A target glider, whose mass m2 is 300 g , is at rest on an air track, a distance

Question

A target glider, whose mass m2 is 300 g, is at rest on an air track, a distance d = 53 cm from the end of the track. A projectile glider, whose mass m1 is 620 g, approaches the target glider with velocity v1i = -75 cm/s and collides elastically with it (see the figure). 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 ( cm)?

How much more time elapses before the gliders collide again?

Explanation / Answer

THIS THE SIMILAR QUESTION YOU JUST HAVE REPLACE VALUES AND YOU R DONE
(sorry for not doing exact calculations)

Elastic collision KE_{i} = KE_{f}
v_{1f} = rac {m_1 - m_2}{m_1 + m_2} v_{1i}
v_{2f} = rac {2 m_2}{m_1 + m_2} v_{1i}

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} = rac {m_1 - m_2}{m_1 + m_2} v_{1i} = rac {.590kg - .350 kg}{.590kg + .350kg} imes -.75m/s = -.19 m/s

v_{2f} = v_{2f} = rac {2 m_2}{m_1 + m_2} v_{1i} = rac { 2x.350kg}{.590kg + .350kg} imes -.75m/s = -.55 m/s

t_1 = rac {x}{v_{02}} = rac {.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 = rac {x_{01}}{v_{02} - (-v_{01})} = rac{.53m - .18m}{.55m/s + .19 m/s} = .47s

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

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