Question

Blocks (mass 5.50kg ) and (mass 12.50kg ) move on a frictionless, horizontal surface. Initially, block is at rest and block is moving toward it at 3.00 m/s.
The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after the collision is along a straight line. Let be the direction of the initial motion of .
1.Find the maximum energy stored in the spring bumpers and the velocity of each block at that time.
Find the maximum energy.
2.Find the velocity of A
3.Find the velocity of B
Please explain to me the steps, my teacher is the worst!!!!

Answer 1

Hints: When the springs are fully compressed, the blocks won't be moving relative to each other, but the center of mass of the system will be moving at the same velocity as the center of mass of the system was originally. When you have ideal springs, the relative velocity of approach before the collision is equal to the relative velocity of recession (a consequence of the conservation of energy as it applies to perfectly elastic collisions). As with any collision, momentum is conserved throughout and is the same at all points in time in the collision.

Answer 2

but what is the formulas to use, and did this : m1v1=(m1+m2)vsquared. however, the answer was not correct..:/

Answer 3

Don't square the velocity...you need to set momentum equal to momentum. Try\[\begin{align}\vec p_\text{initial} &=\vec p_\text{final} & \text{Conservation of Momentum} \\
m_1\vec v_{1_i}&=(m_1+m_2)\vec v &\text{Plugging in values...}\end{align}\]Can you take it from here?

Answer 4

ok let me try,..ok :) be right back :)

Answer 5

i get stuck :( what the v1 would be? is not 3m/s right?

Answer 6

yay i got it, for both blocks the velocity is the same when conservation of momentum....now i just need to find the maximum energy stored :)

Answer 7

help!!! I dont get the first question :(

Answer 8

Okay, here's the general scheme. Let the kinetic energy of blocks A and B before the collision be \(K_A\) and \(K_B\) (masses \(m_A\) and \(m_B\), and velocities \(v_A\) and \(v_b\)), and the kinetic energy of the system during the collision be \(K_{AB}\) (I'll just call this velocity \(v\)). Then, the lost energy is what goes into the potential spring energy \(U_S\).\[K_A+K_B=U_S+K_{AB}\]\[\frac{1}{2}m_Av_A^2+\frac{1}{2}m_Bv_B^2=U_S+\frac{1}{2}(m_A+m_B)v^2\]\[\boxed{U_S=\dfrac{1}{2}\left[m_Av_A^2+m_Bv_B^2-(m_A+m_B)v^2\right]}\]Give me another holler if this is not giving you the correct answer...I believe it should.

Answer 9

ok this is what i did I will do it numerically ok...1/2(5.50)(3)squared)=1/2 (5.50)(.92)2+1/2(12.50)(1.83)2

Answer 10

kinetic energy initial is just the 1/2*m1v1 right

Answer 11

\(\frac{1}{2}m_1v_1^2\), I believe is what you intend.

Answer 12

yes :)

Answer 13

That's for one block...you do also have to consider the other block.

Answer 14

yes for that other block is 1/2 m1v1^2f+1/2m2v2f^2

Answer 15

the values i got are as follow: for the velocity when each block is moved apart (after collision) is Block A=-1.16, and Block B=1.83, the velocity for both at momentum is 0.92 since they have the same velocity at that moment right? then i dont know if am mixing the velocities here :(