Question

Blocks A (mass 5.00 kg ) and B (mass 14.50 kg ) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is moving toward it at 9.00 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 +x be the direction of the initial motion ofA.

Part A

Find the maximum energy stored in the spring bumpers and the velocity of each block at that time.
Find the maximum energy.

Ub___ J

Part B

Find the velocity of A

Part C

Find the velocity of B

Part D

Find the velocity of each block after they have moved apart.
Find the velocity of A

vA___m/s

Part E

Find the velocity of B

vB___m/s

some one did it for me like this

initialy the bumpers have maximum energy stores

then maximum energy stored is U = 0.5*5*9^2 = 202.5 J

Part B

VA = 9 m/s

Part C

VB = 0 m/s

Part D

VA = (m1-m2)*u1/(m1+m2)= (5-14.5)*9/(5+14.5) = -4.4 m/s

PartE

V2 = u1+v1 = 9-4.4 = 4.6 m/s

but its incorrect... can anyone help me pleaseeee

Answer 1

First we have to find velocity of its centre of mass

$V_{cm } = \frac{m_1v_1 + m_2 v_2}{(m_1+ m_2)}$

$V_{cm } = \frac{5*9 + 14.5*0}{(5+ 14.5)} = 2.31 m/s$

Now Kinetic energy of Centre of mass of the system

$KE = \frac{1}{2}(m_1+m_2)V_{cm}^2$

$KE = \frac{1}{2}(5 + 14.5)(2.31)^2 = 51.92 J$

Initial total kinetic energy of two blocks

$KE_i = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$

$KE_i = \frac{1}{2}5*9^2 + \frac{1}{2}14.5*0^2 = 202.5 J$

Now the maximum energy stored in spring is given by

PART B)

At the same time when energy stored in the spring is maximum then both the blocks will move with common velocity

PART C)

Similarly as stated above

PART D)

when two blocks are moved apart

$V_{af} = \frac{(m_1 - m_2)u_1}{m_1+m_2}$

$V_{af} = \frac{(5 - 14.5)*9}{5 +14.5} = -4.38 m/s$

PART E)

$V_{bf} = \frac{2m_1v_1}{m_1+m_2}$

$V_{bf} = \frac{2*5*9}{5+14.5} = 4.62 m/s$