The sample data shows a cart that passes s through the first and second photogat

Question

The sample data shows a cart that passes s through the first and second photogates, bounces off the wall, at the end of the track, and passes back through both photogates.

You will measure the distance between the photogates during the lab. You will then use the time data to calcualte the cart'[s velocity for the collision studies.

Calcualte the velocity as it passes through the photogates the first time.

Assume they are 8cm apart.

So Gatestate1 is 1

Gatstate 2 is not given (I assume 0)

And then the time is 1.200125

so how do I calculate the velocity?

It's hitting the wall and then bouncing back.

Explanation / Answer

now, assume the ball is travelling with constant velocity. and also assume the collision is elastic. now, m1v1 = -m2v2. ==> it rebounds with same velocity . now, given, d= 8cm = 0.08 m t= 1.200125 s s= ut v= s/t= 0.08 / 1.200125= 0.0666 m/s = 6.66cm/s

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