Problem 5 At Test Block A compresses a spring k 200 Nm by 0

Problem 5: At Test, Block A compresses a spring (k - 200 N/m) by 0.25 m. One the system is released, Block A slides across a surface with a coefficient of kinetic friction of 4 -0.2 and strikes Black B which is sitting, at rest, 3 meters from Block A\'s starting position. After the impact, Block B slides 1 m until it makes contact with another spring (k- NO Nm) and compresses it 0.2 m before Block B comes to a rest. What is the coefficient of restitution of the impact between Block A and Black B? (Block A weighs 3N, Block B weighs 1.5N) 3-0 25 mm -NË 10 km -- A-0.2 Im3,-0.m

Solution

using the conservation of energy to calculate the velocity with which block A mves once su=ustem is released

1/2 kx² = 1/2 mu²

200 ( 0.25)² = ( 3/9.8) u²

u * 2 = 40.33

retardation due to friction = mu ( g) = 0.2 (9.8) = 1.96 m/s²

the vlocity of block A after travelling 3 m

v² = 40.33 - 2 ( 1.96)(3)

v( velocit of A when it hits block B) = 5.345 m/s

using the conservation of momentum

3( 5.345) = 1.5 ( V_b) + 3 ( V_a) ( where V_a is the velovoty of block A after collsion, and V_b is the velocity with which block B moves after collsion)

using the conservation of energy when block B hits the spring

1/2 ( 1.5/ 9.8) V_b² = 1/2 kx²

0.153 V_b² = 80 ( 0.20²)

V_b= 4.57 m/s

3( 5.345 ) = 1.5 ( 4.57) + 3 ( V_a)

(16.035 - 6.86 )/ 3= V_a

V_a= 3.06 m/s

e =( 3.06- 4.47)/ ( 5.345-0) = 0.264 apprx