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0000 T-Mobile 1:59 PM @ 67%. D spock.physast.uga.edu C Course Contents » »Problem Set 9» (7) 'Timer LI Notes Evaluate tw/ Feedback PrintInfo Two blocks, each with a mass m = 0.157 kg, can slide without friction on a horizontal surface. Initially, block 1 is in motion with a speed V = 1.52 m/s; block 2 is at rest. When block 1 collides with block 2, a spring bumper on block 1 is compressed. Maximum compression of the spring occurs when the two blocks move with the same speed, V/2 = 0.760 m/s. If the maximum compression of the spring is 1.42 cm, what is its force constant? Submit AnswerTries 0/8 Post Discussion Send Feedback In

Explanation / Answer

The change in kinetic energy of the system in the collision will be stored in the spring as restoring energy

Therefore

KEi - KEf = Energy stored in the spring (E) ------------------ (1)

KEi = Intial Kinetic Energy of the system = (1/2) m1 (v1)i2 + (1/2) m2 (v2)i2 = (1/2 ) m v2

here m1 = m2 = m = 0.157 kg  

(v1)i = Initial velocity of the mass m1 = v =1.52 m/s

(v2)i = Initial velocity of the mass m2 = 0 ( Initially at rest )

KEf = Final Kinetic energy of the system = (1/2) m1 (v1)f2 + (1/2) m2 (v2)f2 = (1/2) (2m) (v/2)2 = (1/4) m v2

{ m1 = m2 = m ,  (v1)f = (v2)f = v/2 = 0.760 m/s }

E = Energy Stored in the spring = (1/2) k x 2

{ x = maximum compression in the spring = 1.42 cm = 0.0142 m }

On substituting in (1) we get

(1/2 ) m v2 -  (1/4) m v2 =  (1/2) k x 2

(1/4 )  m v2 =  (1/2) k x 2

k =(m v2 ) / (2 x2 )

k = [ (0.157 kg ) ( 1.52 m/s)2 ] / [ 2 (0.0142 m)2 ]

k = 899.46 N/m

Therefore the force constant is 899.46 N/m

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