"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?"

Solution Preview

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 ...

Solution Summary

The solution gives all steps along with proper explanations so that you can solve similar problems yourself. The compressed spring and conservation of momentum is analyzed.

... The final momentum is: From conservation of momentum we obtain: Then: All that is left is to substitute the numbers: The spring compresses 25 centimeters. ...

... friction is k, whole elastic potential energy initially stored in the compressed spring is utilized ... According to law of conservation of momentum we get. ...

... the plate is resting on the spring and compressing it by x ... K is the force constant of the spring = 200 N ... is given by according to law of conservation of energy ...

... the mass of the block, then according to law of conservation Initial momentum... The combined mass then moves on the table against the spring and compresses it ...

... mass velocity will stay constant due to conservation of momentum... impact, the potential energy in the spring is zero ... 4 M u^2. At maximum compression, the internal ...

... In case the car lands on a huge spring, the spring gets compressed and brings about ... To summarize: with the springs present, the rate of change of momentum...

... If the carts collide elastically, find the maximum compression in the spring. ... According to law of conservation of energy loss of PE of m1 = gain in KE of m1. ...

... By conservation of momentum, equating (1) and (2): (mL2/3)ω = (mvocosβ) x L/2. ... Initial mechanical energy = PE stored in the compressed spring. ...

... after collision be v. Then we can write applying conservation of momentum... k = spring constant, x = extent of stretching (or compression) of the spring). ...