Clay Ball Thrown Agains Door Bouncy Ball

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A homo with a mass of is painting a firm. He stands on a tall ladder of height. He leans over and falls straight down off the ladder. If he is in the air for seconds, what will be his momentum right before he hits the basis?

Correct reply:

(mg)(s)

Explanation:

The trouble tells the states he falls vertically off the ladder (straight down), and so we don't need to worry about motion in the horizontal direction.

The equation for momentum is:

p=mv

Nosotros can presume he falls from balance, which allows us to find the initial momentum.

$m*0\frac{m}{s}=0\frac{kg*m}{s}$ .

From here, we can use the formula for impulse:

$\Delta p=F\Delta t$

$(p_{f}-p_i)=F\Delta t$

We know his initial momentum is zero, and so we can remove this variable from the equation.

$p_f=F\Delta t$

The problem tells us that his change in time is seconds, and then we tin can insert this in identify of the time.

p_f=(F)(s)

The only force acting upon man is the force due to gravity, which will always exist given by the equation F_G=mg .

p_f=(F_G)(s)

p_f=(mg)(s)

A 6kg brawl is thrown due west at $\small 20\frac{m}{s}$ and collides with a 14kg ball while in the air. If the assurance stick together in the crash and fall straight down to the basis, what was the velocity of the second ball?

Correct reply:

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

Explanation:

We know that if the balls fell straight downwardly later the crash, then the full momentum in the horizontal direction is zero. The just motion is due to gravity, rather than any remaining horizontal momentum. Based on conservation of momentum, the initial and final momentum values must exist equal. If the terminal horizontal momentum is zero, and then the initial horizontal momentum must also be zippo.

p_i=p_f

m_1v_1 + m_2v_2 = m_1v_3 + m_2 v_4

In our situation, the final momentum is going to be zero.

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

Employ the given values for the mass of each ball and initial velocity of the first ball to notice the initial velocity of the second.

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

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

(14kg)(v_2) = -120N

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

The negative sign tells us the second brawl is traveling in the opposite direction as the outset, meaning it must be moving e.

A 2000kg car travelling at $25\frac{m}{s}$ rear ends another 2000kg car at rest. The two bumpers lock and the cars motion forward together. What is their last velocity?

Right reply:

$12.5\frac{m}{s}$

Explanation:

This is an example of an inelastic collision, as the 2 cars stick together afterward colliding. We tin presume momentum is conserved.

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

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

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

Since the cars stick together, they will have the aforementioned final velocity. We know the second car starts at residuum, and the velocity of the starting time automobile is given. Plug in these values and solve for the final velocity.

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

$50000\frac{kg*m}{s}=(4000kg)v_{final}$

$\frac{50000\frac{kg*m}{s}}{4000kg}=v_{final}$

$12.5\frac{m}{s}=v_{final}$

A 900kg car strikes a 1000kg car at rest from backside. 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 kickoff car?

Correct answer:

$38\frac{m}{s}$

Explanation:

This is an case of an inelastic standoff, as the ii cars stick together later on colliding. We tin assume momentum is conserved.

To brand the equation easier, let'southward call the showtime motorcar "1" and the second car "2."

Using conservation of momentum and the equation for momentum, p=mv , nosotros tin ready the following equation.

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

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

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

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

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

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

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

A 10kg ball moving at $30\frac{m}{s}$ strikes a 12kg brawl at residue. Later on the collision the 10kg ball is moving with a velocity of $13\frac{m}{s}$ . What is the velocity of the second brawl later the collision?

Correct answer:

$14.2\frac{m}{s}$

Explanation:

Nosotros can utilise the law of conservation of momentum:

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

We know the mass of each ball and their initial velocities.

$(10kg*30\frac{m}{s})+(12kg*0\frac{m}{s})=m_1v_{1final}+m_2v_{2final}$

We also know the last velocity of the offset brawl. This leaves only i variable: the final velocity of the second ball.

$(10kg*30\frac{m}{s})+(12kg*0\frac{m}{s})=(10kg*13\frac{m}{s})+(12kg*v_{2f})$

Solve to isolate the variable.

