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High School Physics : Calculating Momentum

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Example Questions

Example Question #31 : Calculating Momentum

A 2kg hammer moving with a velocity of $12\frac{m}{s}$ strikes a nail. The two are in contact for 0.2s , after which the hammer has a velocity of $0\frac{m}{s}$ . What is the force of the hammer on the nail?

Possible Answers:

-120N

130N

120N

-60N

60N

Correct answer:

120N

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.

$(2kg*0\frac{m}{s})-(2kg*12\frac{m}{s})=F*(0.2s)$

$(-2kg*12\frac{m}{s})=F*(0.2s)$

$(-24\frac{kg *m}{s})=F*(0.2s)$

$\frac{-24\frac{kg *m}{s}}{0.2s}=F$

-120N=F

This equation solves for the force of the nail on the hammer, as we were looking purely at the momentum of the hammer.

According to Newton's third law, $F_{hammer}=-F_{nail}$ . This means that if the nail exerts -120N of force on the hammer, then the hammer must exert 120N of force on the nail.

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Example Question #52 : Momentum

A 1.8kg hammer moving with a velocity of $11\frac{m}{s}$ strikes a nail, after which the hammer has a velocity of $0\frac{m}{s}$ . If the hammer strikes the nail with 160N of force, how long were the two in contact?

Possible Answers:

0.12s

1.1s

0.06s

0.11s

0.33s

Correct answer:

0.12s

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. This allows us to set up the left side of our equation.

$(1.8kg*0\frac{m}{s})-(1.8kg*11\frac{m}{s})=F*\Delta t$

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 for our equation since 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 160N of force on the nail, then the nail must exert -160N of force on the hammer.

We can plug that value in for force and solve our equation for time.

$(0)-(1.8kg*11\frac{m}{s})=(-160N)(\Delta t)$

$(-1.8kg*11\frac{m}{s})=(-160N)(\Delta t)$

$-19.8\frac{kg*m}{s}=-160N*\Delta t$

$\frac{-19.8\frac{kg*m}{s}}{-160N}=\Delta t$

$0.12s=\Delta t$

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Example Question #53 : Momentum

A 0.5kg 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 0.1s , what is the force exerted by the wall on the ball?

Possible Answers:

-20N

-5N

-100N

-10N

-200N

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.

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

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.

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

$(-5\frac{kg*m}{s})-(5\frac{kg*m}{s})=F(0.1s)$

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

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

-100N=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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Example Question #31 : Calculating Momentum

A 13kg object moving at $2\frac{m}{s}$ strikes a 7kg object at rest. After the collision, the first object is motionless. What is the velocity of the second object?

Possible Answers:

$1.3\frac{m}{s}$

$0.27\frac{m}{s}$

$3.71\frac{m}{s}$

$1.1\frac{m}{s}$

$182\frac{m}{s}$

Correct answer:

$3.71\frac{m}{s}$

Explanation:

We need to solve this question be using conservation of momentum. (Note that the question can also be solved by using conservation of kinetic energy).

m_1v_1+m_2v_2=m_1v_3+m_2v_4

In this problem, the first object transfers all of its momentum to the second object. The initial velocity of the second object is zero, and the final velocity of the first object is zero. This means that:

m_1v_1=m_2v_2

Plug in the given values for the masses and initial velocity of the first mass to solve for the final velocity of the second mass.

m_1v_1=m_2v_2

$(13kg)(2\frac{m}{s})=(7kg)v_2$

$26\frac{kg*m}{s}=(7kg)v_2$

$\frac{26\frac{kg*m}{s}}{7kg}=v_2$

$3.7\frac{m}{s}=v_2$

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Example Question #327 : Motion And Mechanics

A cat bats at a 0.08kg toy mouse. The mouse starts at rest and, after being struck, has a velocity of $12.1\frac{m}{s}$ . The cat's paw is in contact with the mouse for 0.22s . How much force did the cat exert on the mouse?

Possible Answers:

4.4N

0.23N

We need to know the acceleration of the mouse in order to solve

0.97N

0.21N

Correct answer:

4.4N

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 mouse starts at rest, the initial velocity must be zero. Using this information, the time of contact, the final velocity, and the given mass, we can solve for the force of the cat's paw on the mouse.

$(0.08kg)(12.1\frac{m}{s})-(0.08kg)(0\frac{m}{s})=F(0.22s)$

$0.968\frac{kg\cdot m}{s}-0\frac{kg\cdot m}{s}=F(0.22s)$

$0.968\ \frac{kg\cdot m}{s}=F(0.22s)$

$\frac{0.968\frac{kg\cdot m}{s}}{0.22s}=F$

4.4N=F

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Example Question #32 : Calculating Momentum

A 0.25kg ball hits a brick wall with a velocity of $30\frac{m}{s}$ and bounces back at the same speed. If the ball is in contact with the wall for 0.1s , what is the value of the force exerted by the wall on the ball?

Possible Answers:

100N

-150N

-100N

-300N

Correct answer:

-150N

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 this understanding we can solve for the force in our equation.

$(0.25kg* -30\frac{m}{s})-(0.25kg* 30\frac{m}{s})=F(0.1s)$

$(-7.5\frac{kg\cdot m}{s})-(7.5\frac{kg\cdot m}{s})=F(0.1s)$

$(-15\frac{kg\cdot m}{s})=F(0.1s)$

$\frac{(-15\frac{kg\cdot m}{s})}{0.1s}=F$

-150N=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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High School Physics : Calculating Momentum

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #31 : Calculating Momentum

Example Question #52 : Momentum

Example Question #53 : Momentum

Example Question #31 : Calculating Momentum

Example Question #327 : Motion And Mechanics

Example Question #32 : Calculating Momentum

All High School Physics Resources

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