Inelastic collision and Hook’s law

In summary: You can't cancel it with the one outside of the square root.Try solving for m2 by isolating it on one side of the equation.
  • #1
yellowmax31
3
0

Homework Statement


a bullet of mass 0.005kg traveling at 120m/s hits a block at rest that falls and compresses a spring (k=99n/m) to a maximum compression of 0.121m what is the mass of the block? (solve using momentum)
m1=0.005kg
m2=?
v1=120m/s
v`=?
k=99n/m
x=0.121m

Homework Equations


m1v1=(m1+m2)v`

Ee=Ek

Et=Et`

The Attempt at a Solution


attempt one
Ee=Ek
1/2kx^2=1/2m1v^2
v=squareroot((kx^2)/m1)
v=17m/s
m1v1=(m1+m2)v`
0.005(120)= (0.005+ m2)17
0.005(120)/17=(0.005+ m2)
0.03529=(0.005+ m2)
0.03529-0.005=m2
0.030=m2

I got it wrong and my teacher will not tell me the answer I can't see were i went wrong
 
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  • #2
Does the bullet stick to the block? If so, ½kx² should be equal to ½(m1 + m2)v² (instead of ½m1v²).
 
  • #3
yes the bullet sticks to the block and i am trying to figure out what m2 is so I am not sure how to incorporate ½(m1 + m2)v² (instead of ½m1v²).
 
Last edited:
  • #4
Solve for v in terms of m1 and m2 and substitute it back in the momentum equation.
 
  • #5
v=squareroot((kx^2)/(m1+m2)

m1v1=(m1+m2)v

m1v1=(m1+m2)squareroot((kx^2)/(m1+m2)

m1^2v1^2=m1+m2((kx^2)/(m1+m2))

then the m1+m2 cancel iam sorry i not getting this i just can't see where iam going wrong

Or did you mean
m1v1=(m1+m2)v

v=m1v1/(m1+m2) & v=squareroot((kx^2)/(m1+m2)
so
m1v1/(m1+m2)=squareroot((kx^2)/(m1+m2)

(m1v1)^2/(m1+m2)^2 = (kx^2)/(m1+m2)

(m1v1)^2/(m1+m2) = (kx^2) (brought m1+m2 over)

(0.005*120)^2/(m1+m2) = 1.449459

0.36=1.4 (m1+m2) (mutiplied both sides by (m1=+m2))

0.36= 0.007247295 + 1.4m2

0.352752705= 1.4m2
0.2519=m2
? is that it or did i mess up again
 
Last edited:
  • #6
yellowmax31 said:
v=squareroot((kx^2)/(m1+m2)

m1v1=(m1+m2)v

m1v1=(m1+m2)squareroot((kx^2)/(m1+m2)

m1^2v1^2=m1+m2((kx^2)/(m1+m2))

then the m1+m2 cancel iam sorry i not getting this i just can't see where iam going wrong

They don't cancel. One of the (m1 + m2)'s is under a square root.
 

FAQ: Inelastic collision and Hook’s law

What is an inelastic collision?

An inelastic collision is a type of collision in which kinetic energy is not conserved. This means that the total kinetic energy before and after the collision is not equal. In an inelastic collision, some of the kinetic energy is converted into other forms of energy, such as heat or sound.

What is Hook’s law?

Hook’s law states that the force required to extend or compress a spring is directly proportional to the distance the spring is extended or compressed. This means that the more you stretch or compress a spring, the greater the force required to do so.

How does Hook’s law relate to inelastic collisions?

Inelastic collisions often involve objects colliding and causing a spring to compress or stretch. Hook’s law can be used to calculate the force exerted on the spring during the collision, as well as the amount of compression or extension of the spring.

What are some real-life examples of inelastic collisions?

Some examples of inelastic collisions include a car crash, a tennis ball hitting the ground, and a baseball hitting a bat. In each of these situations, kinetic energy is not conserved and some of it is converted into other forms of energy, such as heat and sound.

How is the coefficient of restitution related to inelastic collisions?

The coefficient of restitution, or COR, is a measure of the elasticity of a collision. In inelastic collisions, the COR is less than 1, meaning that the objects involved do not bounce off of each other with the same velocity as before the collision. The lower the COR, the more inelastic the collision.

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