Ball of mass 450g lands on a spring with k = 5.0E3

In summary, a ball of mass 450g, dropped from a height of 2.50m, lands on a spring with a spring constant of 5.0E3. The maximum compression of the spring is x=mg/k and the velocity it will leave with after bouncing off the spring can be found using the equation 1/2*k2 = 1/2*m*v2."
  • #1
DrVirus
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A ball of mass 450g lands on a spring with k = 5.0E3 . The ball was originally dropped from a height of 2.50 m. Determine the maximum compression of the spring AND the velocity it will leave with after bouncing of the spring.

Thank u for helping
 
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  • #2
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Let F[/f]be the force exerted by the falling ball of mass m on the spring with spring constant k. Since the ball is under free fall, the force wih which the ball falls on the spring is its weight mg. Hence,

mg=-kx, where xis the displacement of the spring.
Therefore, x=mg/k. After the spring has been compressed, force generated is transferred to the ball. Thus the Potential energy of the spring is converted to the kinetic energy of the ball.

Thus,
1/2*k2 = 1/2*m*v2

Find out the velocity of ball using the above equation.


Sridhar
 
  • #3
me with my physics problem!

To determine the maximum compression of the spring, we can use the equation for potential energy, PE = 1/2kx^2, where k is the spring constant and x is the maximum compression. We know that the initial potential energy of the ball is equal to its gravitational potential energy, PE = mgh, where m is the mass of the ball, g is the acceleration due to gravity, and h is the initial height. Setting these two equations equal to each other and solving for x, we get:

1/2kx^2 = mgh
x = √(2mgh/k)

Plugging in the values given, we get:

x = √(2(0.450 kg)(9.8 m/s^2)(2.50 m)/(5.0E3 N/m))
x = 0.094 m

Therefore, the maximum compression of the spring is 0.094 m.

To determine the velocity of the ball after bouncing off the spring, we can use the equation for conservation of energy, E = KE + PE, where E is the total energy, KE is the kinetic energy, and PE is the potential energy. We know that the total energy before and after the bounce is the same, so we can set the two equations equal to each other:

mgh = 1/2mv^2 + 1/2kx^2

Solving for v, we get:

v = √(2gh + kx^2/m)

Plugging in the values given, we get:

v = √(2(9.8 m/s^2)(2.50 m) + (5.0E3 N/m)(0.094 m)/(0.450 kg))
v = 5.07 m/s

Therefore, the ball will leave the spring with a velocity of 5.07 m/s.
 

1. What is the equation used to calculate the period of oscillation for a ball on a spring?

The equation used to calculate the period of oscillation for a ball on a spring is T = 2π√(m/k), where T is the period, m is the mass of the ball, and k is the spring constant.

2. How does the mass of the ball affect the period of oscillation?

The mass of the ball has a direct relationship with the period of oscillation. As the mass increases, the period of oscillation also increases. This means that a heavier ball will take longer to complete one full cycle of oscillation on the spring.

3. What does the spring constant represent in this scenario?

The spring constant, denoted as k, represents the stiffness of the spring. A higher spring constant means that the spring is stiffer and will require more force to stretch or compress it. In this scenario, a higher spring constant, k = 5.0E3, means that the spring is relatively stiff.

4. How does the spring constant affect the period of oscillation?

The spring constant has an inverse relationship with the period of oscillation. This means that as the spring constant increases, the period of oscillation decreases. This can be seen in the equation T = 2π√(m/k), where a higher k value will result in a smaller T value.

5. Is the mass of the ball the only factor that affects the period of oscillation in this scenario?

No, the period of oscillation is also affected by the spring constant and the amplitude of the oscillation. A higher spring constant or a larger amplitude will result in a shorter period of oscillation, while a lower spring constant or a smaller amplitude will result in a longer period of oscillation.

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