Determine the amount of kinetic energy lost due to the collision

  • #1
ClintStibbard
2
0

Homework Statement


So I am faced with a theoretical equation which uses all variables in terms of Mass,Length,Coefficient of friction which i will just write as u, radius R and gravity g. Block A of mass M is released from rest at the top of an arc ramp which is frictionless it contacts block B of mass M at the bottom inelastically. the Two blocks move together to the sliding past a distance L along a flat track of friction u. Express answers in terms of M,L,u,R and g

a determine the speed of block A before it hits block B

b determine the speed of the combined blocks immediately after collision

c determine the amount of Kinetic energy lost due to the collision.

d the specific heat of the the blocks is c. determine the temperature rise that results from the collision in terms of c and other give quatities if no energy is transferred to the track or air.

e determine additional thermal energy that is generated as the blocks move from y to p



Homework Equations





The Attempt at a Solution



a mgh=1/2mv^2 v=(2gh)^(1/2)

b determine the speed of the combined blocks after collision. m(2gh)^(1/2)=2m(2gh)^(1/2)/2

c this is where i get lost, i understand Ke is lost to an inelastic collision, but i don't understand how to determine how much?
 

Answers and Replies

  • #2
Doc Al
Mentor
45,447
1,907
a mgh=1/2mv^2 v=(2gh)^(1/2)
Seems reasonable. What's h in terms of R?

b determine the speed of the combined blocks after collision. m(2gh)^(1/2)=2m(2gh)^(1/2)/2
Not sure what you're doing here. Call the speed after the collision "V". Set up your equation and solve for V.

c this is where i get lost, i understand Ke is lost to an inelastic collision, but i don't understand how to determine how much?
What's the total KE of the blocks before the collision? After the collision? Subtract!
 
  • #3
ClintStibbard
2
0
I realized that R could be substituted in for h, and I seem to have ended up with the right answer for c. as to part b i was conserving momentum and with twice the mass the velocity had to be halved. Thank you for your help!
 

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