Collision is perfectly elastic?

In summary, the conversation is about solving a problem where a bullet with a speed of 5560 cm/s hits an 8.45 kg block on a table and bounces back with a speed of 1260 cm/s. The question is to find the speed of the block immediately after the collision. It is suggested to use the conservation of kinetic energy and momentum to solve the problem. The person is unsure of how to approach the problem and asks for help. It is then advised to assume that energy is conserved and write down equations for the conservation of momentum and energy, using "m" as the mass of the bullet and "V" as the final speed of the block. This approach helps the person solve the problem.
  • #1
dphin
11
0
Trying to figure this problem...
A bullet moves with a speed 5560 cm/s, strikes an 8.45 kg block resting on the table, and bounces straight back with a speed of 1260 cm/s. Find the speed of the block immediately after collision.

P before = P after
I'm unsure of how to solve without knowing the mass of the bullet.
 
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  • #2
Does the problem tell you that the collision is perfectly elastic? If so, then you will get a second independent equation from conservation of kinetic energy.
 
  • #3
No it doesn't, it just has a diagram. With the bullet being fired into the block from a distance, on a horizontal plane.
 
  • #4
If that is all that is given, then it seems you will have to assume kinetic energy conservation, because that is the only way you are going to generate a second equation.
 
  • #5
I feel like I have just been staring at this problem and getting know where...

So, would I need to find the distance or acceleration first and then use the work energy theorem??
 
  • #6
Think of what role the (elastic) change in velocities for the particular masses has to do with the initial and final momenta and energies.
 
  • #7
I'm just getting more confused.
Ek1 + Ek2 =0 to determine if inelastic or elastic right? I'm unsure of where to go, I'm not getting how to approach the question. We've dealt with inelastic problems, determining whether it is or not in the beginning and using the above equation. However, if never approached a question that way. Help??
 
  • #8
Do as Tom says: assume that energy is conserved.

Now write down the equations for 1) conservation of momentum & 2) conservation of energy. (Hint: call the mass of the bullet "m" and the final speed of the block "V"; those are your unknowns.)
 
  • #9
Thank you, this really helped!
 

1. What does it mean for a collision to be perfectly elastic?

When a collision is perfectly elastic, it means that the total kinetic energy of the system is conserved. This means that after the collision, the objects involved will have the same total kinetic energy as they did before the collision.

2. How is a perfectly elastic collision different from an inelastic collision?

In an inelastic collision, some of the kinetic energy is lost and converted into other forms of energy, such as heat or sound. In a perfectly elastic collision, there is no loss of kinetic energy and it is all conserved.

3. Can any type of collision be perfectly elastic?

No, not all collisions can be perfectly elastic. In order for a collision to be perfectly elastic, the objects involved must be able to bounce off each other without any loss of kinetic energy. This typically occurs with objects that are very hard and do not deform upon collision.

4. What are some real-life examples of perfectly elastic collisions?

Some examples of perfectly elastic collisions include billiard balls colliding on a pool table, two atoms colliding in a gas, or two rubber balls colliding with each other.

5. How is the elasticity of a collision calculated?

The elasticity of a collision is calculated using the coefficient of restitution, which is the ratio of the relative velocity after the collision to the relative velocity before the collision. A perfectly elastic collision has a coefficient of restitution of 1, while a completely inelastic collision has a coefficient of restitution of 0.

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