How Do Wood Blocks Behave in a Collision on a Frictionless Surface?

AI Thread Summary
In a collision involving two wood blocks on a frictionless surface, the first block (mass m1 = 0.15 kg) collides with a stationary second block (mass m2 = 0.30 kg) attached to a spring. The total momentum before the collision remains unchanged, resulting in a final velocity of 8.46 m/s for both blocks after they stick together. The initial kinetic energy of the first block is calculated to be 48.4 J, while the second block has zero kinetic energy before the collision. The collision results in a decrease in total kinetic energy of -32.3 J due to the compression of the spring. The discussion highlights the application of momentum and energy conservation principles in solving collision problems.
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A block of wood of mass m1 = 0.15 kg slides to the right on a frictionless horizontal table with an unknown speed. It collides with a second piece of wood of mass m2 = 0.30 kg which is stationary and attached to a spring with a spring constant of k = 159.00 N/m. The two pieces of wood stick together, and the spring compresses by an amount of 0.45 m.


(a) What is change in total momentum of the two masses as a result of the collision,
(before the spring compresses)?




ANSWER: __________________ _____

(b) What is the velocity of each mass just after collision (and before the spring
compresses)?





ANSWER: v1f = _______________________ _____
ANSWER: v2f = _______________________ _____


(c) What is the kinetic energy of each mass before collision?




ANSWER: K1i = _______________________ _____
ANSWER: K2i = _______________________ _____

(d) What is the change in total kinetic energy of the two masses as a result of the
collision (before the spring compresses)?




ANSWER: __________________ _____

Answers are suppose to be: A1) 0 A2) 8.46 m/s, 8.46 m/s A3) 48.4 J, 0J A4) -32.3 J
 
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Show your work please. We are not going to do your homework for you.
 


Its not Homework. It test review but i understand what you are saying. I am just trying to get a starting point since i looked at these problems for about 10 min each. But i have three other tests to study for and all four of them are on the same day so i was trying to get a boost or something. Ill reply in a little with my attempt.
 


refresh. Just putting the forums on my subscription. not trying to get people to do it for me
 


Ok can anyone help me with part B with finding the initial velocity of mass 1.

I have the formula to solve for the speeds.

M1V1 + M2V1 = M1V2 + M2V2

which the gets simplified to

.15 kg (v1) +0 = ( .15+.30)V2
 


refresh
 


refresh
 


Someone please help
 


Ok to solve B i did.

W = 1/2 K (X2-X1) = .5 (159.00)(.45)=35.775

w = (change in K) = .5 (ma + Mb)V2^2

V2 = (35.775/ (1/2 * .45))^(1/2)

however i got 12.6 m/s^2

what was i suppose to do instead?
 
  • #10


ok i forgot the ^2 in theW = 1/2 K (X2-X1)^2

so i did get the right answer but how do i go about getting part C first part.

The second part is clearly zero because it doesn;t have any velocity so there for no kinetic energy.
 
  • #11


ok wow i keep finding how to do stuff.

Ok to get part C. I realized that i had solved for Vf.

So therefore you can use this to find Vi

M1V1+M2V1= (M1+M2)V2

then you can use

K = 1/2 * m1 * V1^2

= 48.4J
 
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