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natita [175]
3 years ago
8

A ball is projected at an immovable wall with a speed vi and bounces back the wall in such a manner that it only has 1/3 of its

original linear momentum. a) Determine what fraction of the kinetic energy is lost in the collision.
Physics
1 answer:
sergey [27]3 years ago
5 0

The fraction of the kinetic energy of the ball lost during the collision is \frac{1}{3}.

The given parameters;

  • <em>initial speed of the ball, = vi</em>
  • <em>final momentum of the ball, Pf = ¹/₃Pi</em>

The initial and final momentum of the ball is calculated as;

P_i = m_ivi

P_f = m_fv_f = \frac{1}{3} m_iv_i

The initial and final kinetic energy of the ball is calculated as;

K.E_i = \frac{1}{2} m_iv_i^2 = \frac{1}{2} P_iv_i\\\\K.E_f = \frac{1}{2} m_fv_f^2= \frac{1}{2} (\frac{1}{3} P_iv_i)= \frac{1}{6} P_iv_i

The change in the kinetic energy is calculated as;

\Delta K.E = K.E_f - K.E_i \\\\\Delta K.E = \frac{1}{6} P_iv_i - \frac{1}{2} P_iv_i = \frac{1}{3} P_iv_i

Thus, the fraction of the kinetic energy of the ball lost during the collision is \frac{1}{3}.

Learn more here:brainly.com/question/18566218

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What is storage of energy​
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Hey,

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I hope it helped.

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3 0
3 years ago
How does torque of rotating body relate to velocity
Fudgin [204]

Answer:

We know that the torque can be calculated as follows:

T = rpsinα

With r being the distance of the body from the center of the circumference he has as trajectory, p being the momentum of the body and sinα being the sine of the angle between the 2 vectors: r and p.

It's pretty obvious that T is directly proportional to the momentum, that can be written as p = m·v, with m being the mass of the object and v the velocity of the object.

8 0
3 years ago
a 40 kg block is being pulled at constant velocity across a horizontal surface by a 150 N force at an angle 60 above the horizon
k0ka [10]

Answer:

0.19

Explanation:

mass of block, m = 40 kg

F = 150 N

Angle make with the horizontal, θ = 60 degree

Let μ be the coefficient of kinetic friction

The component of force along horizontal direction  is F Cos θ

                                                                    = 150 cos 60 = 75 N

As it is moving with constant velocity it mean the acceleration of the block is zero.

Applied force in horizontal direction = friction force

75 = μ x Normal reaction

75 = μ x m x g

75 = μ x 40 x 9.8

μ = 0.19

Thus, the coefficient of kinetic friction is 0.19.

8 0
3 years ago
An object of mass 2kg is on an incline where there is an applied force of 15N. The
seraphim [82]

a) See free-body diagram in attachment

b) The acceleration is 2.46 m/s^2

Explanation:

a)

The free-body diagram of an object is a diagram representing all the forces acting on the object. Each force is represented by a vector of length proportional to the magnitude of the force, pointing in the same direction as the force.

The free-body diagram for this object is shown in the figure in attachment.

There are three forces acting on the object:

  • The weight of the object, labelled as mg (where m is the mass of the object and g is the acceleration of gravity), acting downward
  • The applied force, F_a, acting up along the plane
  • The force of friction, F_f, acting down along the plane

b)

In order to find the acceleration of the object, we need to write the equation of the forces acting along the direction parallel to the incline. We have:

F_a - F_f - mg sin \theta = ma

where:

F_a = 15 N is the applied force, pushing forward

F_f = 5 N is the frictional force, acting backward

mg sin \theta is the component of the weight parallel to the incline, acting backward, where

m = 2 kg is the mass of the object

g=9.8 m/s^2 is the acceleration of gravity

\theta=15^{\circ} is the angle between the horizontal and the incline (it is not given in the problem, so I assumed this value)

a is the acceleration

Solving for a, we find:

a=\frac{F_a - F_f - mg sin \theta}{m}=\frac{15-5-(2)(9.8)(sin 15^{\circ})}{2}=2.46 m/s^2

Learn more about inclined planes:

brainly.com/question/5884009

#LearnwithBrainly

8 0
3 years ago
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