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

Answer:

$E_k = 0.55v_1^2 J$

$h = 0.0128v_1^2$

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

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

$E_k = \frac{(m_1 + m_2)v^2}{2} = \frac{(2.2 + 2.2)(0.5v_1)^2}{2} = \frac{4.4*0.25v_1^2}{2} = 0.55v_1^2 J$

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:

$gh = v^2/2$

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