Moving mass strikes a spring attached to a second mass, what is maximum spring compression?

Question

A block of mass M1 slides along a frictionless tabletop at a speed of v0 toward a second block of mass M2. A coil spring with spring constant k is attached to the second block in such a way that it will be compressed when struck by the moving block. What is the maximum compression of the spring?

This is the drawing given with the problem:
http://img43.imagevenue.com/img.php?image=28138_mass_spring_drawing_122_1014lo.jpg

Thanks in advance,
-tauntingpillow

Thanks 4Brain for the prompt response but the tip you gave is actually false.

"hint: Force = (1/2)(k)(x^2) where x is the change in length of the spring."

Force = (k)(x) which is Hooke's law.

Potential energy of the spring = (1/2)(k)(x^2).

Answer 1

TE0 = KE(M1); total energy before impact is M1's kinetic energy.
TE1 = KE(M1 + M2) + PE(S); where KE is kinetic energy of the combined masses and PE is potential energy in the the spring.

From the conservation of energy, we have KE(M1) = KE(M1 + M2) + PE(S); so PE(S) = KE(M1) - KE(M1 + M2) = k delX^2; where delX is the compression you are looking for.

From the conservation of momentum, we have M1 v0 = (M1 + M2) v1; where v1 is the velocity of the two masses after impact. Thus v1 = M1 v0/(M1 + M2)

Putting this all together, we have 1/2 [M1 v0^2 - M1 (M1/(M1 + M2))^2 v0^2 - M2 (M1/(M1 + M2))^2 v0^2] = k delX^2 = 1/2 M1 v0^2 [1 - M1/(M1 + M2) - M2/(M1 + M2)] = 0.

Thus, delX = 0, the spring will not compress. This is intuitively pleasing because M2 is also on the same frictionless tabletop. Which means there is no resistive force that the spring can push back against as it is compressing. That is, as soon as block M1 strikes that spring, M2 will become part of the system and all the energy will be distributed between the two blocks and none of it will go into the spring.

Answer 2

The ratio of force to stretch is called the spring constant (k). The units are (N/cm).

Hook's Law: F = k(deltax), where x is the change in length (cm).

PE of the spring is 1/2kx^2.




X everything below the line of X's
XXXXXXXXXXXXXXXXXXXXXXXX

KE = 1/2mv^2

PE = KE

1/2kx^2 = 1/2mv^2

kx^2 = mv^2

x^2 = mv^2/k

x = square root of (mv^2/k)

Answer 3

hint: Force = (1/2)(k)(x^2) where x is the change in length of the spring.