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alexandr1967 [171]
3 years ago
15

Given two identical 2.2 kg masses. The first mass is moving with a velocityFind the kinetic energy of the compound system immedi

ately 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
Physics
1 answer:
deff fn [24]3 years ago
8 0

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

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Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2

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Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.30)}{((0.30)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{202500}{0.0316}\\\\E_x=6408227.848\ N/C

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Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(1)}{((1)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{675000}{1.015}\\\\E_x=665024.6305\ N/C

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