Question

A block slides along a frictionless floor and into a stationary second block with the same mass. Figure 9-29 shows four choices for a graph of the kinetic energies Kof the blocks. (a) Determine which represent physically impossible situations. Of the others, which best represents (b) an elastic collision and (c) an inelastic collision?

Short answer

1. Case (c) and (d) represent a physically impossible situation.
2. Case (a) represents an elastic collision.
3. Case (b) represents an inelastic collision.

Step-by-step solution

The given data

Four choices for a graph of the kinetic energy of the blocks, one slides along a frictionless floor and hits the block of the same mass which is on rest.

Understanding the concept of the kinetic energy before and after collision

From the graph, we can see the kinetic energy before the collision and kinetic energy after the collision. We know that the kinetic energy before and after the collision is conserved. Using this we can determine the different situations.

a) Calculation of the given cases from the given graphs

For Case (c):

From the graph, we can say that the kinetic energy of the moving block becomes negative after the collision.

But, the kinetic energy of any particle can’t be negative

So we can say that Case (c) is a physically impossible situation.

For Case (d):

From the graph, we can say that the kinetic energy of both the blocks increases after the collision.

But according to the law, the energy is conserved after collision. It can’t be increased.

So we can say that Case (d) is a physically impossible situation.

b) Calculation of the case that represents the elastic collision

For Case (a):

From the graph, we see that the amount of kinetic energy lost by the moving block is equal to the amount of kinetic energy gained by the block which is at rest.

So the kinetic energy is conserved after the collision, but they have different velocities.

So we can say that Case (a) shows an example of elastic collision.

c) Calculation of the case that represents the inelastic collision

For Case (b):

The amount of kinetic energy lost by the moving block is not equal to the amount of kinetic energy gained by the block which is at rest.

But still, the kinetic energy is conserved after the collision

So we can say that Case (b) shows an example of inelastic collision.