# Compressed Springs and Conservation of Momentum

"A spring of negligible mass is compressed between two masses on a frictionless table with upward sloping ramps at each end. The masses are released simultaneously. The mass of M1 is less than the mass of M2.

A) Is the force exerted by the spring on M1 greater than, equal to, or less than the force it exerts on M2?

B) Is the final height up the ramp reached by M1 greater than, equal to, or less than the height reached by M2?

C) Is the speed of M2 greater than, equal to, or less than the speed of M1 once they both lose contact with the spring?

D) Is the momentum of M2 greater than, equal to, or less than the momentum of M1 once they both lose contact with the spring?

E) Is the kinetic energy of M1 greater than, equal to, or less than the kinetic energy of M2 once they both lose contact with the spring?

F) Is the duration of the force exerted by the spring on M1 greater than, equal to, or less than the time the force acts on M2?"

https://brainmass.com/physics/conservation-of-energy/compressed-springs-conservation-momentum-10255

## SOLUTION This solution is **FREE** courtesy of BrainMass!

a)The force exerted by the spring is given by, F = k dx

where k is the spring constant and dx is the distance through which the spring moves from the equilibrium position.

The spring generates a force of this magnitude at each end of the spring, so if we take a spring of equilibrium length x,the force acting on each mass is +kdx and -kdx (so that there is no net force and the system is in equilibrium)

b)The energy stored in the compressed spring is (1/2)k dx^2, when released, this much energy is transferred to the masses.

At the maximum height

m *g*H = (1/2)k dx^2

The height reached on the ramp is given by the expression,

H = ((1/2)k dx^2)/mg

since, M1 < M2

H(M1) > H(M2)

That is the height reached by M1 will be greater than that reached by M2.

c)The velocity of recoil of the two springs is connected by the expression, (see answer (d) also)

v1 = (M2/M1)*v2

(M2/M1) > 1

Therefore, v1 > v2

That is the velocity of M1 will be greater than that of the velocity of M2.

d)The initial momentum of the system is zero. So momentum conservation implies that the momenta of the two bodies after the release of the spring must cancel each other.

That is, final momentum = initial momentum

=>M2v2 - M1v1 = 0.....(v1 and v2 are in opposite direction)

=> M2v2 = M1v1

That is the momentum of the two bodies will be equal once they lose contact with the spring.

e)Since the velocity of M1 is greater than that of M2, kinetic energy, (1/2)Mv^2, of M1 will be greater than that of M2.

f)The impulse (duration of force force*time) is equal to the change in momentum

for mass m1, the change in momentum is 0 - m1v1

and for m2, momentum change is 0 - m2 v2

but for momentum conservation, these two quantities must be equal (and opposite), thus the duration of the force will also be the same.

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