Question

Two blocks of mass 1 and 3 Kg are moving with velocities 2 and 1 m/s respectively as shown.If the spring constant is 75 ,the maximum compression of the spring is

Answer 1

Explanation

Conservation of momentum

m₁v₁ +m₂v₂ = (m₁+m₂)v

1x2 + 3x1  = 4v

            v = 5/4

Conservation of energy

1/2m1 v1²  + 1/2 m2 v2² = 1/2kx² +1/2 (m1+m2)v²

4 + 3 = 75 x² +4 x 25/16

    x² =1/100

   x =0.01 m = 10 cm

                    Hope you understand

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Answer 2

Given:

Two masses, of $1kg$ and $3 kg$ moving with velocities $2m/s$ and $1m/s$ .

Spring constant $=75N/m$

To find:

Maximum compression in the springs.

Explanation:

We can see that the two blocks of masses $1kg$ and $3kg$ are connected to a common spring and moving in such a manner that the momentum possessed by one body is passed on to the other body on collision.

After certain time, the two bodies are seen as connected and moving with the same velocity.

This can be shown using the Law of conservation of momentum, which states that of the momentum of two bodies before them getting collided is equal to their momentum after collision .

                     $m_{1} u_{1}+m_{2} u_{2}=m_{1} v_{1}+m_{2} v_{2}$

Solution:

According to the question,

Let the two masses be $m_{1}=1kg$  and $m_{2}=3kg$ moving initially with velocities $u_{1}=2m/s$  and $u_{2} =1m/s$ and after collision they both move together with a velocity $v_{2}$.

Hence,

Applying Law of conservation of momentum to the two masses before and after collision, we get

$1$ ×$2$ $+$ $3$ ×$1$ $=(1+3)v_{}$

$2+3=4v_{}$

$v_{} =\frac{5}{4}m/s$  

Hence, $m_{1}$ and $m_{2}$ after collision, move together with a final velocity of 1.25 m/s.

Now,

If we see from the diagram, the masses on the spring are initially moving with a kinetic energy and after their collision, the energy in the springs is transferred as the compression produced in the spring.

According to law of conservation of energy, energy in a system remains conserved always that is it is never lost anywhere but is transferred from one form of energy to the other.

Δ$Energy=0$

Energy of the spring before collision is equal to the energy in the spring after collision.

$\frac{1}{2}m_{1}u_{1} ^{2} +\frac{1}{2}m_{2}u_{2} ^{2} =\frac{1}{2}(m_{1}+ m_{2} ) v^{2} +\frac{1}{2}kx^{2}$

Substituting the known values in the equation, we get

$\frac{1}{2}(1)(2)^{2} +\frac{1}{2}(3)(1)^{2} =\frac{1}{2}(1+3)(\frac{5}{4} )^{2}+\frac{1}{2}(75)x^{2}$

$4+3=6.25+75x^{2}$

$7-6.25=75x^{2}$

$0.75=75x^{2}$

$x=\frac{1}{100}m$    $;$ or $x=1 cm$

Final answer:

Hence, the maximum compression produced in the spring is 10⁻²m or 1 cm.

Although your question is incomplete, you might be referring to the diagram below.