Question

Blocks A (mass 4.00 kg ) and B (mass 7.00 kg ) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is moving toward it at 4.00 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 A.

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 1

Given that :

mass of block A, m_A = 4 kg

mass of block B, m_B = 7 kg

initial velocity of block A, v_0,A = 4 m/s

initial velocity of block B, v_0,B = 0 m/s

(a) The maximum energy stored in the spring bumpers which is given as :

using conservation of energy, we have

K.E = P.E_spring

(1/2) m_A v_A² + (1/2) m_B v_B² = P.E_spring

P.E_spring = (1/2) m_A v_A²                                                       { eq.1 }

inserting the values in above eq.

P.E_spring = (0.5) (4 kg) (4 m/s)²

P.E_spring = 32 J

(b) The velocity of block A when the energy stored in the spring bumpers is maximum which is given as :

using an equation,   v_A = v_0,A (m_A – m_B)/ (m_B + m_A)                                           { eq.2 }

inserting the values in eq.2,

v_A = (4 m/s) [(4 kg) - (7 kg)] / [(7 kg) + (4 kg)]

v_A = - (12 kg.m/s) / (11 kg)

v_A = -1.09 m/s

(b) The velocity of block B when the energy stored in the spring bumpers is maximum which is given as :

using an equation,      v_B = (v_0,A + v_A)                                                         { eq.3 }

inserting the values in eq.3,

v_B = [(4 m/s) + (-1.09 m/s)]

v_B = 2.91 m/s