momentum question.. no calculation?

Question

a bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of length and negligible mass. What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle?

Answer 1

initial momentum of the bullet = mv
final momentum of the bullet = mv/2
Δmomentrum of the bullet= mv-mv/2 = mv/2
because the conservation of momentum
this whole momentum has been transfered to the pendulum bob
so the momentum gained by the pendulum bob = mv/2
if the mass of pendulum bob = M
and velocity = V
mass * velocity = momentum
M * V = mv/2
V = mv/2M
intial velocity of the pendulum = mv/2M
if the length of rod = l
the elevation of the pendulum at the top position from the initial position = 2l

on the top position centripetal force =mg + T -------- T is tension when bearly completing the circle T = 0
so at the top position to the pendulum bob
F = ma
centripetal acceleration = v²/r
mg = mv² / r
mg = mv² / l
v = √(l.g)
so to complete the circle the velocity at the top should at least be √(l.g)

so the energy of the pendulum at the top = kinetic E + potential E

E = ½mv² + mgh
E = ½M(lg) + Mg(2l)
E = M.g.l [½ + 2]
E = 3Mgl/2

according to conservation of mechanical energy
Energy at the top = Energy at the bottom
so the Energy at the bottom = 3Mgl/2
there is no potential energy = 0
so kinetic energy = 3Mg/2
so ½MV² = 3Mgl/2
V² = 3gl
V = √(3gl)

but V = mv/2M
√(3gl) = mv/2M
v = 2√(3gl)M/m

Answer 2

Well, at this point you have two different answers. Let's try for one more and see if this agrees with either of them.

The work to get the bob M raised h = 2L is exactly PE = Mgh = Mg2L. L is the length of the pendulum. PE represents the potential energy of the bob at the top of its swing. The kinetic energy spent to raise that bob is KE = 1/2 Mv^2; where v is the tangential velocity of the bob as the bullet emerges.

The conserved momenta are mV = mV/2 + Mv; so that m(V - V/2) = Mv and v = m(V/2)/M. V is the initial velocity of the bullet with mass m. The v of the bob M after the impact can be inserted into the KE equation. So combining, we have:

PE = Mg2L = 1/2 Mv^2 = KE and v^2 = 4gL; so that v = sqrt(4gL) = (m/M)(V/2) then V = 2(M/m)sqrt(4gL), which is my answer. The V is your v in the question.

Let's see if V = 2(M/m)sqrt(4gL) has consistent units and if it makes physics sense.

The units on the LHS of the equation are m/sec (velocity). The units on the RHS are sqrt((m/sec^2)m) = m/sec; so the LHS and RHS of the equals sign have consistent units.

Now does it make physics sense? It tells us the minimum velocity of the bullet goes up with a more massive bob or with a lighter bullet. That makes sense, the bullet would have to go faster to raise the bob if the bob is heavier. Also, if the bullet is lighter, it would have to go faster to impart the same amount of KE to the bob.

Finally, the equation says that V must be greater for longer lengths (L). That also makes physics sense because with longer L, the bob will be lifted a greater height; and it will require more KE (more work) to make that greater lift.

So the units and the physics check out OK.

Answer 3

The pendulum bob will barely swing through a complete vertical circle =>
velocity of tope is 0
The pendulum bob is suspended by a stiff rod of length l
assuming length is l.
v(1) => initial velocity of bob at the bottom
then 2Mgl <= 1/2Mv(1)^2
v(1)^2 >= 4gL ---------------(1)
By conservation of momentum
mv = mv/2 + Mv(1)
=> v(1) = mv/(2M)

[mv/(2M)]^2 >= 4gL
v >= 4(M/m) *sqrt(gL)

Min value of v 4(M/m) *sqrt(gL)

Answer 4

2 L g = V^2 / 2

m v = M V + m v / 2

v = 2 M V / m = 4 M/m sqrt( L g)

to narendrino: the problem says "pendulum bob is suspended by a *stiff* rod of length L (missing) and negligible mass." You solved more difficult problem -- well done anyway!

Answer 5

You want the kinetic energy imparted to the pendulum to just equal the potential energy gained at the apex of its rise. But the radius of the circle which will be followed is not given - the length of the rod. So either there is another solution path, or a missing datum in the problem as posed.

not so QED...hmmm....

mv/2 = MV => V = mv/2M ... Ek^M = [M/2].(mv)*2/[4M*2] ...
Ek of M = 1/8.[(mv)*2]/M ...
Ep at height 2r = Mg.2r
So 2gMr = 1/8.[(mv)*2]/M ... hmmm ....
There is a zero (velocity) at the apex ... perhaps that will suffice where r is not given...no calculation per se, but the skeleton of the same!!....
r16g(M/m)*2 = v*[-2] ...
arrggh .. I gibbupp!!...GL.

Answer 6

I assume that you meant for the length of the rod to be "L".

Conservation of Energy:

2MgL = 1/2 Mu^2 (kinetic energy of block after impact is equal to potential energy when it reaches the top of the swing).

4gL = u^2 --> u = 2 sqrt (gL) (u is velocity of block right after impact).

Then conservation of momentum:
mv = 1/2mv + 2M sqrt (gL)

1/2 mv = 2M sqrt (gL)
v = 4 M/m sqrt (gL)