Question

A body of mass 20kg moving with a speed of 10 ms^-1 on a horizontal smooth surface collides with a massles spring of spring constant 5N/m. If it stops after collision, distance of compression of spring will be ?

Answer 1

Solution

Compression in the spring is 20 m

Given

• Mass of the moving object,M = 20 Kg

• Spring Constant,K = 5 N/m

• Velocity ,v = 10 m/s

To finD

Compression In Spring

When the object collides the Kinetic Energy of the object would be equal to Elastic Potential Energy of the string

$\longrightarrow \: \sf \: \dfrac{1}{2} m {v}^{2} = \dfrac{1}{2} k {x}^{2} \\ \\ \longrightarrow \: \sf \: m {v}^{2} = k {x}^{2} \\ \\ \longrightarrow \: \sf \: x = \sqrt{ \dfrac{m {v}^{2} }{k} } \\ \\ \longrightarrow \: \sf \: x = v \times \sqrt{ \dfrac{m}{k} } \\ \\ \longrightarrow \: \sf \: x = 10 \times \sqrt{ \dfrac{20}{5} } \\ \\ \longrightarrow \: \sf \: x = 10 \times \sqrt{4} \\ \\ \huge{ \longrightarrow \: \boxed{ \boxed{ \sf \: x = - 20 \: m}}}$

The negative sign indicates the compression of spring

Answer 2

$\huge{\underline{\underline{\red{\mathfrak{AnSwEr :}}}}}$

Given :

• Mass of body (m) = 20 kg
• Velocity (v) = 10 m/s
• Spring Constant (k) = 5 N/m

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To Find :

• Distance of collision of spring

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Solution :

In case of collision Potential Energy is equal to the Elastic Potential energy.

We have the formulas :

$\large{\boxed{\boxed{\sf{K.E \: = \: \dfrac{1}{2} mv^2}}}} \\ \\ \large{\boxed{\boxed{\sf{K.E_{spring} \: = \: \dfrac{1}{2} kx^2}}}}$

Equate both the above equations

$\implies {\sf{\dfrac{1}{2} mv^2 \: = \: \dfrac{1}{2} kx^2}} \\ \\ \implies {\sf{mv^2 \: = \: kx^2}} \\ \\ \implies {\sf{x^2 \: = \: \dfrac{mv^2}{k}}} \\ \\ \implies {\sf{x^2 \: = \: \dfrac{20 \: \times \: 10^2}{5}}} \\ \\ \implies {\sf{x^2 \: = \: \dfrac{2000}{5}}} \\ \\ \implies {\sf{x^2 \: = \: 400}} \\ \\ \implies{\sf{x \: = \: \sqrt{400}}} \\ \\ \implies {\sf{x \: = \: 20}}$

As the distance is in compression.

So,

⇒ x (distance) = -20 m