Question

Blocks A (mass 5.50 kg ) and B (mass 13.00 kg ) move on a frictionless, horizontal surface. Initially, block B is at rest and block A 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 +x be the direction of the initial motion of A.

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

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.

Part E: Find the velocity of B.

Answer 1

Blocks A (mass 5.50 kg ) = m_A

and

B (mass 13.00 kg ) = m_B

block B is at rest V_B = 0

block A is moving toward it at 3.00 m/s = V_A

Part A )

we have U = KE_total - KE_cm  

KE_total = 1/2 ( m_A V_A² + m_B V_B² )

= 0.5 X 5.5 X 3² - 0

KE_total = 24.75 J

V_cm = m_A V_A + m_B V_B / ( m_A + m_B )

= 5.5 X 3 + 0 / ( 5.5 + 13 )

V_cm = 0.89 m/sec

KE_cm = 1/2 ( m_A + m_B ) V_cm²

KE_cm = 0.5 X ( 5.5 + 13 ) X 0.89²

KE_cm = 7.32 J

U = 24.75 J - 7.32 J

U = 17.43 J

Part B )

V_A = V_cm = 0.89 m/sec

Part C )

V_B = V_cm = 0.89 m/sec

Part D )

3 - 0 = V_B - V_A

V_B = V_A + 3

V_A = V_cm - [ m_B X 3 / ( m_A + m_B ) ]

V_A = 0.89 - [ 13 X 3 / 18.5 ]

V_A = - 1.21 m/s

Part E )

3 - 0 = V_B - V_A

3 - 0 = V_B - (- 1.21)

V_B = 3 - 1.21

V_B = 1.79 m/sec