1. ## physics

i need help on how to solve this problem

a linear air track with length 1.5 m is inclined at an angle of 30.0 degrees with the horizontal. a glider of mass 25.3 g is released from the bottom of the track with an initial velocity of 3.0 m/s upward along the track. after reaching zero velocity, another glider of mass 30.5 g was released but this time from the top end of the air track. the initial velocity of the second glider is 9.0 m/s down along the air track. will the two gliders meet? if they will meet, at what distance from the bottom of the track will they meet? assume gliders as points and the linear air tracks as frictionless.

2. Originally Posted by maeca
i need help on how to solve this problem

a linear air track with length 1.5 m is inclined at an angle of 30.0 degrees with the horizontal. a glider of mass 25.3 g is released from the bottom of the track with an initial velocity of 3.0 m/s upward along the track. after reaching zero velocity, another glider of mass 30.5 g was released but this time from the top end of the air track. the initial velocity of the second glider is 9.0 m/s down along the air track. will the two gliders meet? if they will meet, at what distance from the bottom of the track will they meet? assume gliders as points and the linear air tracks as frictionless.
Please show us what you have done so far and let us know where you get stuck.

I can start you off:

1. The masses do not matter. How heavy a ball is does not change how fast it drops towards the ground.

2. A change of coordinates will make your life a lot easier. Instead of working with the glider at an incline, pretend that you tilt your head so that the glider is horizontal from your point of view.

3. The acceleration for a free-fall drop close to Earth is $\displaystyle g$ (approximately $\displaystyle 9.81 m/s$). However, because the glider is inclined only by 30 degrees instead of the full 90, use sine of 30 as scaling factor in your accelerations.

4. Set up two kinematic equations dictating each glider and calculate for the point where the two equations are equal (i.e. - the gliders meet)

3. acceleration for both gliders will be a constant, $\displaystyle a = g\sin(30) = 4.9 \, m/s^2$ down the track.

the first cart will have a displacement of $\displaystyle \Delta x = \frac{v_f^2 - v_0^2}{2a} = \frac{-9}{-9.8} = .92 \, m$ up the track.

therefore, the second cart will be released when the first cart is $\displaystyle 1.5 - .92 = .58 \, m$ down the track.

taking the position of the second cart at the top of the track as 0, the position of the first cart as a function of time is ...

$\displaystyle x_1 = .58 + 4.9t^2$

the position of the second cart as a function of time is ...

$\displaystyle x_2 = 9.0t + 4.9t^2$

the meeting point will be when $\displaystyle x_1 = x_2$. set the two position functions equal to each other, solve for t, then determine the meeting position relative to the top of the track ... if it is $\displaystyle \leq 1.5$ , then the two carts will meet while they are both still on the track. finally, subtract the determined position from 1.5 to answer the desired question.

4. ## thanks

thanks so much