A lump of putty and a rubber ball have equal mass. Both are thrown with equal speed against a wall. The putty sticks to the wall
1 answer:
Answer:
Option (c). The ball experiences the greater momentum change.
Explanation:
Suppose the initial momentum of the putty and the ball is .
Since the final velocity of the putty is zero, it final momentum must be zero, too. So its momentum was reduced in .
In the case of the ball, we have that its final speed is the same, but its final velocity has opposite direction. As the velocity and the momentum are vectors, the direction is relevant. Taking this into account, the final momentum of the ball is . So, its momentum was reduced in
.
This leads us to the conclussion that the rubber ball experiences a greater momentum change than the lump of putty. This is the option (c).
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where
q is the electron charge
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2) 4.1 mm
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where m is the mass of the electron and r is the radius of the trajectory. Solving the formula for r, we find
3)
The speed of the electron in the circular trajectory is equal to the ratio between the circumference of the orbit, , and the period, T:
Solving the equation for T and using the results found in 1) and 2), we find the period of the orbit:
Answer:
Explanation:
The standard form of the 2nd order differential equation governing the motion of mass-spring system is given by
Where m is the mass, ζ is the damping constant, and k is the spring constant.
The spring constant k can be found by
The damping constant can be found by
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Therefore, the required differential equation is
The initial position is
The initial velocity is