a block of mass M1=1.32kg slides along a frictionless table with a speed of V1=2

Question

a block of mass M1=1.32kg slides along a frictionless table with a speed of V1=23.36 m/s . In front of it is a second block of mass M2=2.5 kg that is moving in the same direction as M1at V2=5.2 m/s. A massless spring of constant 739.2 N/m is attached to M2 on the side toward M1. When the blocks collide , what is the maximun compression of the spring? Hint: At the moment of maximun compression the two blocks move as one; the velocity thus can be found by nothing that the collision is completely inelastic until then

Explanation / Answer

By conservation of momentum :

m1v1 +m2v2 =( m1 +m2) v

1.32*23.36+2.5*5.2=(1.32+2.5)*V

v = 11.47518325 m/s

By conservation of Energy:
1/2 m1v1^2 + 1/2m2v2^2 = 1/2( m1 +m2)v^2 + 1/2 kx^2

m1v1^2 + m2v2^2 =( m1 +m2)v^2 +kx^2
284.893319 = kx^2

0.385408=x^2

X= 0.6208 m

maximun compression of the spring= 0.6208 m

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