A0.450-kg ice puck, moving east with a speed of 5.60m/s , has a head-on collision with a 0.950-kg puck initially at rest. assume that the collision is perfectly elastic. (1)- what is the speed of the 0.450-kg puck after the collision? (2)- what is the direction of the velocity of the 0.450- puck after the collision? (i)- to the east (ii)- to the west (3)- what is the speed of the 0.950-kg puck after the collision? (4)- what is the direction of the velocity of the 0.950-kg puck after the collision? (i)-to the east (ii)- to the west

1) The disc speed of 0.450 kg is vif = -2.0 m / s

2) The disc is receding because the negative bone velocity is heading west

3) Disc speed 0.950 kg V2f = 3.6 m / s

4) The disk follows the initial direction of the light disk, bone moves east

Explanation:

For the problems of crashes we have to define a system in this case the system is formed by the two disks, write the amount of movement before and after the crash

         p₀ = m₁ v₁₀ + 0

         pf = m₁ v₁f + m₂ v₂f

For the system during the crash all forces are internal, so the moment is preserved

        po = pf

        m₁ v₁₀ = m₁ v₁f + m₂ v₂f

Like elastic shock, kinetic energy is also conserved

        Ko = Kf

        ½ m₁ v₁₀² = ½ m₁ v₁f² + ½ m₂ v₂f²

Let's write the system of equations

       m₁ (v₁₀ -v₁f) = m₂ v₂f

       m₁ (v₁₀² - v₁f²) = m₂ v₂f²

We replace the first in the second

       m₁ (v₁₀² - v₁f²) = m₂ [m₁ (v₁₀ -v₁f) / m₂]²

       m₁ (v₁₀² - v₁f²) = [m1² (v₁₀ -v₁f)² / m₂

       (v₁₀² - v₁f²) = [m₁ (v₁₀ -v₁f)² / m₂

       m₁ (v₁₀ - v₁f) (v₁₀ + v₁f) = [(v₁₀ -v₁f)² m₁ / m₂

       (v₁₀ + v₁f) = (v₁₀ -v₁f) m₁ / m₂

       v₁f (1+ m₁ / m₂) = v₁₀ (m₁ / m₂ -1)

       v₁f (m₂ + m₁) / m₂ = v₁₀ (m₁ -m₂) / m₂

       v₁f = v₁₀ [(m₁ -m₂) / (m₁ +m₂)]

We calculate

      V₁f = 5.6 [(0.450 -0.950) / ((0.450+ 0.90)] = 5.6 (-0.500 /1.4)

      V₁f = - 2.0 m / s

The negative sign indicates that the disk goes back

     m₁ (v₁₀ -v₁f) = m₂ v₂f

     v₂f = m₁ / m₂ (v₁₀ -v₁f)

     v₂f = 0.450 / 0.950 (5.6 - (-2))

     v₂f = 3.6 m / s

Let's write the results

1) The disc speed of 0.450 kg is vif = -2.0 m / s

2) The disc is receding because the negative bone velocity is heading west

3) Disc speed 0.950 kg V2f = 3.6 m / s

4) The disk follows the initial direction of the light disk, bone moves east

explanation:

as the height from which the ruler was dropped increased, the speed of the ruler at ryan's hand increased.

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.