# Elastic collision of a glider

1. Apr 21, 2007

### r16

1. The problem statement, all variables and given/known data
A target glider, whose mass $m_2$ is 350g is at rest on an air track, a distance $d =53cm$ from the end of the track. A projectile glider whose mass $m_1$ is 590g approaches the target flider with velocity $v_{1i} = -75 cm/s$ and collides elastically with it. The target glider rebounds elastically from a short spring at the end of the track and meets the projectile glider for a second time. How far from the end of the track does this second collision occur?

2. Relevant equations

Elastic collision $KE_{i} = KE_{f}$
$$v_{1f} = \frac {m_1 - m_2}{m_1 + m_2} v_{1i}$$
$$v_{2f} = \frac {2 m_2}{m_1 + m_2} v_{1i}$$

3. The attempt at a solution

I broke it down into 2 seperate stages, a $t_1$ from when $m_2$ goes from its starting point to the wall (a distance of $d$) and a $t_2$ from when $m_2$ rebounds from the wall and collides with $m_1$ again.

$$v_{1f} = \frac {m_1 - m_2}{m_1 + m_2} v_{1i} = \frac {.590kg - .350 kg}{.590kg + .350kg} \times -.75m/s = -.19 m/s$$

$$v_{2f} = v_{2f} = \frac {2 m_2}{m_1 + m_2} v_{1i} = \frac { 2x.350kg}{.590kg + .350kg} \times -.75m/s = -.55 m/s$$

$$t_1 = \frac {x}{v_{02}} = \frac {.53m}{.55m/s} = .96s$$

$$x_1 = v_{01}t = (.19)(.96) = .18m$$

So in time interval $t_1$ the collision occurs and accelerates $m_2$ from rest to .55 m/s and $m_1$ is still moving at .19 m/s. It takes .96 seconds for $m_2$ to go $d$ and reach the end of the track and in this time $m_1$ moves .18m. Then $m_2$ has an elastic collision with the short spring and now has a velocity of $v_{2f}$.

Now:

$$x_2 = v_{02} t$$
$$x_1 = v_{01}t + x_{01}$$

Setting these equal when they collide and solving for $t$:

$$t_2 = \frac {x_{01}}{v_{02} - (-v_{01})} = \frac{.53m - .18m}{.55m/s + .19 m/s} = .47s$$

$$x_2 = v_{02}t = (.55)(.47) = .26m$$

I feel confident this is the correct answer; however, the book says they collide the second time at .35m. $m_1$ was at .35m when $m_2$ collided with the wall at the end of the track. I think the book may have gotten those answers confused. Or I did something incorrectly, but then I don't know what it is. Is this the correct answer?

Last edited: Apr 21, 2007
2. Apr 21, 2007

### denverdoc

heres how i did: assuming your numbers are right and both moving towards end;

equate times for two sliders at which collision occurs, (I omitted units cm and cm/s)

53/55+(53-Y)/55=y/19 where y is the distance travelled by slower block after collision

solving above, gives approx 27cm, and the distance to end= 53-27=26

Last edited: Apr 21, 2007
3. Apr 21, 2007

### r16

Agreed that the book is wrong again? I'm using halladay resnick walker fundamentals of physics. Does anybody know of more errors in the book? I'm not in an actual physics class, I'm just freelancing doing problems in my spare time so the answers in the back are the only way to know I'm understanding the material.

4. Apr 22, 2007

### denverdoc

Assuming your velocities are right, and thats my recollection of the elastic case w/o looking it up, yea an error. That text has been around since I went to undergrad school, I should think it would be nearly error free, course they muct update their problems periodically.

You may want to pick up the Zen Of Physics, 3000.... for like 15.00 Not a stand alone text, but if you want a lot of problems to check your methods, good investment IMO.

5. Apr 22, 2007

What is the title called exactly? I am interested...:tongue2:

6. Apr 23, 2007

### r16

Fundamentals of physics
Fifth edition

Halliday resnick walker

7. Apr 23, 2007