In the figure, determine the character of the collision. The masses of the blocks, and the velocities before and after are given. The collision is (Show your work-no work shown = ZERO POINTS) 1.8 m/s 0.2 m/s 0.6 m/s 1.4 m/s 4 kg 6 kg 4 kg 6 kg Before After A) perfectly elastic. B) partially inelastic. C) completely inelastic. D) characterized by an increase in kinetic energy E) not possible because momentum is not conserved.

Answers

Answer 1
Answer:

When two bodies come into close touch with one another, a collision occurs. In this instance, the two bodies quickly exert forces on one another. The collision changes the energy and momentum of the bodies that are interacting.

Briefing

the system's initial kinetic energy, KEi, is equal to 0.5 * 4 * 1.8 2 plus 0.5 * 6 * 0.2 2 J.

KEi = 6.6 J

The system's ultimate kinetic energy, KEf

, following the collision is equal to 0.5 * 4 * 0.6 + 0.5 * 6 * 1.4 J.

KEf = 6.6 J

since KEi = KEf

Perfectly elastic is the collision

the appropriate response is A) completely elastic.

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A 50 kg woman and an 80 kg man stand 12.0 m apart on frictionless ice.(a) How far from the woman is their CM?
m

(b) If each holds one end of a rope, and the man pulls on the rope so that he moves 1.3 m, how far from the woman will he be now?
m

(c) How far will the man have moved when he collides with the woman?
m

Answers

Answer:

Given that

m₁ = 50 kg

m₂=80 kg

d= 12 m

a)

We know that center of mass given as

X = (x₁m₁+x₂m₂)/(m₁+m₂)

Lets take distance of CM from woman is X

So now by putting the value

X = (0 x 50+12 x 80)/(50+80)

x=7.38 m

b)

There is no any external force so the CM  will not move.

So we can say that

x₁m₁+x₂m₂ = 0

50(x) - 80(1.3)=0

x=2.08

So the distance move by woman d=12-2.08-1.3=8.62 m

d=8.62 m

c) lets take distance move by man is x

50 (x) - 80 (12-x) =0

x=7.38

So the distance move by woman d=12-7.38

d=4.62 m

A small ball of mass m is held directly above a large ball of mass M with a small space between them, and the two balls are dropped simultaneously from height H. (The height is much larger than the radius of each ball, so you may neglect the radius.) The large ball bounces elastically off the floor and the small ball bounces elastically off the large ball. a) For which value of the mass m, in terms of M, does the large ball stop when it collides with the small ball? b) What final height, in terms of H, does the small ball reach?

Answers

a) The large sphere has 3 times the mass of the small sphere

b)   The final height at which small ball reach y = 4H

What will be the mass of the sphere and height covered by the small ball?

We must start this problem by calculating the speed with which the spheres reach the floor

V_f^2=V_o^2-2gy      

As the spheres are released v₀ = 0

V_f^2=2gH      

V_f=√(2gH)      

The two spheres arrive at the same speed to the floor.

The largest sphere clashes elastically so that with the floor it has a much higher mass, the sphere bounces with the same speed with which it arrived, the exit speed of the  spheres

V_(10)=\sqrt2gH    

The big sphere goes up and the small one down, the two collide, let's form a system that contains the two spheres, let's use moment conservation

Let's call

V_h=\sqrt2gH

Small sphere m₂ and  V_(20)=-\sqrt2gH=-V_h

Large sphere m₁ and  V_(10)=√(2gH)=V_h  

Before crash

P_o=m_1V_(10)+m_2V_(20)      

After the crash

P_f=m_1V_(1f)+m_2V_(2f)    

   

P_o=P_f  

   

m_1V_(10)+m_2V_(20)=m_1V_(1f)+m_2V_(2f)  

The conservation of kinetic energy

K_o=(1)/(2) m_1V_(10)^2+(1)/(2) m_2V_(20)^2      

K_f=(1)/(2) m_1V_(1f^2)+(1)/(2) m_2V_(2f)^2    

K_o=K_f  

K_o=(1)/(2) m_1V_(10)^2+(1)/(2) m_2V_(20)^2  =  K_f=(1)/(2) m_1V_(1f^2)+(1)/(2) m_2V_(2f)^2    

Let's write the values

-m_1V_h+m_2V_h=m_1V_(1f)+m_2V_(2f)    

 m_1V_h^2+m_2V_h^2=m_1V_(1f)^2+m_2V_(2f)^2

 

The solution to this system of equations is

m_t=m_1+m_2    

V_(1f)=((m_1-m_2))/(m_tV_(10)) +(2m_2)/(m_tV_2)      

V_(2f)=(2m_1)/(m_tV_(10)) +(m_2-m_1)/(m_tV_2)        

The large sphere is labeled 1, we are asked for the mass so that  V_(1f) = 0, let's clear the equation

V_(1f)=(m_1-m_2)/(m_tV_(10)) +(2m_2)/(m_tV_(20))      

