# Mechanics Question - Bullet fired into Block

I am studying for the GRE and came across this question (Question 93 from Sample Exam 2 in Kahn and Anderson's Conquering the GRE):

A bullet of mass 5 g is fired at a block of wood of mass 1 kg, which is hanging from a massless rigid rod of length 0.4 m. The block is thick enough to stop the bullet entirely inside. Which of the following is closest to the minimum velocity of the bullet such that the block makes a complete vertical revolution?

(A) 200 m/s

(B) 400 m/s

(C) 800 m/s

(D) 1000 m/s

(E) 1600 m/s

When I tried to solve it, I started by claiming that the kinetic energy of the bullet must be equal to the potential energy of the block-bullet at the top of the loop (since all the kinetic energy of the bullet goes into work done to move the block to the top) (again approximating $$m+M \approx M$$): $$\frac{1}{2}mv^2 = 2Mgl$$. Solving this yields $$v = 2\sqrt{\frac{M}{m}gl}$$.

The solution they present in the back of the book is as follows: momentum is conserved in the collision between the block and the bullet, and since the bullet is much lighter than the block, the velocity of the combination is $$V=\frac{m}{M}v$$ where $$m$$ is the mass of the bullet, $$M$$ is the mass of the block, $$v$$ is the velocity of the bullet, and $$V$$ is the velocity of the combination. They then equate kinetic energy of the block-bullet to its potential energy at the top of the loop: $$\frac{1}{2}(m+M)V^2 = (m+M)g(2l)$$, where $$l$$ is the length of the rod. This yields: $$V = 2\sqrt{gl}$$, and since $$V=\frac{m}{M}v$$, $$v = 2\frac{M}{m}\sqrt{gl}$$.

My question is: why does starting with conservation of energy lead to the incorrect answer?

• Yet another variation on the ballistic pendulum with added loop-de-loop goodness. And I have to say that mixing two common, well-understood, but non trivial exercises together is something I remember well from the subject exam. Mar 21, 2019 at 23:48
• See my last comments below to harshit54. Since the rigid rod for your example is massless, the pendulum can be treated as a simple pendulum, meaning it is no different than if a string suspended it. The only role the rigid rod plays in this example is to resist compressive forces, which a string can’t, so that the mass will not fall down as it approaches the top. In support of my answer, please see the following link: hyperphysics.phy-astr.gsu.edu/hbase/balpen.html Mar 22, 2019 at 14:05

My question is: why does starting with conservation of energy lead to the incorrect answer?

The collision was completely inelastic and therefore kinetic energy is not conserved (energy lost as heat). What is conserved, however, is momentum.

The minimum initial kinetic energy that is needed for the bullet plus wood to make a complete vertical revolution has to equal to the required increase in potential energy to complete the revolution. Since $$m< we can ignore the mass of the bullet and therefore have, where $$l$$ is the length of the rod,

$$\frac {Mv^2}{2}=Mg(2l)$$

$$v=2\sqrt{gl}$$

For conservation of momentum (ignoring $$m$$ on the right side) we have

$$mv_{B}=Mv$$

where $$v_B$$ is the bullet velocity. Combining the two results gives

$$v_{B}=\frac{2M\sqrt{gl}}{m}$$

Hope this helps.

• Wouldn't the rod be hinged at a point and this hinge would provide an impulse to the system. Why should the linear momentum be conserved? The only thing I can think about is conserving angular momentum about the hinge. Mar 21, 2019 at 14:53
• In my opinion Immediately after impact (before the mass starts rising) you can consider linear momentum as being conserved just to determine the kinetic energy upon impact. After that all we care about is conversion of kinetic energy to potential energy. Mar 21, 2019 at 15:58
• But the hinge exerts an external force on the bullet-block system and the force is large enough to give an impulse. I don't think conservation of linear momentum should be valid. Conservation of angular momentum about the hinge however is valid because the hinge force does not exert a torque. Mar 21, 2019 at 16:22
• @harshit54 If you sit down and actually compute the result for the two cases (using a representative mass distribution) you'll probably be surprised at how well they agree. As always doing physics in the world (as opposed to at the board) is a matter of balancing the scale of effects. Mar 22, 2019 at 4:09

The collision between the bullet and the block of wood is completely inelastic, and so energy is not conserved in the collision. The energy lost by the bullet is used to heat up the system and to make deformation work (the bullet digs a hole in the block).

• Comments are not for extended discussion; this conversation has been moved to chat. Mar 22, 2019 at 18:41