Answer :
Refer to the diagram shown below.
Let m = the mass (g) of the door.
Let v = the launch velocity
Let u = the velocity of the door after impact.
Elastic impact (rubber ball):
The rubber ball bounces off the door with presumably elastic impact, which means that both momentum and kinetic energy are conserved.
Conservation of momentum requires that
400v = -400v + mu
Therefore
[tex]u=( \frac{800}{m} )v[/tex]
Inelastic impact (clay):
The clay sticks to the door after impact.
Conservation of momentum requires that
400g = (m+400)u
Therefore
[tex]u=( \frac{400}{m+400} )v[/tex]
When we compare magnitudes of u for the door, we find that
[tex]u_{1}=( \frac{400}{m} )(2v), \,\, elastic \\\\ u_{2}=( \frac{400}{m+400} )v , \,\, inelastic[/tex]
Clearly, the elastic impact creates a greater value of u for the door.
Answer:
The rubber ball creates a larger impulse to the door because the nature of its impact is approximately elastic.
Let m = the mass (g) of the door.
Let v = the launch velocity
Let u = the velocity of the door after impact.
Elastic impact (rubber ball):
The rubber ball bounces off the door with presumably elastic impact, which means that both momentum and kinetic energy are conserved.
Conservation of momentum requires that
400v = -400v + mu
Therefore
[tex]u=( \frac{800}{m} )v[/tex]
Inelastic impact (clay):
The clay sticks to the door after impact.
Conservation of momentum requires that
400g = (m+400)u
Therefore
[tex]u=( \frac{400}{m+400} )v[/tex]
When we compare magnitudes of u for the door, we find that
[tex]u_{1}=( \frac{400}{m} )(2v), \,\, elastic \\\\ u_{2}=( \frac{400}{m+400} )v , \,\, inelastic[/tex]
Clearly, the elastic impact creates a greater value of u for the door.
Answer:
The rubber ball creates a larger impulse to the door because the nature of its impact is approximately elastic.