0=(m_1-m_2)/(m_tV_h) +(2m_2)/(m_t(-V_h))      

 

(m_1-m_2)/(m_tV_h) =(2m_2)/(m_tV_h)      

(m_1-m_2)=2m_2    

m_1=3m_2      

The large sphere has to complete 3 times the mass of the sphere1 because it stops after the crash.

b) Let us calculate with the other equation the speed with which the sphere comes out2 (small)

  V_(2f)=(2m_1)/(m_tV_(10)) +(m_2-m_1)/(m_tV_(20))    

 

V_(2f)=(2m_1)/(m_tV_h) +(m_2-m_1)/(m_t(-V_h))

In addition, we know that m₁ = 3 m₂

m_t=3m_2+m_2  mt = 3m2 + m2

m_t=4m_2    

     

V_(2f)=(2* 3m_2)/(4m_2V_h-(m_2-3m_2)4m_2V_h)  

   

V_(2f)=(3)/(2) V_h+(1)/(2) V_h  

       

V_(2f)=2V_h

V_(2f)=2\sqrt{2gh        

This is the rate of rising of sphere 2 (small. At the highest point, it zeroes velocity V_f= 0

V^2=V_(2f)^2-2gy  

0=(2√(2gh))^2-2gy        

       

y=4H

Thus

a) The large sphere has 3 times the mass of the small sphere

b)   The final height at which small ball reach y = 4H

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Answer:

a) the large sphere has 3 times the mass of the small sphere

b)     y = 4H

Explanation:

We must start this problem by calculating the speed with which the spheres reach the floor

       vf² = vo² - 2g y

As the spheres are released v₀ = 0

      vf² = - 2g H

      vf = √ (2g H)

The two spheres arrive with the same speed to the floor.

The largest sphere clashes elastically so that with the floor it has a much higher mass, the sphere bounces with the same speed with which it arrived, the exit speed of the  spheres

     V₁₀ = √2gH

The big sphere goes up and the small one down, the two collide, let's form a system that contains the two spheres, let's use moment conservation

Let's call vh = √2gH

Small sphere m₂ and v₂o = - √2gH = -vh

Large sphere m₁ and v₁o = √ 2gh = vh

Before crash

        p₀ = m₁ v₁₀ + m₂ v₂₀

After the crash

      pf = m₁ v₁f + m₂ v₂f

      po = pf

      m₁ v₁₀ + m₂ v₂₀ = m₁ v₁f + m₂ v₂f

The conservation of kinetic energy

      Ko = ½ m₁ v₁₀² + ½ m₂ v₂₀²

      Kf = ½ m₁ v₁f² + ½ m₂ v₂f²

      Ko = KF

      ½ m₁ v1₁₀² + ½ m₂ v₂₀² = ½ m₁ v₁f² + ½ m₂ v₂f²

Let's write the values

      -m₁ vh + m₂ vh = m₁ v₁f + m₂ v₂f

       m₁ vh² + m₂ vh² = m₁ v₁f² + m₂ v₂f²

The solution to this system of equations is

       mt = m₁ + m₂

       v1f = (m₁-m₂) / mt v₁₀ + 2m₂ / mt v₂

       v₂f = 2m₁ /mt v₁₀ + (m₂-m₁) / mt v₂

The large sphere is labeled 1, we are asked for the mass so that v1f = 0, let's clear the equation

       v₁f = (m₁-m₂) / mt v₁₀ + 2m₂ / mt v₂₀

        0 = (m₁-m₂) / mt vh + 2 m₂ / mt (-vh)

       (m₁-m₂) / mt vh = 2 m₂ / mt vh

       (m₁-m₂) = 2m₂

        m₁ = 3 m₂

The large sphere has to complete 3 times the mass of the sphere1 because it stops after the crash.

b) Let us calculate with the other equation the speed with which the sphere comes out2 (small)

        v₂f = 2m₁ / mt v₁₀ + (m₂-m₁) / mt v₂₀

        v₂f = 2 m₁ / mt vh + (m₂-m₁) mt (-vh)

In addition, we know that m₁ = 3 m₂

        mt = 3m2 + m2

         mt= 4m2

        v₂f = 2 3m₂ / 4m₂ vh - (m₂-3m₂) 4m₂ vh

        v₂f = 3/2 vh +1/2 vh

        v₂f = 2 vh

        v₂f = 2 √ 2gh

This is the rate of rise of sphere 2 (small. At the highest point its zero velocity vf = 0

        V² = v₂f² - 2 g Y

          0 = (2√2gh)² - 2gy

        2gy = 4 (2gH)

         y = 4H

A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's initial speed is 1.37 m/s and it stops after 2.8 s. Determine the magnitude of the friction force exerted on the box.A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's initial speed is 1.37 m/s and it stops after 2.8 s. Determine the magnitude of the friction force exerted on the box.

