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tino4ka555 [31]
3 years ago
7

Croquet balls must have a mass of 0.50 kg. A red croquet ball is moving at 5 m/s. It strikes an at-rest green croquet ball head-

on and then continues to move in the same direction, but with a speed of 2 m/s. What is the final speed of the green ball?
A.
5 m/s

B.
0.5 m/s

C.
2 m/s

D.
3 m/s
Physics
1 answer:
DochEvi [55]3 years ago
8 0

Answer:

Explanation:

The Law of Momentum Conservation for us has the equation

[m_rv_r+m_gv_g]_b=[m_rv_r+m_gv_g]_a and filling in:

[(.50)(5.0)+(.50)(0)]=[(.50)(2.0)+(.50)v_g] and

2.5 = 1.0 + .50v and

1.5 = .50v so

v = 3 m/s

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The initial kinetic energy imparted to a 0.25 kg bullet is 1066 J. The acceleration of gravity is 9.81 m/s 2 . Neglecting air re
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Answer:

The range of the bullet is 0.435 kilometers.

Explanation:

According to the problem, maximum height is equal to the range of the bullet. That is:

\Delta x = \Delta y

Where:

\Delta x - Range of the bullet, measured in meters.

\Delta y - Maximum height of the bullet, measured in meters.

By the Principle of Energy Conservation, gravitational potential energy reaches its maximum at the expense of the initial kinetic energy. That is to say:

K_{1} = U_{2}

Where:

K_{1} - Kinetic energy at point 1, measured in joules.

U_{1} - Gravitational potential energy at point 2, measured in joules, and:

U_{2} = m\cdot g \cdot \Delta y

Where:

m - Mass of the bullet, measured in kilograms.

g - Gravitational constant, measured in meters per square second.

The maximum height is now cleared:

K_{1} = m\cdot g \cdot \Delta y

\Delta y = \frac{K_{1}}{m\cdot g}

If K_{1} = 1066\,J, m = 0.25\,kg and g = 9.81\,\frac{m}{s^{2}}, the maximum height is now computed:

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