Understanding Conservation of Momentum

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Physics › Understanding Conservation of Momentum

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Two identical billiard balls traveling at the same speed have a head-on collision and rebound. If the balls had twice the mass, but maintained the same size and speed, how would the rebound be different?

They would rebound at a higher speed

They would rebound at a slower speed

No difference

Explanation

Consider the law of conservation of momentum.

$m_{1}v_{i1}+m_{2}v_{i2}=m_{1}f1+m_{2}v_{f2}$

If both balls are identical then we can say that

$m_{1}=m_{2}=M$

Therefore we can state the equation as

$Mv_{i1}+Mv_{i2}=Mf1+Mv_{f2}$

Since is the common factor we can remove it from the equation completely.

$v_{i1+v_{i2}}=f1+v_{f2}$

Since mass factors out of the equation; then it does not matter if the balls increase or decrease in their mass.

You are lying in bed and want to shut your bedroom door. You have a bouncy ball and a blob of clay, both with the same mass. Which one would be more effective to throw at your door to close it?

The bouncy ball

The blob of clay

Both the same

Neither will work

Explanation

A bouncy ball will have an elastic collision with the door, causing the ball to move backward at the same speed it hit the door.

On the other hand, the blob of clay will have an inelastic collision with the door, causing the blob of clay to move with the same speed as the door.

So let us look at the conservation of momentum for an elastic collision.

$m_{1}v_{i1}+m_{2}v_{i2}=m_{1}v_{f1}+m_{2}v_{f2}$

Since the door has no initial velocity we can remove it from the beginning of the equation.

$m_{1}v_{i1}=m_{1}v_{f1}+m_{2}v_{f2}$

Since the ball will rebound with the same amount that it hits the door the velocity at the end is a negative of the velocity at the beginning.

$m_{1}v_{i1}=m_{1}(-v_{i1})+m_{2}v_{f2}$

Rearrange for the final velocity of the door.

$m_{1}v_{i1}+m_{1}(v_{i1})=m_{2}v_{f2}$

$2m_{1}v_{i1}=m_{2}v_{f2}$

$\frac{2m_{1}v_{i1}}{m_{2}}=v_{f2}$

Now let us examine the law of conservation of momentum for the inelastic collision. Again the door has no initial velocity so we can remove it from the beginning of the equation.

$m_{1}v_{i1}=m_{1}v_{f1}+m_{2}v_{f2}$

Since the collision is inelastic, the final velocity of both objects will be the same so we can set them equal to each other.

$m_{1}v_{i1}=m_{1}v_{f2}+m_{2}v_{f2}$

Now solve for the final velocity of the door.

$m_{1}v_{i1}=(m_{1}+m_{2})v_{f2}$

When we compare these two final velocities, it is clear that the elastic collision will create a larger velocity for the door because the top number is twice the value as the inelastic collision, and it is being divided by only the mass of the second object, instead of both objects combined.

A 12kg hammer strikes a nail at a velocity of 7.5m/s and comes to rest in a time interval of 8.0ms What is the average force acting on the nail?

1150N

8750N

11250N

15350N

Explanation

Knowns

$v_{i}=7.5m/s$

$v_{f}=0m/s$

First, let us find the change in the momentum of an object.

$\delta p=m(v_{f}-v_{i})$

$\delta p=(12kg)(0m/s-7.5m/s)$

$\delta p=90kgm/s$

We know that the change in momentum of an object is equal to the impulse of an object. Impulse is equal to the force acting on the object multiplied by the time that force is applied.

$\delta p=I=F\delta t$

$90kgm/s=F\delta t$

We need to convert our time from milliseconds to seconds.

$8ms*\frac{1s}{1000ms}=0.008s$

Solve for the force

$F=\frac{90kgm/s}{0.008s}$

A 0.145kg baseball pitched horizontally at 27m/s strikes a bat and pops straight up to a height of 31.5m. If the contact time between the bat and ball is 2.5ms, calculate the average force between the ball and bat during contact.

1215N

1438N

1566N

2126N

2436N

Explanation

To figure this out we needed to consider the change in momentum of the ball.

We know that the impulse of the ball is equal to the change in momentum.

$I=\delta p$

The impulse is equal to the force times the time the force is applied.

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

$\delta p=m\delta v$

Therefore we can combine these equations to say

$Ft=m\delta v$

So let us look at what is happening in the x-direction.

$F(2.5x10^{-3}s)=(0.145kg)(0m/s-27m/s)$

Rearrange and solve the force in the x-direction.

