A 2.00 kg ball is thrown upward at some unknown angle from the top of a 20.0 m high building. If the initial magnitude of the ve
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Answer:
The total momentum of the system before the collision is 0.0325 kg-m/s due east direction.
Explanation:
Given that,
Mass of the cart, m = 250 g = 0.25 kg
Initial velocity of the cart, u = 0.31 m/s (due right)
Mass of another cart, m' = 500 g = 0.5 kg
Initial velocity of the another cart u' = -0.22 m/s (due left)
Let p is the total momentum of the system before the collision. It is given by :
So, the total momentum of the system before the collision is 0.0325 kg-m/s due east direction.
Answer:
a
b
The value is
Explanation:
From the question we are told that
The mass is
The spring constant is
The instantaneous speed is
The position consider is x = 0.750A meters from equilibrium point
Generally from the law of energy conservation we have that
The kinetic energy induced by the hammer = The energy stored in the spring
So
Here a is the amplitude of the subsequent oscillations
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Generally from the law of energy conservation we have that
The kinetic energy by the hammer = The energy stored in the spring at the point considered + The kinetic energy at the considered point
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