A block sliding along a horizontal frictionless surface with speed v collides with a spring and compresses it by 2.0 cm. What will be the compression if the same block collides with the spring at a speed of 2v?

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oladayoademosu oladayoademosu · Answer 1 · 2020-03-22T06:39:21+01:00

Answer:

4.0 cm

Explanation:

For the compression of the spring, the kinetic energy of the mass equals the elastic potential energy of the spring.

So, 1/2mv² = 1/2kx² ⇒ x = (√m/k)v

Since m and k are constant since its the same spring x ∝ v

If our speed is now v₁ = 2v, our compression is x₁

x₁ = (√m/k)v₁ = (√m/k)2v = 2(√m/k)v = 2x

x₁ = 2x

Since x = 2.0 cm, our compression for speed = 2v is

x₁ = 2(2.0) = 4.0 cm

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nuhulawal20 nuhulawal20 · Answer 2 · 2021-10-28T14:37:55+02:00

If the same block collides with the spring at a speed of 2v, the compression will be 4.0cm.

Given the data in the question;

Using conservation of energy:

Kinetic energy of the mass = Elastic potential energy of the spring

We have:

[tex]\frac{1}{2}kx^2 = \frac{1}{2}mv^2\\\\kx^2 = mv^2[/tex]

"v" is directly proportional to "x"

Hence,

[tex]\frac{x_1}{x_2} = \frac{v_1}{v_2}[/tex]

We substitute in our given values

[tex]\frac{2.0cm}{x_2} = \frac{v}{2v}\\\\x_2 = \frac{v(2.0cm*2)}{v} \\\\x_2 = (2.0cm*2)\\\\x_2 = 4.0cm[/tex]

Therefore, if the same block collides with the spring at a speed of 2v, the compression will be 4.0cm.

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