Blocks A (mass 4.00 kg ) and B (mass 7.00 kg ) move on a frictionless, horizonta
ID: 1458963 • Letter: B
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?
Explanation / Answer
Given that :
mass of block A, mA = 4 kg
mass of block B, mB = 7 kg
initial velocity of block A, v0,A = 4 m/s
initial velocity of block B, v0,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.Espring
(1/2) mA vA2 + (1/2) mB vB2 = P.Espring
P.Espring = (1/2) mA vA2 { eq.1 }
inserting the values in above eq.
P.Espring = (0.5) (4 kg) (4 m/s)2
P.Espring = 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, vA = v0,A (mA – mB)/ (mB + mA) { eq.2 }
inserting the values in eq.2,
vA = (4 m/s) [(4 kg) - (7 kg)] / [(7 kg) + (4 kg)]
vA = - (12 kg.m/s) / (11 kg)
vA = -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, vB = (v0,A + vA) { eq.3 }
inserting the values in eq.3,
vB = [(4 m/s) + (-1.09 m/s)]
vB = 2.91 m/s