Next, let us determine what is happening in the y-direction. We will need to figure out the initial velocity of the ball in the y-direction using the height. At the peak, we also know the velocity is 0m/s. We also know that the acceleration due to gravity is .

$v^{2}_{f}=v^{2}_{i}+2a\delta x$

$(0m/s)^{2}=v^{2}_{i}+2(-9.8m/s/s)(31.5m)$

$0=v^{2}_{i}-617.4$

$617.4=v^{2}_{i}$

$24.8m/s=v^{2}_{i}$

Once we have the initial velocity, we can now determine the force in the y-direction.

$F(2.5x10^{-3}s)=(0.145kg)(24.8m/s-0m/s)$

Now that we have the force in both the x and y direction we can determine the overall resultant force using the Pythagorean Theorem.

$(-1566N)^{2}+(1438.4N)^{2}=F^{2}$

A 950kg sports car collides into the rear end of 2200kg SUV stopped at a red light. The bumper lock, the brakes lock, and the two cars skid forward 3.0m before coming to a stop. The police officer, knowing the coefficient of kinetic friction between the tires and the road is about 0.8. He calculates the speed of the sports car at impact. What was that speed?

6.9m/s

17.4m/s

20.5m/s

22.9m/s

46.7m/s

Explanation

Knowns

$m_{A}=950kg$

$m_{B}=2200kg$

$v_{0B}=0m/s$

$\mu =0.8$

Unknowns

$v_{0A}=?$

To solve the problem we must consider two different situations. The first is when the cars are skidding across the ground. While they are skidding the force of friction is what is resisting their motion and therefore doing work on the cars to slow them down to a stop.

The second situation is the collision itself when the first car has an initial speed and the second car is stopped. After the collision, both cars are traveling at the same speed as their bumpers have been locked together.

To begin, we need to find the speed that the cars are skidding across the ground after the bumpers have been interlocked.

The work-kinetic energy theorem states that the work done is equal to the change in the kinetic energy of the object.

$-W=\Delta KE$

Work is directly related to the force times the displacement of the object.

In this case, the force that is doing the work is friction.

$F_{f}=\mu F_{N}$

Since the cars are on a level surface the normal force is equal to the force of gravity.

$F_{G=}F_{N}$

The force of gravity is directly related to the mass and the acceleration due to gravity acting on an object.

$F_{G}=mg$

When we put all these equations together we get

$W=\mu mgd$

We also know that kinetic energy is related to the mass and velocity squared.

$KE=\frac{1}{2}mv^{2}$

Therefore our final equation should look like

$-\mu mgd=\frac{1}{2}mv{_{f}}^{2}-\frac{1}{2}mv{_{0}}^{2}$

Notice that the mass falls out of the equation since it is in every term. Also, note that the final velocity is 0m/s so this will cancel out as well.

$-\mu gd=-\frac{1}{2}v{_{0}}^{2}$

Since both of these terms have a negative we can cancel this out as well.

$\mu gd=\frac{1}{2}v{_{0}}^{2}$

We can now substitute our variables to determine the velocity of both cars just after the crash.

$(0.8)(9.8m/s^{2})(3.0m)=\frac{1}{2}v{_{0}}^{2}$

$47.04=v{_{0}}^{2}$

$v_{0}=6.9m/s$

This is the velocity of the cars after the collision. We must now consider our second situation of the collision itself. During this collision, momentum must be conserved. The law of conservation of momentum states

$m_{A}v_{0A}+m_{B}v_{0B}=m_{A}v_{fA}+m_{B}v_{fB}$

We can substitute our values for the masses of the cars, the final velocity of the cars after the collision (which we just found), and the initial value of the stopped car.

$(950kg)v_{0A}+(2200kg)(0m/s)=(950kg)(6.9m/s)+(2200kg)(6.9m/s)$

Now we can solve for our missing variable.

$(950kg)v_{0A}=21735kgm/s$

$v_{0A}=22.9m/s$

A car strikes a car at rest from behind. The bumpers lock and they move forward together. If their new final velocity is equal to $18\frac{m}{s}$ , what was the initial speed of the first car?

$38\frac{m}{s}$

$40\frac{m}{s}$

$12\frac{m}{s}$

$44\frac{m}{s}$

$36\frac{m}{s}$

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, , 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 is given. Plug in these values and solve for the initial velocity of the first car.

