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have a look at the diagram

A physical system consists of wood block B and mass M. B is at rest on frictionless horizontal table. A small vertical wall W is place near one of the ends of B. The wall is fastened to B.A spring with spring constant k is attached to the wall and is connected tp a block C. of mass m.

No friction between C and B. A bullet of mass mu [itex] \mu [/itex] is shot is shot into block C. The velocity v is parallel to the table. The spring does not bend.

Assume that the bullet is stopped in such a short time taht t is negligible. The spring mass and the wall mas are neglected

a) find the maximum compression d of the spring, in terms of v, k, M , m and [itex] \mu [/itex]

bullet goes into the block inelastically so

[tex] \mu v = (\mu + m) v_{f} [/tex]

and [tex] v_{f} = \frac{\mu v}{\mu + m} [/tex]

this vf (kinetic energy) is converted to spring energy

[tex] \frac{1}{2} (\mu + m) (\frac{\mu v}{\mu + m})^2 = \frac{1}{2} kx^2 [/tex]

and thus [tex] x = \mu v \sqrt{\frac{1}{k(\mu + m)}} [/tex]

this is assuming that [tex] x = \Delta L + D [/tex]

and D is converted to the enrgy that makes this thing go forward

so [tex] \frac{1}{2} k D^2 = \frac{1}{2} ( \mu + m + M ) v_{f}^2 [/tex]

[tex] D = v_{f} \sqrt{\frac{\mu + m + M}{k}} [/tex]

now here's the dilemma, what is v?? i cannot find D without using some unknown velocity

how would i use the conservation of momentum here?

A physical system consists of wood block B and mass M. B is at rest on frictionless horizontal table. A small vertical wall W is place near one of the ends of B. The wall is fastened to B.A spring with spring constant k is attached to the wall and is connected tp a block C. of mass m.

No friction between C and B. A bullet of mass mu [itex] \mu [/itex] is shot is shot into block C. The velocity v is parallel to the table. The spring does not bend.

Assume that the bullet is stopped in such a short time taht t is negligible. The spring mass and the wall mas are neglected

a) find the maximum compression d of the spring, in terms of v, k, M , m and [itex] \mu [/itex]

bullet goes into the block inelastically so

[tex] \mu v = (\mu + m) v_{f} [/tex]

and [tex] v_{f} = \frac{\mu v}{\mu + m} [/tex]

this vf (kinetic energy) is converted to spring energy

[tex] \frac{1}{2} (\mu + m) (\frac{\mu v}{\mu + m})^2 = \frac{1}{2} kx^2 [/tex]

and thus [tex] x = \mu v \sqrt{\frac{1}{k(\mu + m)}} [/tex]

this is assuming that [tex] x = \Delta L + D [/tex]

and D is converted to the enrgy that makes this thing go forward

so [tex] \frac{1}{2} k D^2 = \frac{1}{2} ( \mu + m + M ) v_{f}^2 [/tex]

[tex] D = v_{f} \sqrt{\frac{\mu + m + M}{k}} [/tex]

now here's the dilemma, what is v?? i cannot find D without using some unknown velocity

how would i use the conservation of momentum here?