How Fast Must a Bullet Travel to Tip a Block on Its Edge?

[Krappy]

Homework Statement

A block of wood, of side 2a and mass M is on an horizontal plane. When it turns, it does it over the edge AB. A bullet of mass m<<M and velocity v hits on the opposite face to ABCD, at a height of 4/3*a. The bullets gets stuck on the block. Find the minimum value of v so that the cube turns over AB and falls on the face ABCD.

Homework Equations

$$\frac{d\vec{L}}{dt} = \vec{\tau}$$
$$\vec{L} = I\vec{\omega} = \vec{r} \times \vec{p}$$
$$\vec{\tau} = \vec{r} \times \vec{F} = I\vec{\alpha}$$
$$\Delta p = 0$$

The Attempt at a Solution

I've tried conservation of energy $$1/2(M+m)(\frac{m}{2(M+m)}v)^2 = (m+M)a(\sqrt(2)-1)g$$ but this doesn't depend on the height at which the bullet hits the block.

I've also thought of integrating $$\frac{d\vec{L}}{dt} = \vec{\tau}$$
but the Torque is not constant and depends on the angle between the diagonal of the cube and the gravity.

I've ran out of ideas.

[PLAIN]http://img855.imageshack.us/img855/3033/picture1ic.png

Regards
Johnny

 

[Quinzio]

You have to know the value of $m$. When they say m<<M they make a mistake because $m$ cannot be ignored. If you wish to continue, you'll have to assign $m$ a value, e.g. $M/1000$

 

[TROGLIO]

jonny
you haven't taken into account that fact that the bullet will get deformed
so energy will be used to do that hence the energy conservation idea is tough
as v do not know how much energy is used in deforming the bullet


tried balancing torque?

the bullet wil produce a torque about ab
and in order to keep the block straight the normal reaction given to the block from the ground wil shiftby a certain distance to produce an equal and opposite torque as that of the bullet


but the normal force is constant in magnititude
and it can shift a maximum distance of "a" or the normal reaction wil be appleid outside the block which would be wrong
hence the normal torque has a maximim value
if the torque by the bullet exceds that value then the block wil tip over

note that when you take the moment of inertia of the block be sure to take it about ab as that is the axis of rotation

 

[tiny-tim]

Hi Johnny!

I've tried conservation of energy …

This is a collision, and energy is never conserved at the instant of collision unless the question says so (for example, by calling it elastic).

But momentum and angular momentum are always conserved at the instant of collision

(and energy will be conserved after that).

Try again! ​