Answers

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

\Sigma F = -f = m\cdot a(Eq. 1)

Where:

f - Kinetic friction force, measured in newtons.

m - Mass of the box, measured in kilograms.

a - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight (W = m\cdot g) and uniform accelerated motion (v = v_(o)+a\cdot t), we expand the previous expression:

-f = \left((W)/(g) \right)\cdot \left((v-v_(o))/(t)\right)

And the magnitude of the friction force exerted on the box is calculated by this formula:

f = -\left((W)/(g) \right)\cdot \left((v-v_(o))/(t)\right)(Eq. 1b)

Where:

W - Weight, measured in newtons.

g - Gravitational acceleration, measured in meters per square second.

v_(o) - Initial speed, measured in meters per second.

v - Final speed, measured in meters per second.

t - Time, measured in seconds.

If we know that W = 52.4\,N, g = 9.807\,(m)/(s^(2)), v_(o) = 1.37\,(m)/(s), v = 0\,(m)/(s) and t = 2.8\,s, the magnitud of the kinetic friction force exerted on the box is:

f = -\left((52.4\,N)/(9.807\,(m)/(s^(2)) ) \right)\cdot \left((0\,(m)/(s)-1.37\,(m)/(s)  )/(2.8\,s) \right)

f = 2.614\,N

The magnitude of the friction force exerted on the box is 2.614 newtons.

Final answer:

The magnitude of the friction force acting on the box is determined by calculating the box's acceleration, establishing its mass based on its weight information, and applying these values in Newton's second law. The calculated value is 2.62 N.

Explanation:

To determine the magnitude of the friction force, we first have to compute the acceleration of the box. Acceleration (a) can be found using the formula 'final velocity - initial velocity / time'. Since the final velocity is 0 (the box stops), and the initial velocity is 1.37 m/s, and the time is 2.8 s, we get: a = (0 - 1.37) / 2.8 = -0.49 m/s^2. The negative sign indicates deceleration.

Next, we use Newton's second law, which states that the net force acting on an object is equal to its mass times its acceleration. The net force in this case is the frictional force because there is no other force acting on the box in the horizontal direction. However, we do not know the mass of the box, but we do know its weight, and weight = mass x gravitational acceleration (g). So mass = weight/g = 52.4N / 9.8m/s^2 = 5.35 kg.

Lastly, we substitute the mass and deceleration into Newton's second law to find the frictional force (f): f = mass x deceleration = 5.35kg x -0.49m/s^2 = -2.62 N. Again, the negative sign indicates that the force acts opposite to the direction of motion. Thus, the frictional force magnitude is 2.62 N.

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A 500-gram mass is attached to a spring and executes simple harmonic motion with a period of 0.25 second. If the total energy of the system is 4J, find the force constant of the spring?

Answers

Answer:

315.5 N/m

Explanation:

m = 500 g = 0.5 kg

T = 0.25 second

Total energy, E = 4 J

Let K be the spring constant.

The formula for the time period is given by

T = 2\pi \sqrt{(m)/(K)}

0.25 = 2* 3.14* \sqrt{(0.5)/(K)}

0.0398=\sqrt{(0.5)/(K)}

1.585* 10^(-3)={(0.5)/(K)}

K = 315.5 N/m

" A sound wave moving through air consists of alternate regions of high pressure and low pressure. If the frequency of the sound is increased, what happens, if anything, to the distance between successive high-pressure regions, and why

Answers

Answer: wavelength will reduce

Explanation: The region of low pressure is know as the rarefraction region while the region of high pressure is the compression region.

The distance between 2 successive rarefraction or compression region is known as the wavelength.

Now the question is concerned about what an increase in frequency will cause to wavelength.

The speed of sound in air is a constant and it is approximately 343 m/s.

But recall that v = fλ

By assuming a fixed value for speed (v), we have that

k = fλ

Hence, f = k/ λ

This implies that at a fixed wave speed, the wavelength and frequency have an inverse relationship.

An increase in frequency will bring about a reduction in wavelength.

What is the kinetic energy k of an electron with momentum 1.05×10−24 kilogram meters per second?

Answers

Momentum = mv
where m is the mass of an electron and v is the velocity of the electron.

v = momentum ÷ m
   = (1.05×10∧-24)÷(9.1×10∧-31) = 1,153,846.154 m/s

kinetic energy = (mv∧2)÷2
                       = (9.1×10∧-31 × 1,153,846.154∧2) ÷2
                      = (1.21154×10∧-18) ÷ 2
                      = 6.05769×10∧-19 J

Answer:

K = 6.02 × 10⁻¹⁹ J

Explanation:

The momentum (p) of an electron is its mass (m) times its speed (v).

p = m × v

v = p / m = (1.05 × 10⁻²⁴ kg.m/s) / 9.11 × 10⁻³¹ kg = 1.15 × 10⁶ m/s

We can find the kinetic energy (K) using the following expression.

K = 1/2 × m × v²

K = 1/2 × 9.11 × 10⁻³¹ kg × (1.15 × 10⁶ m/s)²

K = 6.02 × 10⁻¹⁹ J