Determine the amount of kinetic energy lost due to the collision

AI Thread Summary
The discussion focuses on calculating kinetic energy loss during an inelastic collision between two blocks on a frictionless ramp. The speed of block A before the collision is determined using the equation v = (2gh)^(1/2), where h is expressed in terms of radius R. After the collision, the combined speed of the blocks is derived by conserving momentum, leading to a halving of the velocity due to their combined mass. The kinetic energy lost during the collision is calculated by finding the difference in total kinetic energy before and after the collision. The conversation also touches on the subsequent thermal energy generated and temperature rise, assuming no energy loss to the environment.
ClintStibbard
Messages
2
Reaction score
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?
 
Physics news on Phys.org
ClintStibbard said:
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!
 
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!
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
I was thinking using 2 purple mattress samples, and taping them together, I do want other ideas though, the main guidelines are; Must have a volume LESS than 1600 cubic centimeters, and CAN'T exceed 25 cm in ANY direction. Must be LESS than 1 kg. NO parachutes. NO glue or Tape can touch the egg. MUST be able to take egg out in less than 1 minute. Grade A large eggs will be used.
Back
Top