Please Solve for all parts! Two blocks move towards each other on a frictionless

Question

Please Solve for all parts!

Two blocks move towards each other on a frictionless surface, the one on the left having a velocity of v_1 = 13 m/s and mass m_1 = 5kg, the one on the right having a velocity of v_2 = 12 m/s and mass m_2 = 13 kg. However the one on the left has a spring attached to it of spring constant 1251 N/m. The figure shows the initial configuration of the system in purple. The blocks in the middle (green and red) are the same two blocks but at a later time. This configuration shows the spring at maximum compression when they're in contact. What is the initial total momentum? kg m/s At the instant when the spring is at maximum compression, the two blocks are moving at the same velocity. What is their velocity at this instant? Since the surface is frictionless, you can assume conservation of momentum. m/s What is the maximum compression from equilibrium of the spring? Assume conservation of energy. m

Explanation / Answer

Given: v1= 13 m/s , v2= -12 m/s

m1= 5 kg, m2= 13 kg

k = 1251 N/m

Solution:

Initial Momentum can be find out as:

Pi= m1v1+m2v2 = [5×13] + [13×(-12)]

= -91 kgm/s

Final velocity:

Momentum is conserved

Pi = Pf

-91 = (m1+m2)vf

-91 = (5+13)vf

vf= -5.05 m/s

Energy of harmonic motion Evib = KEi- KEf

KEi = (1/2) × 5× (13)2+ (1/2)×13×(-12)2

= 1358.5 J

KEf= (1/2) × (5+13)×(-5.05)2

= 229.52 J

Evib = 1358.5-229.52 = 1128.98

Evib = (1/2)kxmax2

xmax = [(2×1128.98)/1251]1/2

= 1.34 m

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