A block of mass m1 = 2.4 kg initially moving to the right with a speed of 3.3 m/s on a frictionless, horizontal track collides with a spring attached to a second block of mass m2 = 4.1 kg initially moving to the left with a speed of
1.5
m/s as shown in figure (a). The spring constant is 514 N/m.
What will be the individual velocities of the two masses (v1 and v2) after the spring extended fully again?
To find the individual velocities (v1 and v2) of the two masses after the spring extended fully again, we can use the principle of conservation of linear momentum and the conservation of mechanical energy.
1. Conservation of Linear Momentum:
The principle of conservation of linear momentum states that the total linear momentum before the collision is equal to the total linear momentum after the collision (assuming no external forces act on the system).
In this case, the collision between the two blocks is a one-dimensional elastic collision. Therefore, we can use the equation:
m1v1i + m2v2i = m1v1f + m2v2f
where:
- m1 and m2 are the masses of the two blocks,
- v1i and v2i are the initial velocities of the two blocks, and
- v1f and v2f are the final velocities of the two blocks after the collision.
2. Conservation of Mechanical Energy:
The principle of conservation of mechanical energy states that the total mechanical energy of the system before the collision is equal to the total mechanical energy after the collision.
In this case, the only form of potential energy involved is the potential energy of the spring. The potential energy stored in the spring is given by:
PE = (1/2)kx^2
where:
- PE is the potential energy,
- k is the spring constant, and
- x is the displacement of the spring from its equilibrium position.
Since the spring extends fully again, the displacement of the spring (x) is equal to the maximum displacement, which we can denote as x_max. Therefore, the potential energy before and after the collision is the same:
(1/2)kx_max^2 = 0
Solving for x_max, we can find that x_max equals 0 since the potential energy is zero.
By using the equations of conservation of linear momentum and conservation of mechanical energy, we can calculate the individual velocities (v1f and v2f) of the two masses after the spring extended fully again.