Conservation of momentum & energy

[Jngo22]

Homework Statement

A block of mass m1 = 2 kg slides along a frictionless table with speed 10 m/s. Directly in front of it, and moving in the same direction with a speed of 3 m/s is a block of mass m2 = 5kg. A massless spring with spring constant k = 1120 N/m is attached to the second block. The spring is compressed a maximum amount x. What is the value of x?

Homework Equations

m1v1 + m2v2 = m1vf1 + m2vf2
KEi = KEf + 1/2kx^2

The Attempt at a Solution

I knew that there would be a maximum compression of the spring when the first block has velocity = 0 m/s. I used cons. of momentum :
(2 kg)(10 m/s) + (5 kg)(3 m/s) = (5 kg)(Vf) to find Vf = 7 m/s
then used cons. of energy:
(0.5)(5 kg)(7 m/s)^2 = 1/2(1120)x^2
and got x = 0.4677 m

but the correct answer is 0.250 m
can anyone help me find the correct answer

 

[rl.bhat]

After collision both the masses move together until the compression is maximum. At that instant the common velocity is given by
m1v1 + m2v2 = (m1+m2)vf.
Applying the conservation of energy, you can write
m1v1^2 + m2v2^2 = (m1+m2)vf^2 +1/2*k*x^2.
Now proceed to find x.

 

[kerimek]

?

As a scientist, it is important to always double check your equations and calculations to ensure that your answer is accurate. In this case, it seems that there may have been an error in your calculation for the final velocity (Vf). It should be noted that the conservation of momentum equation only applies to systems without any external forces acting on them, and in this scenario, the spring is an external force. Therefore, the equation should be modified to include the force from the spring, which is equal to kx. This would result in the following equation:

m1v1 + m2v2 = m1vf1 + m2vf2 + kx

Using this equation and your values for the masses and initial velocities, we can solve for the final velocity and then plug it into the conservation of energy equation to find the correct value for x. Additionally, be sure to use the correct units throughout your calculations to avoid any errors.