Question

Need help with conversation of momentum!
Q: A block of mass m1 = 2.0 kg slides along a friction less table with a speed of 10 meters/sec. Directly in front of it, and moving in the same direction, is a block of mass m2 = 5.0 kg moving at 3.0 meters/sec. A mass-less spring with spring constant k = 1120 nt/meter is attached to the backside of m2 . When the blocks collide, what is the maximum compression of the spring? Assume that the spring does not bend and always obeys Hooke's law

Answer 1

Dont want it solved, just want some advice on how to approach the problem or how to start it.

Answer 2

i'd say that at that point the blocks have the same velocity.....

Answer 3

So would the first step be
M1*V1=(M1+M2)*V2 and solve for V2
Then plug V2 in
1/2(M1+M2)(V2)^2=I/2KX^2 and solve for x?

Answer 4

@IrishBoy123

Answer 5

certainly agree in principle with your approach, but a few little bits missing i think

eg momentum before collision = M1*U1 + M2*U1

[changed letter V -> U for initial velocity, as otherwise numbering system gets pretty tricky]

so M1*U1 + M2*U2 = (M1 + M2)*V2 -- as per your equation, V2 is when spring max compressed, M! & M@ moving at same velocity.

then make same adjustment for your second equation IE include ALL the kinetic energy before collision and make that equal to (A) ALL the KE at point spring is max compressed PLUS (B) spring energy at that point.