Max Shortening of Spring for Elastic Collision 3

[Karol]

Homework Statement

2 masses moving on a friction less surface. one of 2 kg moving at a speed of 10 and one of 6 kg moving at the speed of 4. at the back of the heavier one is a spring with a constant k=800.
What is the maximum shortening of the spring

Homework Equations

Conservation of momentum: m_1v_1+m_2v_2=m_1u_1+m_2u_2
Conservation of energy: \frac{1}{2}mv^2=mgh
Elastic energy in s spring: E=\frac{1}{2}kx^2

The Attempt at a Solution

At the maximum shortening the velocities are equal:
Conservation of energy:
\frac{1}{2}\cdot 2 \cdot 100=\frac{1}{2}\cdot 800 \cdot x^2+\frac{1}{2}\cdot 8 \cdot u^2
Conservation of momentum:
2 \cdot 10=8 \cdot u
Those give x=0.43 while it should be 0.259

 

[Tanya Sharma]

At the maximum shortening the velocities are equal:
Conservation of energy:
\frac{1}{2}\cdot 2 \cdot 100=\frac{1}{2}\cdot 800 \cdot x^2+\frac{1}{2}\cdot 8 \cdot u^2

6 kg block is in motion .You have not considered its kinetic energy.

Conservation of momentum:
2 \cdot 10=8 \cdot u

Same mistake .You have not included momentum of the 6 kg block.

 

[vela]

It would help if you would write the equations down in terms of variables first. You should avoid plugging numbers in until later in the exercise-solving process.

 

[Karol]

I took in consideration the 6 kg block in both cases. the number 8 in the kinetic energy and the momentum equation is (2+6)

 

[Tanya Sharma]

I took in consideration the 6 kg block in both cases. the number 8 in the kinetic energy and the momentum equation is (2+6)

You have incorrectly presumed that 6 kg block is initially stationary .Look carefully in the picture .The 6 kg block is moving with 4m/s before the 2 kg block comes in contact with the spring.

Include the momentum of 6 kg block in the initial momentum i.e in the left hand side .Same with energy conservation .

 

[Karol]

Equations

$\frac{1}{2}m_1v_1=\frac{1}{2}kx^2+\frac{1}{2}(m_1+m_2)u^2$
$\Rightarrow \frac{1}{2}\cdot 2 \cdot 100=\frac{1}{2}\cdot 800 \cdot x^2+\frac{1}{2}\cdot 8 \cdot u^2$
$m_1v_1=(m_1+m_2)u$
$\Rightarrow 2 \cdot 10=8 \cdot u$

 

[vela]

Have you read at all what Tanya has written?

 

[Tanya Sharma]

$\frac{1}{2}m_1v_1=\frac{1}{2}kx^2+\frac{1}{2}(m_1+m_2)u^2$
$\Rightarrow \frac{1}{2}\cdot 2 \cdot 100$$=\frac{1}{2}\cdot 800 \cdot x^2+\frac{1}{2}\cdot 8 \cdot u^2$
$m_1v_1=(m_1+m_2)u$
$\Rightarrow 2 \cdot 10$$=8 \cdot u$

Items in red are incorrect.

 

[Karol]

Equations

$\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2=\frac{1}{2}kx^2+\frac{1}{2}(m_1+m_2)u^2$
$\Rightarrow \frac{1}{2}\cdot 2 \cdot 100+\frac{1}{2}\cdot 6 \cdot 16=\frac{1}{2}\cdot 800 \cdot x^2+\frac{1}{2}\cdot 8 \cdot u^2$

$m_1v_1+m_2v_2=(m_1+m_2)u$
$\Rightarrow 2 \cdot 10 +4 \cdot 6=8 \cdot u \rightarrow u=6$

They give x=0.1, wrong

 

[Tanya Sharma]

$\Rightarrow 2 \cdot 10 +4 \cdot 6=8 \cdot u \rightarrow$ u=6

Be careful with your calculations .$u = \frac{44}{8} ≠ 6$