a bullet of mass m with velocity u strikes a suspended wooden block of mass M . if the block rises to a height h the initial velocity is given by...

Question

a bullet of mass m with velocity u strikes a suspended wooden block of mass M . if the block rises to a height h the initial velocity is given by...

Answer 1

Hi Vishal,

 

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 v_f (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 = KE_f + PE_f

 

    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 KE_f=0.

 

    KE₀ = PE_f

 

   (1/2)(m+M)v_f² = (m+M)gh

 

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

 

    v_f = √(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 = p_f

 

    m_bullet*v_0,bullet + M_block*v_0,block = (m_bullet + M_block)*v_f

 

    Since the block is not moving before the collision, v_0,block = 0

 

    m_bullet*v_0,bullet = (m+M)*v_f

 

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

 

    v₀ = (m+M)*v_f

                 m

 

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

 

    v₀ = (m+M)*√(2gh)

                   m

 

I hope this helps!

 

Andy