Blocks A (mass 4.00 kg ) and B (mass 6.00 kg ) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is moving toward it at 5.00 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 of block Find the maximum energy stored in the spring bumpers. Find the velocity of block A when the energy stored in the spring bumpers is maximum. Find the velocity of block B when the energy stored in the spring bumpers is maximum. Find the velocity of block A after they have moved apart. Find the velocity of B after they have moved apart.

Answer and Explanation:

mass of the bock A is, m1=4 kg

mass of the bock B is, m2=6 kg

velocity of the block A is u1=5 m/sec

velocity of the block B is u2=0

a)

by using law of conservation of energy,

m1*u1+m2*u2=(m1+m2)*v

4*5+0=(4+6)*v

final velocity v=2 m/sec ( when the spring gets compression)

by using law of conservatio of enegry,

maximum energy stored U=K1-K2

U=1/2*m1*u1^2-1/2*(m1+m2)v^2

U=1/2*4*5^2-1/2*(4+6)*2^2

U=30 J

b)

if energy stored in the bumper is maximum,

velocity of block A is, v=2 m/sec

velocity of block B is, v=2 m/sec

c)

after moving back,

velocity of the block A is,

v1=((m1-m2)/(m1+m2))*u1

v1=((4-6)/(4+6))*5

v1=-1 m/sec

and

velocity of the block B is,

v2=((2*m1)/(m1+m2))*u1

v2=((2*4)/(4+6))*5

v2=4 m/sec