Asked by cake
Use the equations below:
P=mv
Ek=1/2mv^2
2) A linear air track can be used to investigate collisions. Two trolleys or “gliders” are supported on a cushion of air. A glider of mass 0.30kg is stationary in the middle of the track. A second glider of mass 0.25kg and velocity of 0.20ms-1 collides with the first glider and they stick together.
You may assume that this collision is perfectly elastic.
A) What is meant by a “perfectly elastic” collision?
B) Calculate the velocity of the glider combination immediately after the collision.
6 marks
P=mv
Ek=1/2mv^2
2) A linear air track can be used to investigate collisions. Two trolleys or “gliders” are supported on a cushion of air. A glider of mass 0.30kg is stationary in the middle of the track. A second glider of mass 0.25kg and velocity of 0.20ms-1 collides with the first glider and they stick together.
You may assume that this collision is perfectly elastic.
A) What is meant by a “perfectly elastic” collision?
B) Calculate the velocity of the glider combination immediately after the collision.
6 marks
Answers
Answered by
cake
please help with B
Answered by
cake
is it 0.5 * 0.2= 0.11
then 0.11=1/0.55*0.20^2 though i dont think that makes much sense
then 0.11=1/0.55*0.20^2 though i dont think that makes much sense
Answered by
Henry
Conservation of kinetic Energy.
0.5*M2*V^2 = 0.5*M1*V^2 + 0.5*M2*V^2.
0.5*0.25*0.2^2 = 0.5*0.3V^2+0.5*0.25V^2
0.005 = 0.15V^2 + 0.125V^2,
=.005 = 0.275V^2,
V^2 = 0.0182,
V = 0.135 m/s.
0.5*M2*V^2 = 0.5*M1*V^2 + 0.5*M2*V^2.
0.5*0.25*0.2^2 = 0.5*0.3V^2+0.5*0.25V^2
0.005 = 0.15V^2 + 0.125V^2,
=.005 = 0.275V^2,
V^2 = 0.0182,
V = 0.135 m/s.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.