What was the speed of the bullet before it hit the block?

Question

A 2.00-g bullet hits and becomes embedded in a 5.00kg wood block which is hanging from a 1.4m long string. This causes the block to swing through an arc of 4.5 degrees. What was the speed of the bullet before it hit the block?

 

 

Answer 1

Hi Iva,

Let's break the problem into two parts - in the first part, the bullet strikes and becomes embedded in the block of wood (an inelastic collision). In the second part, this bullet/block combination swings from its initial hanging position through an arc of 4.5º.

I think it will be easiest if we work backwards, looking first at the "swinging" part of the problem.

After 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 through an arc of 4.5º. 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

At this point, the mass we are talking about is the combined mass of the bullet "m" and the block "M".

    (1/2)(m+M)v_f² = (m+M)gh = (m+M)g(L-Lcos(Θ))

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

    v_f = √(2g(L-Lcos(Θ))

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), but the kinetic energies will NOT be equal.

    p₀ = p_f

    m*v_0,bullet + M*v_0,block = (m + M)*v_f

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

    m*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 "vf" that we solved for earlier:

    v₀ = (m+M)*√(2g(L-Lcos(Θ))
                         m

 

Now we can plug in the numbers that we have for m,M,g,L, and Θ (I will leave out the units for clarity):

 

    v₀ =  (.002 + 5.00)*√(2*9.8*(1.4-1.4cos(4.5º)))

                                 .002

 

    v₀ = 727 m/s (rounded to three digits)

I hope this helps!

Andy

Answer 2

For this problem, you must use conservation of energy, where the change in KE of the bullet as it goes from velocity v to 0, is all converted into PE of the wood block as it is raised through a change in height (h).

 

The block is raised an amount h= L - Lcos(4.5deg), where L= 1.4m. You can deduce this by drawing a diagram.

 

Calculate the inital KE of the bullet using 1/2 m v^2, but watch your units of grams (block is in Kg).

Then calculate the change in PE from Mgh= (5000 grams ) * 10 m/s/s * h

 

Now set the 2 equal and solve for the velocity v.