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## Homework Statement

A target glider, whose mass [itex]m_2[/itex] is 350g is at rest on an air track, a distance [itex] d =53cm [/itex] from the end of the track. A projectile glider whose mass [itex] m_1 [/itex] is 590g approaches the target flider with velocity [itex] v_{1i} = -75 cm/s [/itex] 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?

## Homework Equations

Elastic collision [itex] KE_{i} = KE_{f} [/itex]

[tex]v_{1f} = \frac {m_1 - m_2}{m_1 + m_2} v_{1i}[/tex]

[tex]v_{2f} = \frac {2 m_2}{m_1 + m_2} v_{1i}[/tex]

## The Attempt at a Solution

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

[tex]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 [/tex]

[tex] 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 [/tex]

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

[tex] x_1 = v_{01}t = (.19)(.96) = .18m [/tex]

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

Now:

[tex] x_2 = v_{02} t [/tex]

[tex] x_1 = v_{01}t + x_{01} [/tex]

Setting these equal when they collide and solving for [itex] t [/itex]:

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

[tex] x_2 = v_{02}t = (.55)(.47) = .26m [/tex]

I feel confident this is the correct answer; however, the book says they collide the second time at .35m. [itex]m_1[/itex] was at .35m when [itex]m_2[/itex] 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?

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