Elastic collision 3

1. May 24, 2014

Karol

1. The problem statement, all variables and given/known data
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

2. Relevant 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: $E=\frac{1}{2}kx^2$

3. 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

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2. May 24, 2014

Tanya Sharma

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

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

3. May 24, 2014

vela

Staff Emeritus
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.

4. May 24, 2014

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)

5. May 24, 2014

Tanya Sharma

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 .

6. May 24, 2014

Karol

Equations

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

7. May 24, 2014

vela

Staff Emeritus
Have you read at all what Tanya has written?

8. May 24, 2014

Tanya Sharma

Items in red are incorrect.

9. May 24, 2014

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

Last edited: May 24, 2014
10. May 24, 2014

Tanya Sharma

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

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