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? (That is, when the two masses separate from each other after the collision is complete.)
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To find the individual velocities of the two masses after the spring extends fully again, we can use the principles of conservation of momentum and conservation of mechanical energy.
First, let's calculate the initial momentum of the system:
Initial momentum of the system = momentum of mass 1 + momentum of mass 2
Momentum (p) = mass (m) × velocity (v)
For mass 1 (m1 = 2.4 kg):
Initial momentum of mass 1 = m1 × v1
For mass 2 (m2 = 4.1 kg):
Initial momentum of mass 2 = m2 × v2
Since the masses are initially moving towards each other, the velocity of mass 1 (v1) is positive (rightward), and the velocity of mass 2 (v2) is negative (leftward).
Using conservation of momentum, we can write:
Initial momentum of the system = final momentum of the system
m1 × v1 + m2 × v2 = 0 (since the final momentum of the system is zero)
Now, let's consider the conservation of mechanical energy. In an ideal system with no external forces or friction, the mechanical energy is conserved.
The mechanical energy of the system includes both the kinetic energy (KE) and the potential energy stored in the spring (PE_spring).
Initial mechanical energy of the system = final mechanical energy of the system
Initial KE + Initial PE_spring = final KE + final PE_spring
Since the block initially moving to the right compresses the spring, the initial potential energy of the spring is zero. Therefore, the equation becomes:
Initial KE = final KE
Using the kinetic energy equation, KE = (1/2) × mass × (velocity)^2, we can write:
(1/2) × m1 × v1^2 = (1/2) × m1 × v1_final^2 + (1/2) × k × x^2
where v1_final is the final velocity of mass 1 and x is the displacement/stretch of the spring.
Now, let's analyze the movement of the masses after the collision. Since the system is frictionless, the total mechanical energy is conserved.
When the spring extends fully, the potential energy of the spring is at a maximum, and the kinetic energy is zero:
Final KE = 0
Final PE_spring = maximum (when the spring is fully extended)
Using this information, we can revise the previous equation:
(1/2) × m1 × v1^2 = (1/2) × m1 × v1_final^2 + (1/2) × k × x_max^2
Simplifying the equation, we have:
v1_final^2 = ((m1 × v1^2) - (k × x_max^2)) / m1
Now, we need to calculate the maximum displacement of the spring (x_max). The maximum displacement occurs when the total kinetic energy is converted to potential energy stored in the spring. At this point, the two masses move as one unit:
Initial KE (mass 1) + Initial KE (mass 2) = Maximum PE_spring
(1/2) × m1 × v1^2 + (1/2) × m2 × v2^2 = (1/2) × k × x_max^2
Simplifying the equation, we have:
x_max^2 = ((m1 × v1^2) + (m2 × v2^2)) / k
Finally, substitute the values of m1, v1, m2, v2, and k into the equations to find the individual velocities of the two masses (v1_final and v2_final) after the spring extends fully again.