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A 10 g piece of sticky clay moving horizontally with a speed of 12 m/s strikes a pendulum bob and sticks to the bob. The pendulum bob has a mass of 50 g and it is suspended from a 0.7 m string.

a. Find the momentum of the clay before the collision.

b. *Find the momentum of the clay-bob system after the collision.

c.*Find the velocity of the clay-bob system after the collision.

d. *Find the kinetic energy of the clay-bob system after the collision.

e. *Find the maximum height of the clay-bob system that they deflect after the collision.

f. *What velocity should the clay have before the collision in order for the clay-bob system to complete one circle?​

Respuesta :

Answer:

it is d that my answer to the question

a. The momentum of the clay before the collision is 0.12m/s

b. The momentum of the clay-bob system after the collision is 0.12m/s

c. The velocity of the clay-bob system after the collision is 2m/s

d. The kinetic energy of the clay-bob system after the collision is 0.12m/s

e. The maximum height of the clay-bob system that they deflect after the collision is 0.204m

f. The velocity  clay should have  before the collision in order for the clay-bob system to complete one circle is 35.13m/s

What is momentum?

Momentum is the product of mass and velocity

It is given by ,

p= mv

where, p = momentum

m = mass

v = velocity

What is velocity?

velocity is the rate of change of displacement.

To solve , we use Law of conservation of momentum and Law of conservation of  mechanical energy

What is Law of conservation of momentum?

Law of conservation of momentum states that  in absence of any external forces , the momentum of a system always remains conserved.

i.e,

The momentum  of the system before collision = momentum of the system after collision.

[tex]m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}[/tex]

Where,

m₁ = mass of first body.

m₂ = mass of the second body.

u₁ = velocity of first body before collision,

u₂ = velocity of the second body before collision

v₁ = velocity of first body after collision

v₂ = velocity of first body after collision.

v₂ = velocity of second body after collision.

What is law of conservation of mechanical energy

The Law states that the total mechanical energy of a system remains conserved and energy can be converted from one form to another form.

The sum of kinetic energy an potential energy is equal to total mechanical energy which is a constant.

So, K.E + P.E = constant

a, Here mass of clay m₁ = 10g = 0.01kg,

velocity of the clay = 12m/s.

therefore, momentum  p = mv.

p = 0.01 x 12 = 0.12kg m/s.

Hence momentum of clay is 0-12kg m/s.

b. The momentum of clay - bob system after collision is

From law of conservation of momentum ,

final momentum = initial momentum

since initial momentum  = 0.12kg m/s

therefore momentum of the clay-bob system after collision is 0.12kg m/s

c. To find the velocity of the body we apply law of conservation of momentum

[tex]m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}[/tex]

Here  m₁u₁ = 0.12kgm/s

m₂u₂ = 0 (since bob is at rest initally)

m₁ = 10g = 0.01kg

m₂ =  50g = 0.05kg

after collision the system moves with a common velocity v

So,

m₁u₁ = (m₁ + m₂)v

0.12 = (0.01 + 0.05 ) v

So, v = 0.12/0.06 = 2m/s

Hence velocity of clay -bob system after collision is = 2m/s.

d. The kinetic energy of the clay-bob system after the collision.

kinetic energy = 1/2 mv²

Here. m = 0.06kg

v = 2m/s

therefore K.E = (1/2)x 0.06 x 2²

K.E = 0.12J

Hence Kinetic energy of the clay-bob system after collision is 0.12J

e. The maximum height of the clay-bob system that they deflect after the collision.

From Law of Conservation of mechanical energy we have,

The kinetic energy = The potential energy

so, 1/2 mv² = mgh

Here, m = 0.06kg

g= 9,8m/s²

h = height.

So,  h =  2/9.8 = 0.204m

Hence height to which the clay-bob system will rise after collision is 0.204m

e.  To find the velocity that clay should possess  before collision in order for the clay- bob   system to gain the velocity so that it completes the circle -

The velocity that the particle should have at the bottom of the circle in order to complete the circle is v = [tex]\sqrt{5rg\\}[/tex]

Where, r = radius of the circle

g = acceleration due to gravity

Here,

r = 0.7m

g = 9.8m/s

So, v = [tex]\sqrt{5 X0.7 X 9.8 }[/tex] = 5.85m

So clay-bob system should have a velocity of 5.85m at the bottom of the circle so that it can complete the circle

Now,  applying Law of conservation of momentum

m₁u₁ = (m₁ + m₂)v

Here, m₁ = 0.01kg

m₂ = 0.05kg

v = 5.85m/s

So, 0.01u₁ = 0.06 x 5.85  

Hence u₁ = 35.13m/s

So velocity that clay should have is 35.13m/s.

Therefore  the solution of the question is

. The momentum of the clay before the collision is 0.12m/s

b. The momentum of the clay-bob system after the collision is 0.12m/s

c. The velocity of the clay-bob system after the collision is 2m/s

d. The kinetic energy of the clay-bob system after the collision is 0.12m/s

e. The maximum height of the clay-bob system that they deflect after the collision is 0.204m

f. The velocity  clay should have  before the collision in order for the clay-bob system to complete one circle is 35.13m/s

To know more on momentum here

https://brainly.com/question/24030570

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