$(900kg*v_{1i})+(1000kg*0\frac{m}{s})=(900kg+1000kg)*18\frac{m}{s}$

$900kg*v_{1i}=(1900kg)*18\frac{m}{s}$

$900kg*v_{1i}=34,200\frac{kg*m}{s}$

$v_{1i}=\frac{34,200\frac{kg*m}{s}}{900kg}$

$v_{1i}=38\frac{m}{s}$

Two bumper cars at an amusement park collide elastically as one approaches the other directly from the rear. The car in front (CarA) has a mass of 550kg and the car behind it (CarB) has a mass of 450kg. The car in front was traveling at 3.70m/s while the car behind hit him with a velocity of 4.50m/s. What are their final velocities after the collision?

$v_{Af}=4.4m/s$

$v_{Bf}=3.6m/s$

$v_{Af}=3.6m/s$

$v_{Bf}=4.4m/s$

$v_{Af}=4.5m/s$

$v_{Bf}=3.7m/s$

$v_{Af}=8 m/s$

$v_{Bf}=0m/s$

$v_{Af}=5.4 m/s$

$v_{Bf}=2.3m/s$

Explanation

Knowns

$m_{A}=550kg$

$m_{B}=450kg$

$v_{A0}=3.7m/s$

$v_{B0}=4.5m/s$

Unknowns

$v_{Af}=?$

$v_{Bf}=?$

For elastic collisions, we know that the initial and final velocities are related by the equation

$v_{A0}+v_{Af}=v_{B0}+v_{Bf}$

We also know that the momentum is conserved meaning that

$mv_{A0}+mv_{B0}=mv_{Af}+mv_{Bf}$

Since we have two missing variables and two equations, we can now solve for one of the variables using a system of equations

Let’s get the final velocity of car A by itself from the first equation

$v_{Af}=v_{B0}+v_{Bf}-v_{A0}$

We can now substitute this equation into our momentum equation.

$m_{A}v_{A0}+m_{B}v_{B0}=m_{A}(v_{B0}+v_{Bf}-v_{A0})+m_{B}v_{Bf}$

In our new equation, we only have one missing variable, so let's substitute in values and solve.

$(550kg)(3.7m/s)+(450kg)(4.5m/s)=(550kg)(4.5m/s+v_{Bf}-3.7m/s)+(450kg)(v_{Bf})$

$4060=550(0.8+v_{Bf})+450(v_{Bf})$

$4060=440+550v_{Bf}+450v_{Bf}$

$3620=1000v_{Bf}$

$v_{Bf}=3.6m/s$

We can now plug this value back into our equation for $v_{Af}$

$v_{Af}=v_{B0}+v_{Bf}-v_{A0}$

$v_{Af}=4.5m/s+3.6m/s-3.7m/s$

$v_{Af}=4.4m/s$

A car with mass and initial velocity strikes a car of mass , which is at rest. If the two cars stick together after the collision, what is the final velocity?

$\frac{m_1v_1}{m_1+m_2}$

$\frac{v_1}{m_1+m_2}$

$\frac{m_1+m_2}{v_1+v_2}$

$\sqrt{m_1v_1}$

Explanation

We know that the cars stick together after the collision, which means that the final velocity will be the same for both of them. Using the formula for conservation of momentum, we can start to set up an equation to solve this problem.

First, we will write the initial momentum.

We know that the second car starts at rest, so this equation can be simplified.

$p_1=m_1v_1+m_2(0\frac{m}{}s)=m_1v_1$

Now we will write out the final momentum. Keep in mind that both cars will have the same velocity!

Set these equations equal to each other and solve to isolate the final velocity.

$\frac{m_1v_1}{m_1+m_2}=v_3$

This is our answer, in terms of the given variables.

A ball is thrown west at $\small 20\frac{m}{s}$ 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?

$8.6\frac{m}{s}\ \text{east}$

$9.8\frac{m}{s}\ \text{west}$

$11.2\frac{m}{s}\ \text{downward}$

$2.3\frac{m}{s}\ \text{east}$

$4.7\frac{m}{s}\ \text{east}$

Explanation

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.

$m_1v_1+m_2v_2=0\frac{m}{s}$

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.

$(6kg)(20\frac{m}{s}) + (14kg)v_2 = 0\frac{m}{s}$

$(120N) + (14kg)(v_2)= 0\frac{m}{s}$

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

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

A car travelling at $25\frac{m}{s}$ rear ends another car at rest. The two bumpers lock and the cars move forward together. What is their final velocity?

$12.5\frac{m}{s}$

$25\frac{m}{s}$

$125\frac{m}{s}$

$1.25\frac{m}{s}$

$2.5\frac{m}{s}$

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, , 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.

$(2000kg*25\frac{m}{s})+(2000kg*0\frac{m}{s})=(2000kg+2000kg)v_{final}$