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.
1
Expert's answer
2011-06-23T17:19:51-0400
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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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