PS115L Technical Physics Lab HW#9 Due At Beginning ✓ Solved

PS115L Technical Physics Lab HW#9 Homework 9 Due at begi

A mass of .6 kg moving with a velocity of 2.4 m/s in the positive direction experiences a totally inelastic collision with a mass of 1.1 kg. What is the final velocity of the masses?

A mass of 7.2 kg moving with a velocity of .8 m/s experiences an elastic collision with a mass that is stationary. After the collision, the velocity of the second mass is 10.286 m/s while the velocity of the first mass is 0 m/s. What is the mass of the second object?

A mass of .43 kg moving with a velocity of 1.5 m/s in the positive direction experiences an elastic collision with another mass of .22 kg moving with a velocity of 3.4 m/s in the negative direction. What will the velocities of the two masses be after the collision?

Remember that in elastic collisions, both kinetic energy and momentum are conserved.

Paper For Above Instructions

The study of collisions is fundamental in physics, illustrating principles of momentum and energy conservation. This paper addresses three scenarios involving elastic and inelastic collisions pertinent to PS115L Technical Physics Lab homework.

1. Inelastic Collision

In the first scenario, we have a mass (m₁ = 0.6 kg) moving at a velocity (v₁ = 2.4 m/s) colliding inelastically with another mass (m₂ = 1.1 kg) initially at rest. An inelastic collision characteristically conserves momentum but not kinetic energy.

Using the conservation of momentum formula: m₁v₁ + m₂v₂ = (m₁ + m₂)v_f, where v₂ is 0, we can calculate the final velocity (v_f) of the combined masses:

0.6 kg 2.4 m/s + 1.1 kg 0 m/s = (0.6 kg + 1.1 kg) * v_f

This simplifies to:

1.44 kg·m/s = 1.7 kg * v_f

Therefore, v_f = 1.44 kg·m/s / 1.7 kg ≈ 0.847 m/s.

In conclusion, after the inelastic collision, the final velocity of the two combined masses is approximately 0.847 m/s.

2. Elastic Collision

The second problem presents an elastic collision between a moving mass (m₁ = 7.2 kg) at a velocity (v₁ = 0.8 m/s) and a stationary mass (m₂ with unknown mass) where m₁ comes to rest, and m₂ moves at 10.286 m/s. Elastic collisions conserve both momentum and kinetic energy.

The momentum conservation equation here becomes:

m₁v₁ + m₂v₂₂ = m₁v₁' + m₂v₂', where v₁' = 0 and v₂' = 10.286 m/s.

Therefore:

7.2 kg 0.8 m/s + m₂ 0 = 0 + m₂ * 10.286 m/s

This simplifies to:

5.76 kg·m/s = m₂ * 10.286 m/s.

So, m₂ = 5.76 kg·m/s / 10.286 m/s ≈ 0.56 kg.

Hence, the mass of the second object is approximately 0.56 kg.

3. Elastic Collision with Two Moving Bodies

In the third case, we look at an elastic collision between a mass (m₁ = 0.43 kg) moving at v₁ = 1.5 m/s, and another mass (m₂ = 0.22 kg) moving at v₂ = -3.4 m/s.

By using the equations derived from the conservation of momentum and kinetic energy for elastic collisions, we find:

The conservation of momentum gives us:

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'.

The conservation of kinetic energy yields:

(1/2)m₁v₁² + (1/2)m₂v₂² = (1/2)m₁v₁'² + (1/2)m₂v₂'².

To solve these equations for the final velocities is more complex; however, using the derived formulas for one-dimensional elastic collisions, we state:

v₁' = (v₁(m₁ - m₂) + 2m₂v₂) / (m₁ + m₂)

v₂' = (v₂(m₂ - m₁) + 2m₁v₁) / (m₁ + m₂).

Substituting the known values, we calculate:

v₁' = [(1.5 (0.43 - 0.22) + 2 0.22 * (-3.4)) / (0.43 + 0.22)] ≈ -1.99 m/s.

v₂' = [(-3.4 (0.22 - 0.43) + 2 0.43 * (1.5)) / (0.43 + 0.22)] ≈ 4.75 m/s.

Thus, the velocities after the collision will be approximately -1.99 m/s for m₁ and 4.75 m/s for m₂.

Conclusion

This assignment illustrates the fundamental principles of momentum and energy conservation within elastic and inelastic collisions. The calculated final velocities help understand physical interactions and the outcomes in collision scenarios.

References

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