Question

Given two identical 2.2 kg masses. The first mass is moving with a velocityFind the kinetic energy of the compound system immediately after the collision. Answer in units of J.v1 immediately before colliding with the second mass, which is suspended by a string of length 0.82 m . The two masses are stuck together as a result of the collision. The compound system then swings to the right and rises to the horizontal level. The acceleration of gravity is 9.8 m/s 2

Answer 1

The change in kinetic energy of a system during a collision is found using the conservation of momentum and the work-energy theorem. The kinetic energy of each mass before the collision is compared to the kinetic energy of the combined mass after the collision to calculate the change.

Step-by-step explanation:

To determine the change in kinetic energy of the system as a result of a collision, one can apply the principles of conservation of momentum and the work-energy theorem. Since the two objects stick together after the collision, the situation represents a perfectly inelastic collision. The initial and final kinetic energies can be calculated using the formula KE = 0.5 × m × v² where m is the mass and v is the velocity.

For example, in a scenario where a 2.0 kg mass moving at 15 m/s collides with a 5.0 kg mass initially at rest and they stick together, the change in kinetic energy can be calculated as follows:

• Initial kinetic energy of mass A: KEAinitial = 0.5 × 2.0 kg × (15 m/s)² = 225 J
• The momentum is conserved, so the final velocity vfinal of the combined mass can be determined using mA × vAinitial + mB × vBinitial = (mA + mB) × vfinal.
• Final kinetic energy of the combined mass: KEfinal = 0.5 × (2.0 kg + 5.0 kg) × vfinal²
• The change in kinetic energy is ΔKE = KEfinal - KEAinitial.

In the above example, assuming the final velocity calculated was 4.286 m/s, the final kinetic energy would be 64 J and the change in kinetic energy is -161 J (indicating a loss).

Answer 2

`E_k = 0.55v_1^2 J`

`h = 0.0128v_1^2`

where
`v_1` is the initial velocity of the moving block

Step-by-step explanation:

We can apply the law of momentum conservation to calculate the compund velocity post-collision:

`m_1v_1 +m_2v_2 = (m_1 + m_2)v`

where
`m_1 = m_2 = 2.2 kg` are the masses of block 1 (the moving block) and block 2.
`v_1 , v_2` are the initial velocities before collision of block 1 and 2, and
`v_2 = 0m/s` as it's at rest initially.

`2.2v_1 + 0 = (2.2 + 2.2)v`

`v = 2.2v_1/4.4 = 0.5v_1`

So the kinetic energy right after the collision is

`E_k = ((m_1 + m_2)v^2)/(2) = ((2.2 + 2.2)(0.5v_1)^2)/(2) = (4.4*0.25v_1^2)/(2) = 0.55v_1^2 J`

To calculate the horizontal level where the compound rises to, we need to apply the law energy conservation. As the compound rises its kinetic energy is converted to potential energy:

`E_p = E_k`

`mgh = mv^2/2`

where m is the mass and h is the vertical distance traveled, v is the velocity at the bottom, and g = 9.8 m/s2 is the gravitational acceleration.We can divide both sides by m:

`gh = v^2/2`

`h = v^2/(2g) = (0.5v_1)^2/(2g) = v_1^2/(8*9.8) = 0.0128v_1^2`