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Homework Statement
A spring of negligible mass is compressed between two masses on a frictionless table with sloping ramps at each end. The masses are released simultaneously. The masses have the same volume, but the density of M1 is greater than that of M2.
G-Greater than, L-Less than, E-Equal to.
A) The duration of the force exerted by the spring on M2 is ... the time the force acts on M1.
B) The momentum of M2 is ... the momentum of M1 once they both lose contact with the spring.
C) The final height up the ramp reached by M1 is ... the height reached by M2.
D) The force exerted by the spring on M2 is ... the force it exerts on M1.
E) The speed of M2 is ... the speed of M1 once they both lose contact with the spring.
F) The kinetic energy of M2 is ... the kinetic energy of M1 once they both lose contact with the spring.
The picture is attached below.
The attempt at a solution
This is how I attempted to answer this question. There are flaws in my logic, and I don't know where or what they are.
A) L - M1 has the same volume but is less dense, therefore it is lighter. It would take less force for the spring to return to equilibrium send end it out at a higher velocity.
B) E - Momentum is mv, right? So despite their difference in weight, shouldn't that cancel each other out? M2 would have a larger weight and less velocity, M1 would have a small weight and a heigher velocity?
C) G - M1 would be moving faster, therefor it would go further up the ramp.
D) G - M2 is larger, therefore it would need more force exerted against it to move.
E) L - I think I've beaten my thought on this into the ground. M1 is lighter, so shouldn't it be moving faster?
F) L - Kinetic energy is (1/2)mv^2. M1 may be lighter, but it would be moving at a higher velocity.
A spring of negligible mass is compressed between two masses on a frictionless table with sloping ramps at each end. The masses are released simultaneously. The masses have the same volume, but the density of M1 is greater than that of M2.
G-Greater than, L-Less than, E-Equal to.
A) The duration of the force exerted by the spring on M2 is ... the time the force acts on M1.
B) The momentum of M2 is ... the momentum of M1 once they both lose contact with the spring.
C) The final height up the ramp reached by M1 is ... the height reached by M2.
D) The force exerted by the spring on M2 is ... the force it exerts on M1.
E) The speed of M2 is ... the speed of M1 once they both lose contact with the spring.
F) The kinetic energy of M2 is ... the kinetic energy of M1 once they both lose contact with the spring.
The picture is attached below.
The attempt at a solution
This is how I attempted to answer this question. There are flaws in my logic, and I don't know where or what they are.
A) L - M1 has the same volume but is less dense, therefore it is lighter. It would take less force for the spring to return to equilibrium send end it out at a higher velocity.
B) E - Momentum is mv, right? So despite their difference in weight, shouldn't that cancel each other out? M2 would have a larger weight and less velocity, M1 would have a small weight and a heigher velocity?
C) G - M1 would be moving faster, therefor it would go further up the ramp.
D) G - M2 is larger, therefore it would need more force exerted against it to move.
E) L - I think I've beaten my thought on this into the ground. M1 is lighter, so shouldn't it be moving faster?
F) L - Kinetic energy is (1/2)mv^2. M1 may be lighter, but it would be moving at a higher velocity.