Analyzing Collisions and Energy Transfer in AP Mechanics Problems

  • Thread starter nns91
  • Start date
  • Tags
    Ap
  • #1
nns91
301
1

Homework Statement



1. A 2 kg block and 8kg block are both attached to an ideal spring (k= 200N/m) and both are initially at rest on a horizontal frictionless surface. In an initial experiment, a 0.1kg ball of clay is thrown at the 2 kg block. The clay is moving horizontally with speed v when it hits and sticks to the block. The 8kg block is held still by a removable stop. As a result, the spring compresses a maximum distance of 0.4m.

In a second experiment, an identical ball of clay is thrown at another identical 2kg blcok, bu this time the stop is removed so that the 8kg block is free to move.

d. State whether the maximum compression of the spring will be greater than, equal to , or less than 0.4m. Explain briefly.

e. State the principle(s) that can be used to calculate the velocity of the 8kg block at the instant that the spring regains it original length. Write the appropriate equations and show the numerical substitutions, but do not solve for the velocity

2. A 5kg ball initially rests at the edge of a 2m long, 1.2m high frictionless table. A hard plastic cube of mass 0.5kg slides across the table at a speed of 26 m/s and strikes the ball, causing the ball to leave in the direction in which the cube was moving.

b. Determine the horizontal velocity of the ball immediately after the collision.
c. Determine the cube's speed and direction of travel (if moving) immediately after the collision.
d. Determine the kinetic energy dissipated in the collision.
e. Determine the distance between two points of impact of the objects with the floor


Homework Equations



Conservation of energy, momentum

The Attempt at a Solution


1.I solved part a,b,c already.

In part d, I guess it will less than 0.4 but don't know how to explain that.

I also need some hints on part e too

2. I solved part a which was a graph question.

Part b,c since I don't know the type of collision ( I think it is elastic) so I can't use conservation of momentum. ANy hint ??

Part d: I guess I will calculate KE of the cube initially and KE of the cube after collision and their difference is the answer right ??

Part e: I don't understand the question, can anyone explain for me ??
 
Physics news on Phys.org
  • #2
D. well, no calculation. Here is what I think. Since the stop is remove, then the block can move. Well, then there will be kinetic energy. So rather than having all the energy transfer to potential energy of the spring, it gets diverted to both the kinetic energy of the block and the potential energy. So, by intuition, it must be smaller.

And you can ALWAYS use conservation of momentum. momentum is conserved in both elastic and inelastic collision (or any motion for that matter). Energy is also conserve, however, in inelastic collsion, energy becomes heat...so teachers just tell you that it's not conserve. After all, no one want to calculate the heat loss. ._.

So, if it's elastic --> use energy and momentum conservation
Inelastic --> momentum conservation should be suffice for the AP curriculum.Edit: If it said frictionless, pretty much it means elastic for AP curriculum.
b. Use momentum conservation.
C. well, same as above
D. well, they kinda told you it's inelastic. So use b and c (the velocity you calculated) and calculate the difference in energy. (the final E - inital E is the Energy "dissipated"..or in words that human can understand..become heat)
E. umm use kinematic?
 
  • #3
For b, it will be something like m(cube)*v(cube,i) = m(cube)v(cube,f) + m(ball)v(ball)

but I don't know the final velocity of the cube which is after collision, how can I use conservation of momentum ??

Can I use conservation of Energy by using: mgh=(1/2) mv^2 ?
 
  • #4
nns91 said:
For b, it will be something like m(cube)*v(cube,i) = m(cube)v(cube,f) + m(ball)v(ball)

but I don't know the final velocity of the cube which is after collision, how can I use conservation of momentum ??

Can I use conservation of Energy by using: mgh=(1/2) mv^2 ?

Part D ask you to solve for the energy lost. So that's a hint that energy isn't conserve. So prob don't want to use energy conservation (or else they won't have part d)

And I just realize that it said cube and ball. SO I guess you need to take into consideration of rotation energy? o_O?
 
  • #5
Pard D, they ask for lost of KINETIC ENERGY only not the whole ENERGY. So what is your suggestion for part b without knowing velocity of cube after collision ??
 
  • #6
For number 1, part e, is the principle that I can use conservation of energy ??

For number 2, part b, can I use mgh=1/2 mv^2 ?? If not, what other way can I use ??
 
  • #7
Anyone figure anything out yet ??
 
  • #8
so I have figured out some parts

which left me with part e, question 2 and part e of question 1.

Part e question 2 is a projectile motion problem. Any hint on that ?? I cannot think of any way right now. Can I use the formula for Range to calculate ?? something like R= (v^2*sin2[tex]\theta[/tex]) / g

Part e question 1, can I use conservation of momentum. If so, what will be the final momentum ?
 
Back
Top