Two blocks of masses M and 3M are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them as shown in the figure below. A cord initially holding the blocks together is burned; after that happens, the block of mass 3M moves to the right with a speed of 1.85 m/s.
(a) Is the original energy in the spring or in the cord? Explain your asnwer.
(b) Is momentum of the system conserved in the bursting-apart process? How can it be, with large forces acting? How can it be, with no motion beforehand and plenty of motion afterward?
a) The original energy is in the cord. The cord is holding the blocks together, so the energy is stored in the tension of the cord.
b) Momentum is conserved in the bursting-apart process. The large forces acting on the blocks cause them to accelerate, which results in a change in momentum. However, the total momentum of the system remains the same since the blocks move in opposite directions with equal and opposite velocities.
(a) To determine whether the original energy is in the spring or in the cord, we need to consider the nature of the energy stored in each component.
The energy stored in a spring is potential energy. When a spring is compressed or stretched, it stores potential energy that can be released as kinetic energy when it returns to its original position.
On the other hand, the energy stored in a cord is not potential energy but rather potential elastic energy. It is the energy that is stored in an elastic cord due to its ability to stretch and return to its original length.
In this scenario, when the blocks are pushed together with the spring between them, the energy stored in the spring is potential energy. When the cord is burned, the spring releases that potential energy, converting it into kinetic energy and propelling the block of mass 3M to the right.
Therefore, the original energy is in the spring, not in the cord.
(b) The principle of conservation of momentum states that the total momentum of an isolated system remains constant before and after any interaction, provided no external forces are acting on the system.
In the bursting-apart process described, although it may seem that large forces are acting due to the sudden acceleration of the 3M block, the system is still considered isolated because there are no external forces acting horizontally (assuming there is no friction). Only internal forces between the blocks and the spring are present, which do not affect the overall momentum of the system.
Before the cord burns, the system is at rest, and hence its momentum is zero. However, the momentum of the system after the cord burns changes due to the motion of the 3M block. But the total momentum of the system (considering both blocks) remains conserved. The momentum gained by the 3M block is precisely balanced by the momentum lost by the M block, ensuring overall momentum conservation.
It is important to note that momentum can be transferred between objects within the system, resulting in a change in their individual velocities, but the total momentum of the system as a whole remains unchanged.
(a) The original energy is stored in the spring. When the blocks are pushed together, the spring is compressed. The compression of the spring stores potential energy, which is the original energy in the system.
(b) The momentum of the system is conserved in the bursting-apart process. While there are large forces acting during the bursting apart, the duration of these forces is very short. The forces act over a very small time interval, causing a large impulse on the system. However, since the time interval is small, the change in momentum is also small, leading to a relatively small change in velocity.
Before the bursting-apart, the system is at rest, meaning the initial momentum is zero. After the bursting-apart, the block of mass 3M moves to the right with a speed of 1.85 m/s. The momentum of the 3M block is (3M) * (1.85 m/s) = 5.55M kg·m/s to the right.
To conserve momentum, the other block (with mass M) must move to the left with a speed such that its momentum balances out the momentum of the 3M block. Given that the ratio of masses is 3:1, the velocity of the M block will be three times the velocity of the 3M block, but in the opposite direction. Therefore, the momentum of the M block will be (-1.85 m/s) * (M) = -1.85M kg·m/s to the left. The total momentum of the system is zero before and after the bursting-apart, satisfying the principle of conservation of momentum.