Question

By calculating numerical quantities for a multiparticle system, one can get a concrete sense of the meaning of the relationships

p with arrowsys = Mtotv with arrowCM

and Ktot = Ktrans + Krel.
Consider an object consisting of two balls connected by a spring, whose stiffness is460 N/m. The object has been thrown through the air and is rotating and vibrating as it moves. At a particular instant the spring is stretched 0.37 m, and the two balls at the ends of the spring have the following masses and velocities:
• 1: 8 kg, ‹ 4, 11, 0 › m/s
• 2: 4 kg, ‹ −3, 10, 0 › m/s
(a) For this system, calculate p with arrowsys = ??????? kg · m/s
(b) Calculate v with arrowCM ????????? = m/s
(c) Calculate Ktot ?????????= J
(d) Calculate Ktrans ????????? = J
(e) Calculate Krel ????????? = J

Answer 1

Answer:

Explanation:

Given that,

Stiffness K=460N/m

Extension e=0.37m

Two balls attached to a string

First ball has mass and velocity

M1 =8kg and V1~ =4i +11j +0k

For the second ball

M2 =4kg and V2~ =-3i+10j +0k

a. Momentum of system?

Momentum is given as

P = m1•V1~ +m2•V2~

P= 8(4i +11j +0k) +4(-3i+10j +0k)

P=32i + 88j + 0k -12i +40j + 0k

P= 20i + 128j + 0k kgm/s

2. Magnitude of each velocity is given as

V1²=Vx² +Vy² + Vz²

V1² =4²+11²+0²

V1² =137

V1 =√137

V1=11.705m/s

Also V2

V2²= Vx² +Vy² +Vz²

V2²=(-3)²+(10)² +0²

V2²=109

V2 = √109

V2 =10.44m/s

Ktot= k(trans) +k(rel)

d. K(trans) =½m1•V1²

K(trans) =½×8×11.705²

K(trans) = 548.03J

e. K(rel) =½m2•V2²

K(rel) =½ ×4×10.44²

K(rel) = 217.99J

c. Then, K(tot)=K(trans)+k(rel)

K(tot) =548.03+217.99

K(tot) = 766.02J

Answer 2

The definitions of momentum and energy we can find the results for the questions about the momentum and the kinetic energy of the system are:  

      a) The total momentum is: $p_{total}$ =  (20 i + 128 j + 0 k)   $kg \frac{m}{s}$  

      b) The velocity of the center of mass is: $v_{cm}$ = 1.67 i + 10.67j + 0k) m/s.

      c) The total kinetic energy of the center of mass is: $k_{cm}$ = 699.8 J

      d) The translational kinetic energy is: $k_{traslation}$ = 766 J

      ed) The rotational kinetic energy   is:  $K_{rotation}$ = 66.2 A

Given parameters

• Mass 1  is:  m₁ = 8 kg

• Mass 2 is:  m₂ = 4 kg

• Mass velocity 1 is:  v₁ = {4, 11, 0> m / s

• Mass velocity 2 is: v₂ = <-3, 10.0> m / s

• Spring constant k = 460 N / m

• Spring displacement x = 0.37 m

To find

     a) The moment of the system

     b) The velocity of the center of mass

     c) Total energy

     d) Translational energy

     e) Rotational energy

In a multi-particle system the quantities can be calculated for each part and then added correctly, the physical quantities can be:

• Scalars They are quantities that only have a modulus and are added using the algebraic sum

• Vector. They are quantities that have modulus and direction, vector algebra must be used for their sum.

The system is made up of two spheres and a spring.

a) The momentum is defined by the product of the mass and the velocity of the body.

In the system the total momentum is the sum of the moment of each particle

         $p_{tota} = p_1+p_2 \\p_{total} = m_1 v_1 + m_2v_2$

Let's calculate

         $p_{total}$  = 8 (4 i + 11j + 0 k) + 4 (-3 i + 10 j + 0k)

         $p_{total}$ = i (32- 12) + j (88 + 40) + k 0

         $p_{total}$  = (20 i + 128 j + 0 k)  kg m / s

b) The center of mass of a body is the point where all external forces are applied, its speed is:

          $v_{cm} = \frac{1}{M_{total}} \sum m_i v_i$

          $v_{cm} = \frac{1}{m_1+m_2 } \ (m_1 v_1 + m_2 v_2)$

Let us conclude

         $v_{cm}$ = $\frac{1}{8+4}$  [ 8 (4 i + 11j + 0 k) + 4 (-3 i + 10 j + 0k)]

        $v_{cm}$  = 1.67 i + 10.67j + 0k) m / s

c) Total kinetic energy is the kinetic energy of the center of mass.

         $K_{total = \frac{1}{2} (m_1+m_2) \ v_{cm}^2$  

         

We use the Pythagorean theorem to find the modulus of the velocity.

          $v_{cm}^2 = v_x^2 + v_y^2$  

          $v_{cm}^2$  = 1.67² + 10.67²

          $v_{cm}^2$  = 116.6   m²/s²

 

         $K_{total}$  = ½ (8 +4) 116.6

         $k_{total}$  = 699.8 J

d) Translational kinetic energy

We use the Pythagorean theorem to find the modulus of the velocity.

         $v_1^2 = v_{1x}^2 6 v{1y}^2 + v_{1z}^2 \\v_1^2 = 4^2 +11^2 +0^2$  

          $v_1^2$ = 137    (m/s)²

          $v_2^2 = v_{2x}^2 + v_{2y}^2 + v_{2z}^2 \\v_2^2 = 3^2 +10^2 + 0^2$

           $v_2^2$ = 109

Let's calculate

            $K_{traslation}$  = K₁ + K₂

            $K_{traslation}$  = ½ m₁ v₁² + ½ m₂ v₂²

 

Let's calculate the kinetic energy

           $K_{traslation}$ = ½ 8 137 + ½ 4 109

          $K_{traslation}$ = 766 J

d) Calculate the. kinetic energy of rotation

Let us use that the kinetic energy is formed by a traslational part and a rotational part, the traslation energy is the  energy of the center of mass since, therefore,  does not rotate.

         $K_{total} = K_{traslation} + K_{rotation}$  

         $K_{rotation} = K_{total} - k_{traslation}$

Let's calculate

         Kr = 766 - 699.8

         Kr = 66.2 J

In conclusion, using the definitions of momentum and energy we can find the results for the questions about the moment and the kinetic energy of the system are:

      a) The total momentum is: $p_{total}$ =  (20 i + 128 j + 0 k)   $kg \frac{m}{s}$  

      b) The velocity of the center of mass is: $v_{cm}$ = 1.67 i + 10.67j + 0k) m/s.

      c) The total kinetic energy of the center of mass is: $k_{cm}$ = 699.8 J

      d) The translational kinetic energy is: $k_{traslation}$ = 766 J

      e) The rotational kinetic energy   is: $K_{rotation}$  = 66.2 A

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