# How do I find the speed of a block at the bottom in a half circle cavity?

#### Chemist116

The problem is as follows:

The figure from below shows a block which is shaded with blue color and whose mass is $9\,kg$ and of radius equal to $2\,m$ is put over a frictionless table. Then a small block labeled $A$ is put on $A$ and released so it slides in the circular cavity of the bigger block, there is no friction between them. Find the speed of the block in $\frac{m}{s}$ when the block reaches the lowest point of the half circle cavity. The alternatives given are as follows:

$\begin{array}{ll} 1.&2\,\frac{m}{s}\\ 2.&3\,\frac{m}{s}\\ 3.&4\,\frac{m}{s}\\ 4.&6\,\frac{m}{s}\\ 5.&9\,\frac{m}{s}\\ \end{array}$

I'm lost at this problem. How should I exactly relate the conservation of momentum?.

I recall that the energy will be preserved as follows:

$E_u=E_k$

$mgR=\frac{1}{2}mv^2$

$\frac{mv^2}{R}=mg$

$v^2=Rg$

But if I equate these two expressions I will get to an inconsistency. What part did I got wrong?. Can someone help me here please?.

#### skeeter

Math Team
I assume the mass of the large blue block is 9 kg ... is the mass of the small yellow block given?

#### DarnItJimImAnEngineer

I agree, we seem to be missing some important information. Since conservation of momentum must be conserved in the horizontal direction, the blue block will move a lot faster if $\displaystyle \frac{m_B}{m_A}$ is a small value than if it is a large value.

#### topsquark

Math Team
You are right, we need the mass of the (blue) block.

@Chemist116: I just have to ask.. Where are you getting these problems from? This almost begs for a simple Lagrangian methods problem. It is not obvious about how to do it otherwise. Is your instructor a sadist or something?

This is essentially an energy methods problem. I'm going to lable the yellow block b and the blue block B. Since there is no friction in the problem we can get away with using conservation of energy. But we need to be careful... I'm going to measure all speeds in lab coordinates. The speed of b with respect to B can be calculated but I'm not going to do it.

Some notation: I'm choosing +y upward. I'm setting the lowest point of the block to be the zero for gravitational potential energy. The speed of b at the bottom is v and the speed of B is V.
Energy is conserved so
$$\displaystyle V_i + KE_i = V_f + KE_f$$

$$\displaystyle (V_{bi} + KE_{bi} ) + (V_{Bi} + KE_{Bi}) = (V_{bf} + KE_{bf} ) + (V_{Bf} + KE_{Bf})$$

$$\displaystyle (mgR + 0) + (0 + 0) = \left ( 0 + \dfrac{1}{2} mv^2 \right ) + \left ( 0 + \dfrac{1}{2} MV^2 \right )$$

Momentum is conserved here but that has to be justified. Briefly speaking, the only net external force on anything in the problem is gravity and that won't be an issue in this case since we only need the momentum component in the horizontal direction.

Okay, so point made there. I'm now going to assume we can go ahead and use conservation of momentum. I'm choosing +x to the right.

$$\displaystyle P_i = P_f$$

$$\displaystyle P_{bi} + P_{Bi} = P_{bf} + P_{Bf}$$

$$\displaystyle 0 + 0 = mv - MV$$

So you have the two equations:
$$\displaystyle mgR = \dfrac{1}{2}mv^2 + \dfrac{1}{2}MV^2$$
$$\displaystyle 0 = mv - MV$$

Two equations and three unknowns (M, v, and V) so we can't finish it.

-Dan

#### Chemist116

You are right, we need the mass of the (blue) block.

@Chemist116: I just have to ask.. Where are you getting these problems from? This almost begs for a simple Lagrangian methods problem. It is not obvious about how to do it otherwise. Is your instructor a sadist or something?

This is essentially an energy methods problem. I'm going to lable the yellow block b and the blue block B. Since there is no friction in the problem we can get away with using conservation of energy. But we need to be careful... I'm going to measure all speeds in lab coordinates. The speed of b with respect to B can be calculated but I'm not going to do it.

Some notation: I'm choosing +y upward. I'm setting the lowest point of the block to be the zero for gravitational potential energy. The speed of b at the bottom is v and the speed of B is V.
Energy is conserved so
$$\displaystyle V_i + KE_i = V_f + KE_f$$

$$\displaystyle (V_{bi} + KE_{bi} ) + (V_{Bi} + KE_{Bi}) = (V_{bf} + KE_{bf} ) + (V_{Bf} + KE_{Bf})$$

$$\displaystyle (mgR + 0) + (0 + 0) = \left ( 0 + \dfrac{1}{2} mv^2 \right ) + \left ( 0 + \dfrac{1}{2} MV^2 \right )$$

Momentum is conserved here but that has to be justified. Briefly speaking, the only net external force on anything in the problem is gravity and that won't be an issue in this case since we only need the momentum component in the horizontal direction.

Okay, so point made there. I'm now going to assume we can go ahead and use conservation of momentum. I'm choosing +x to the right.

$$\displaystyle P_i = P_f$$

$$\displaystyle P_{bi} + P_{Bi} = P_{bf} + P_{Bf}$$

$$\displaystyle 0 + 0 = mv - MV$$

So you have the two equations:
$$\displaystyle mgR = \dfrac{1}{2}mv^2 + \dfrac{1}{2}MV^2$$
$$\displaystyle 0 = mv - MV$$

Two equations and three unknowns (M, v, and V) so we can't finish it.

-Dan

The mass of the blue block is $9\,kg$

The mass of the yellow block is $1\,kg$.

When I copied it down the problem in a rush I forgot to add these quantities. Perhaps adding this information will solve the problem?.

Actually I feel that the course in physics I'm taking is too strong for me, (even after all these years when I thought I knew something about it). Yet, I hope that my insistense in some questions is not to bother but really to say that I need assistance in those, which I appreciate your help, because alone I just can't. #### Chemist116

@topsquark I'm still stuck at why did you had to add K.E for the blue block to the K.E of the yellow block?.

I mean the justification of the terms in the right in this equation: I mean the conservation of momentum in the horizontal component.

$mRh=\frac{1}{2}mv^2+MV^2$

$M$ is the mass of the blue block, or is it the mass of the combined masses (the mass of the blue block and the yellow block?). Can you help me with this?.

#### DarnItJimImAnEngineer

In the above equation, we are taking the system to be both blocks. The blue block does not move up or down, so its potential energy does not change. The potential energy lost by the small block ($mRh$) must equal to the combined kinetic energy gained by the system ($\frac{1}{2}[mv^2+MV^2]$).

• Chemist116 and skeeter

#### skeeter

Math Team
${\color{red}mgR} = \dfrac{1}{2}[mv^2 + MV^2]$

also, $\vec{p_0} = \vec{p_f} = 0$

• Chemist116 and DarnItJimImAnEngineer

#### DarnItJimImAnEngineer

• Chemist116 and skeeter

#### skeeter

Math Team
そうだ Mea culpa.
..it happens. 