# Homework Help: 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 .$u = \frac{44}{8} ≠ 6$