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Answer to Question #3217 in Mechanics | Relativity for ZherVin

Question #3217
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.
Expert's answer
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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:

Ek=(M+m)*v22/2=((M+m)* ( (m*v)/(M+m )2 )/2=(m*v)2/(2*(M+m))


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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