K1=initial kinetic energy
K2=kinetic energy after collision
K1 = (1/2)(5.2kg)(9.4m/s)^2
K2 = (1/2)(5.2kg)(1.80m/s)^2
kinetic energy lost is:
K1 - K2
Ohmygoodness, I am not having a good time tonight.
"A 5.2kg block sliding at 9.4 m/s across a horizontal frictionless surface collides head on with a stationary 8.6kg block. The 5.20kg block rebounds at 1.80 m/s. How much kinetic energy is lost during this collision?"
My FBD just isn't working; after I calculate momentum and attempt a vector diagram, my work falls apart. I'm really starting to panic!
4 answers
Use conservation of momentum to compute the velocity of the block that was initially stationary. Make sure you get the directions right. They will go off in opposite directions.
Once you have all the final velocities, compute how much the total kinetic energy decreased after the collision.
Once you have all the final velocities, compute how much the total kinetic energy decreased after the collision.
Quidditch's answer assumes block 2 remains stationary. I do not think that is what they want to to assume.
Disregard my answer. I assumed that the stationary block was fixed. I don't think that is the case. DRWLS has the correct approach. He is very good.
0 answers