$300\frac{kg*m}{s}+0=(130\frac{kg*m}{s})+(12kg*v_{2f})$

${300\frac{kg*m}{s}}-{130\frac{kg*m}{s}}=12kg*v_{2f}$

${170\frac{kg*m}{s}}=12kg*v_{2f}$

$\frac{170\frac{kg*m}{s}}{12kg}=v_{2f}$

$14.2\frac{m}{s}=v_{2f}$

A 11kg ball moving at $33\frac{m}{s}$ strikes a second ball at rest. Afterward the collision the 11kg ball is moving with a velocity of $13\frac{m}{s}$ and the second ball is moving with a velocity of $8\frac{m}{s}$ . What is the mass of the second brawl?

Correct reply:

27.5kg

Explanation:

This is an case of an elastic standoff. We beginning with two masses and stop with two masses with no loss of energy.

We can use the law of conservation of momentum to equate the initial and concluding terms.

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

Plug in the given values and solve for the mass of the second ball.

$(11kg*33\frac{m}{s})+(m_2*0\frac{m}{s})=(11kg*13\frac{m}{s})+(m_2*8\frac{m}{s})$

$(363\frac{kg*m}{s})+(0)=(143\frac{kg*m}{s})+(m_2*8\frac{m}{s})$

$(363\frac{kg*m}{s})-(143\frac{kg*m}{s})=(m_2*8\frac{m}{s})$

$(220\frac{kg*m}{s})=(m_2*8\frac{m}{s})$

$\frac{220\frac{kg*m}{s}}{8\frac{m}{s}}=m_2$

27.5kg=m_2

A 0.5kg ball strikes a second 1.5kg ball at residual. Subsequently the collision the 0.5kg ball is moving with a velocity of $13\frac{m}{s}$ and the second ball is moving with a velocity of $8\frac{m}{s}$ . What is the initial velocity of the get-go ball?

Correct reply:

$37\frac{m}{s}$

Caption:

This is an instance of an elastic standoff. Nosotros start with 2 masses and end with two masses with no loss of energy.

We tin can use the police 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 the initial velocity of the first ball.

$(0.5kg*v_{1i})+(1.5kg*0\frac{m}{s})=(0.5kg*13\frac{m}{s})+(1.5kg*8\frac{m}{s})$

$(0.5kg*v_{1i})+(0)=(6.5\frac{kg*m}{s} )+(12\frac{kg*m}{s})$

$(0.5kg*v_{1i})=(18.5\frac{kg*m}{s} )$

$v_{1i}=\frac{18.5\frac{kg*m}{s}}{0.5kg}$

$v_{1i}=37\frac{m}{s}$

A car with mass m_1 and initial velocity v_1 strikes a car of mass m_2 , which is at remainder. If the two cars stick together after the collision, what is the final velocity?

Right answer:

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

Explanation:

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

p_1=p_2

p=mv

First, we will write the initial momentum.

p_1=m_1v_1+m_2v_2

We know that the second motorcar starts at remainder, 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 concluding momentum. Keep in mind that both cars will have the same velocity!

p_2=m_1v_3+m_2v_3

p_2=v_3(m_1+m_2)

Ready these equations equal to each other and solve to isolate the terminal velocity.

p_1=p_2

m_1v_1=v_3(m_1+m_2)

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

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

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?

Possible Answers:

They would rebound at a higher speed

They would rebound at a slower speed

No deviation

Correct answer:

No difference

Caption:

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 tin can say that

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

Therefore we tin country 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; so information technology does non matter if the balls increment or decrease in their mass.

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

Possible Answers:

The hulk of clay

Neither will work

Both the same

The bouncy ball

Correct reply:

The bouncy ball

Explanation:

A bouncy ball will accept an elastic collision with the door, causing the ball to movement backward at the aforementioned speed it hit the door.

On the other hand, the hulk of clay will accept an inelastic collision with the door, causing the blob of clay to motion with the same speed equally the door.

Then let us expect 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 tin remove it from the get-go 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 first.

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

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 the states examine the police of conservation of momentum for the inelastic collision. Again the door has no initial velocity and so we can remove information technology from the get-go of the equation.

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

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

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

Now solve for the terminal velocity of the door.

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

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

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Clay Ball Thrown Agains Door Bouncy Ball

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