A block of mass M, at rest on a horizontal, frictionless surface, is attached to a rigid support by a spring of force constant k. A bullet of mass m and velocity v strikes the block and remains embedded in the block. Determine, a.) the velocity of the block immediately after the collision and b.) the amplitude of the resulting simple harmonic motion.
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Expert's answer
2011-06-23T17:19:51-0400
To find the velocity of the block immediately after the collision we can use the law of conservation of linear momentum: p2=pblock+pbullet
Where p2 is of linear momentum of block and bullet embedded in the block. According to the definition of linear momentum:
p2=(M+m)*v2 Where v2 is the velocity of the block immediately after the collision;
pbullet=m*v And pblock=0, because block was at rest before collision;
So (M+m)*v2=m*v ; v2=(m*v)/(M+m)
Answer: v2=(m*v)/(M+m)
To find the amplitude of the resulting simple harmonic motion we can use the law of conservation of energy:
Ekis kinetic energy of of the block immediately after the collision; Potential energy of the spring is Ep = -k*x; Where x is displacement of the block; Amplitude is maximum displacement so Ep max=-k*A; A is amplitude; Block is in frictionless surface so the law of conservation of energy can be written as: Ek=Ep max So A=(m*v)2/(2*(M+m)*k) Answer : A=(m*v)2/(2*(M+m)*k)
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