Momentum Conservation One object m1 has a mass of 2 kg and moves with a speed of 9 m/s to the right. Another object m2 has a mass of 3 kg and moves to the left with a speed on 3 m/s. The two objects collide. After the collision, the first object m1 moves to the right with a speed of 1.5 m/s. The second object m2 moves to the right at 2 m/s. Item at position 5 5 1 point What type of collision is this? Inelastic Super-elastic Perfectly inelastic Elastic

see that this shock is completely elastic

Explanation:

In this case we have the collision of two bodies, so that the moments conserve we must define a system in such a way that the forces during the crash are internal, this system is the one formed by the two bodies

let's write the moment in two moments

initial instant. Before the crash

      p₀ = m₁ v₁_i + m₂ v₂_i

final moment. Right after the crash

        pf = m₁ v₁_f + m₂ v₂_f

to define what kind of shock we have we use the recovery constant

      po = e pf

where the term e, called the recovery constant, determines the type of shock we have, for a perfectly elastic shock it is worth 1 and for an inelastic shock it is worth 0

let's substitute the values

      2 9 - 3 3 = e (2 1,5 + 3 2)

          9 = e 9

           e = 1

therefore we see that this shock is completely elastic

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.