Momentum dealing with 1D Collisions

[maniacp08]

A block of mass m1 = 1.0 kg slides along a frictionless table with a velocity of +10 m/s. Directly in front of it, and moving with a velocity of +3.0 m/s, is a block of mass m2 = 9.0 kg. A massless spring with spring constant k = 1120 N/m is attached to the second block as in the figure below.

(b) After the collision, the spring is compressed by a maximum amount Δx. What is the value of Δx?
cm
(c) The blocks will eventually separate again. What is the final velocity of each block measured in the reference frame of the table?
m/s (for m1)
m/s (for m2)


I am having distinguishing the type of collision.
Is this inelastic since the first block will "stick" compressing the spring?
but it will soon separate so I am not sure if this is inelastic or elastic collsion.

Momentum for M1 = m1 * v1 = 1kg * 10m/s
Momentum for M2 = m2 * v2 = 3kg * 9m/s

Can someone clarify if this is elastic or inelastic? and how would I approach this problem
Thanks.

 

[alphysicist]

Hi maniacp08,

A block of mass m1 = 1.0 kg slides along a frictionless table with a velocity of +10 m/s. Directly in front of it, and moving with a velocity of +3.0 m/s, is a block of mass m2 = 9.0 kg. A massless spring with spring constant k = 1120 N/m is attached to the second block as in the figure below.

(b) After the collision, the spring is compressed by a maximum amount Δx. What is the value of Δx?
cm
(c) The blocks will eventually separate again. What is the final velocity of each block measured in the reference frame of the table?
m/s (for m1)
m/s (for m2)


I am having distinguishing the type of collision.
Is this inelastic since the first block will "stick" compressing the spring?

You have to be careful here with what "collision" means. The collision begins when the objects start putting forces on each other and ends when they stop. So we would not say the first block sticks to the second (as in a completely inelastic collision), because here that occurs in the middle of the collision.

but it will soon separate so I am not sure if this is inelastic or elastic collsion.

What exactly does it mean for a collision to be elastic? And what is happening to the energy during and after the collision?

 

[baba89]

I would approach this problem by first identifying the type of collision to determine the equations and principles that can be applied. In this case, it is a one-dimensional collision between two objects, so we can use the principles of conservation of momentum and conservation of energy.

To determine the type of collision, we can look at the behavior of the objects after the collision. If the objects stick together and move as one, it is an inelastic collision. If they separate and continue moving independently, it is an elastic collision.

In this case, the blocks will eventually separate again, indicating an elastic collision. However, during the collision, the spring will be compressed, which could make it seem like an inelastic collision. Therefore, we need to consider both the conservation of momentum and conservation of energy equations to fully analyze the situation.

Using the conservation of momentum equation, we can calculate the initial momentum of the system, which is equal to the final momentum after the collision.

Initial momentum = m1 * v1 + m2 * v2
= (1kg)(10m/s) + (9kg)(3m/s)
= 37 kg*m/s

To find the final velocity of each block, we can use the conservation of energy equation, which states that the total kinetic energy before the collision is equal to the total kinetic energy after the collision.

Initial kinetic energy = final kinetic energy
(1/2)(m1)(v1)^2 + (1/2)(m2)(v2)^2 = (1/2)(m1)(vf1)^2 + (1/2)(m2)(vf2)^2

Substituting the values and solving for the final velocities, we get:
vf1 = 8.2 m/s
vf2 = 5.8 m/s

To find the maximum compression of the spring, we can use the formula for potential energy stored in a spring:
PE = (1/2)kx^2

Setting the initial kinetic energy equal to the final potential energy, we can solve for x:
(1/2)(m2)(v2)^2 = (1/2)kx^2
x = √(m2v2^2/k)
= √(9kg * (3m/s)^2 / 1120 N/m)
= 0.03 m or 3 cm

Therefore, the maximum compression of the spring is