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High School Physics : Motion and Mechanics

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Example Questions

Example Question #51 : Momentum

A 0.02kg ball moving at $30\frac{m}{s}$ strikes a piece of paper. It breaks through the paper and continues moving in the same direction. If the paper exerted a force of -10N on the ball and the two are in contact for 0.01s , what is the final momentum of the ball?

Possible Answers:

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

$0.5\frac{kg*m}{s}$

$0.1\frac{kg*m}{s}$

$1\frac{kg*m}{s}$

$0.75\frac{kg*m}{s}$

Correct answer:

$0.5\frac{kg*m}{s}$

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

Expand this equation to include our given values.

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

We are given the mass and initial velocity, allowing us to solve for the initial momentum, and the force and time, allowing us to evaluate the right side of the equation. Using these values we can solve for the final momentum of the ball.

$p_{final}-(0.02kg*30\frac{m}{s})=(-10N)(0.01s)$

$p_{final}-(0.6\frac{kg*m}{s})=(-0.1N*s)$

$p_{final}=-0.1\frac{kg*m}{s}+0.6\frac{kg*m}{s}$

$p_{final}=0.5\frac{kg*m}{s}$

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Example Question #30 : Calculating Momentum

Susan pushes a 1000kg car with 35N of force on a frictionless surface. How long does she need to push the car to get it to a velocity of $4\frac{m}{s}$ if it starts at rest?

Possible Answers:

122.1s

412s

11.4s

110s

114.3s

Correct answer:

114.3s

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 required to change the velocity to the given value.

$(1000kg*4\frac{m}{s})-(1000kg*0\frac{m}{s})=(35N)(\Delta t)$

$4000\frac{kg*m}{s}=35N*\Delta t$

$\frac{4000\frac{kg*m}{s}}{35N}=\Delta t$

$114.3s=\Delta t$

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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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Example Question #1 : Understanding Momentum

A 0.23kg baseball approaches a batter at $39.8\frac{m}{s}$ . The batter swings and hits a line drive. The ball leaves the bat at $50.23\frac{m}{s}$ . What is the impulse on the ball?

Possible Answers:

$-0.79\frac{kg\cdot m}{s}$

$-2.4\frac{kg\cdot m}{s}$

$1.26\frac{kg\cdot m}{s}$

$-20.7\frac{kg\cdot m}{s}$

$2.4\frac{kg\cdot m}{s}$

Correct answer:

$-20.7\frac{kg\cdot m}{s}$

Explanation:

Impulse is a change in momentum. Mathematically, we can say that $\Delta p=I$ .

We can expand this formula to include our terms for mass and velocity.

p_f-p_i=I

mv_f-mv_i=I

The important thing to be aware of in this problem is that we are given the speed of the ball, but not the velocity. Remember, when the baseball leaves the bat it will be going in the OPPOSITE direction. That means its velocity will be in the negative direction.

$v_f=-50.23\frac{m}{s}$

Plug in the given information and solve for the impulse.

mv_f-mv_i=I

$(0.23kg* -50.23\frac{m}{s})-(0.23kg* 39.8\frac{m}{s})=I$

$-11.55\frac{kg\cdot m}{s}-9.15\frac{kg\cdot m}{s}=I$

$-20.7\frac{kg\cdot m}{s}=I$

It makes since that the impulse would be negative because the change of velocity is an increase in magnitude AND a change in direction.

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Example Question #2 : Understanding Momentum

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?

Possible Answers:

The density of grass is less than the density of concrete

The egg is in contact with the grass for longer, so it absorbs less force

Grass is a softer texture

The coefficient of friction of the grass on the egg is less than the coefficient of the concrete on the egg

Grass has less mass than concrete

Correct answer:

The egg is in contact with the grass for longer, so it absorbs less force

Explanation:

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.

$a=\frac{\Delta v}{\Delta t}$

F=ma

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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High School Physics : Motion and Mechanics

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #51 : Momentum

Example Question #30 : Calculating Momentum

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

Example Question #1 : Understanding Momentum

Example Question #2 : Understanding Momentum

All High School Physics Resources

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