Question

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.

Answer 1

To find the velocity of the block immediately after the collision we can use the law of conservation of linear momentum:
p₂=p_block+p_bullet

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;

p_bullet=m*v
And p_block=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:

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


E_kis kinetic energy of of the block immediately after the collision;
Potential energy of the spring is
E_p = -k*x;
Where x is displacement of the block;
Amplitude is maximum displacement so E_{p max}=-k*A; A is amplitude;
Block is in frictionless surface so the law of conservation of energy can be written as:
E_k=E_{p max}
So A=(m*v)²/(2*(M+m)*k)
Answer :
A=(m*v)²/(2*(M+m)*k)