Question

A block of mass m is hanging by a long massless string from a rigid support. A bullet of same mass m strikes with a velocity v0 horizontally and gets embedded. The block deflects to one side. The rise of the block from the ground level is

Answer 1

Answer:

The problem does not specifically state this, but I'm going to assume that the bullet becomes embedded in the block of wood when it strikes, making this an "inelastic collision". If that's not the case, then we aren't given enough information to solve the problem.

 

Af the bullet strikes and embeds in the wooden block, the block/bullet combined mass is going to have some velocity - I'll call this vf (velocity after the collision). The block/bullet then swings from the rope up to a height of "h". For this motion, the conservation of energy applies:

 

    The initial total energy (kinetic + potential) = the final total energy (kinetic + potential)

 

    KE0 + PE0 = KEf + PEf

 

    Right after the bullet hits the block, the initial height is 0, so PE0=0. When the bullet/block reaches its maximum height, it is at rest, so KEf=0.

 

    KE0 = PEf

 

   (1/2)(m+M)vf2 = (m+M)gh

 

We can use this equation to solve for "vf", the velocity of the bullet/block after the collision:

 

    vf = √(2gh)

 

Now let's go back in time to look at the collision itself. For an inelastic collision (where the objects become stuck together), the initial momentum = final momentum (just like ANY other collision). 

 

    p0 = pf

 

    mbullet*v0,bullet + Mblock*v0,block = (mbullet + Mblock)*vf

 

    Since the block is not moving before the collision, v0,block = 0

 

    mbullet*v0,bullet = (m+M)*vf

 

Rearranging to solve for in initial velocity of the bullet, "v0":

 

    v0 = (m+M)*vf

                 m

 

Substituting in the expression for "vf" that we solved for earlier:

 

    v0 = (m+M)*√(2gh)

                   m