momentum

Susan pushes a car 1412kg with 18N of force along a frictionless surface. How long does she need to push the car to get it to a velocity of $2\frac{m}{s}$ , if it starts at rest?

Possible Answers:

111.23s

156.89s

33.45s

96.23s

301.22s

Correct answer:

156.89s

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

$(1412kg* 2\frac{m}{s})-(1412kg* 0\frac{m}{s})=(18N)(\Delta t)$

$2824\frac{kg\cdot m}{s}=18N* \Delta t$

$\frac{2824\frac{kg\cdot m}{s}}{18N}=\Delta t$

$156.89s=\Delta t$

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Example Question #381 : High School Physics

A 0.12kg tennis ball strikes a racket, moving at $35\frac{m}{s}$ . After striking the racket, it bounces back at a speed of $45\frac{m}{s}$ . What is the change in momentum?

Possible Answers:

$-19.2\frac{kg*m}{s}$

$1.2\frac{kg*m}{s}$

$4.8\frac{kg*m}{s}$

$-0.6\frac{kg*m}{s}$

$-9.6\frac{kg*m}{s}$

Correct answer:

$-9.6\frac{kg*m}{s}$

Explanation:

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:

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

$\Delta p =(0.12kg*-45\frac{m}{s})-(0.12kg*35\frac{m}{s})$

$\Delta p=-5.4\frac{kg*m}{s}-4.2\frac{kg*m}{s}$

$\Delta p = -9.6\frac{kg*m}{s}$

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Example Question #24 : Calculating Momentum

A 0.12kg tennis ball strikes a racket, moving at $35\frac{m}{s}$ . After striking the racket, it bounces back at a speed of $45\frac{m}{s}$ . If the ball is in contact with the racket for 0.12s , how much force did the racket exert on the ball?

Possible Answers:

1.152N

40N

u -80N

810N

Correct answer:

u -80N

Explanation:

The change in momentum is equal to the force exerted on the ball times time, or $\Delta p = F*\Delta t$ .

It is also equal to 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 "bounces back," it is moving in the opposite direction, so its velocity at this point will be negative.

Plug in our values to solve:

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

$\Delta p =(0.12kg*-45\frac{m}{s})-(0.12kg*35\frac{m}{s})$

$\Delta p=-5.4\frac{kg*m}{s}-4.2\frac{kg*m}{s}$

$\Delta p = -9.6\frac{kg*m}{s}$

Plug that into our first equation to solve for the force the racket exerted on the ball:

$\Delta p = F*\Delta t$

$-9.6\frac{kg*m}{s} = F*0.12s$

$\frac{-9.6\frac{kg*m}{s}}{0.12s}{} = F$

$-80\frac{kg*m}{s^{2}}=F$

Since $N = \frac{kg*m}{s^{2}}$ , we can rewrite our answer as:

-80N=F

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Example Question #25 : Calculating Momentum

A 0.12kg tennis ball moving at $35\frac{m}{s}$ strikes a racket. After striking the racket, it bounces back at a speed of $45\frac{m}{s}$ . If the racket exerts 160N of force on the ball, how long were they in contact?

Possible Answers:

0.06s

0.12s

150.4s

16.67s

9.6s

Correct answer:

0.06s

Explanation:

The change in the ball's momentum is equal to the force exerted on the ball times time, or $\Delta p = F*\Delta t$ .

It is also equal to 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 is moving in the opposite direction, and its velocity at this point will be negative.

Plug in our values to solve:

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

$\Delta p =(0.12kg*-45\frac{m}{s})-(0.12kg*35\frac{m}{s})$

$\Delta p=-5.4\frac{kg*m}{s}-4.2\frac{kg*m}{s}$

$\Delta p = -9.6\frac{kg*m}{s}$

Plug that into our first equation:

$\Delta p = F*\Delta t$

$-9.6\frac{kg*m}{s}=-160N*\Delta t$

Since our force is going in the same direction as our change in momentum, make both negative.

$\frac{-9.6\frac{kg*m}{s}}{-160N}=\Delta t$

Since $N = \frac{kg*m}{s^{2}}$ , we can rewrite our equation's units as:

$\frac{-9.6 \frac{kg*m}{s}}{-160\frac{kg*m}{s^{2}}}=\Delta t$

Then, we can simplify the equation's units to get our final answer:

$0.06s=\Delta t$

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Example Question #26 : Calculating Momentum

Possible Answers:

$-80\frac{m}{s^2}$

$-9.6\frac{m}{s^2}$

$-666.67\frac{m}{s^2}$

$-1.2\frac{m}{s^2}$

$-1200\frac{m}{s^2}$

Correct answer:

$-666.67\frac{m}{s^2}$

Explanation:

First, we need to find the force applied to the ball since F=ma . We know the mass, but now we need the force.

The change in momentum is equal to the force exerted on the ball times time, or $\Delta p = F*\Delta t$ .

The change in momentum is also equal to 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 "bounces back," is moving in the opposite direction, and its velocity will be negative at this point.

Plug in our values to solve:

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

$\Delta p =(0.12kg*-45\frac{m}{s})-(0.12kg*35\frac{m}{s})$

$\Delta p=-5.4\frac{kg*m}{s}-4.2\frac{kg*m}{s}$

$\Delta p = -9.6\frac{kg*m}{s}$

Plug that into our other equation:

$\Delta p = F*\Delta t$

$-9.6\frac{kg*m}{s} = F*0.12s$

$\frac{-9.6\frac{kg*m}{s}}{0.12s}{} = F$

$-80\frac{kg*m}{s^{2}}=F$

Since $N=\frac{kg*m}{s^{2}}$ , we can rewrite our answer as:

-80N=F

Plug that into our first equation:

F=ma

-80N=0.12kg*a

$\frac{-80N}{0.12kg}=a$

$-666.67\frac{N}{kg}=a$

Since $N=\frac{kg*m}{s^{2}}$ , we can rewrite our answer as:

$-666.67\frac{\frac{kg*m}{s^{2}}}{kg}=a$

Then, we can reduce our answer's units to:

$-666.67\frac{m}{s^2}=a$

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Example Question #27 : Calculating Momentum

A 100kg 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?

Possible Answers:

-300N

-30N

-10N

-100N

10N

Correct answer:

-100N

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.

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

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

$(-300\frac{kg*m}{s})=F(3s)$

$\frac{(-300\frac{kg*m}{s})}{3s}}=F$

-100N=F

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Example Question #41 : Momentum

A 300kg crate slides along a floor with a starting velocity of $11\frac{m}{s}$ . If the force due to friction is -35N , how long will it take before stopping?

Possible Answers:

35s

942.9s

35.5s

-90s

94.29s

Correct answer:

94.29s

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 that it is in motion.

$(300kg*0\frac{m}{s})-(300kg*11\frac{m}{s})=-35N*\Delta t$

$(-300kg*11\frac{m}{s})=-35N*\Delta t$

$(-3300\frac{kg*m}{s})=-35N*\Delta t$

$\frac{(-3300\frac{kg*m}{s})}{-35N}= \Delta t$

$94.29s= \Delta t$

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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 #11 : Calculating Momentum

Example Question #41 : Momentum

Example Question #21 : Calculating Momentum

Example Question #41 : Momentum

Example Question #381 : High School Physics

Example Question #24 : Calculating Momentum

Example Question #25 : Calculating Momentum

Example Question #26 : Calculating Momentum

Example Question #27 : Calculating Momentum

Example Question #41 : Momentum

All High School Physics Resources

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