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