# Two blocks of wood colliding

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)?

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

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

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

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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hage567
Homework Helper

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

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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?

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.

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