Motion of an object inside an open box

Question:
As shown in the figure, an open box is put on frictionless floor; then, an object is placed in the middle of the box. There is no friction between the object and the surface of the box. The masses of the object and the box are equal. The coefficient of restitution between the object and inside box is $e \ (0<e<1)$.
(1) Let the object move at $v_i$ in the positive $x$ direction. Find the velocity of the object related to the floor after colliding the inside box.
(2) What is the position of the middle of the box after the second collision in the other side?

Answer:
(1) There is no external force; therefore, we can consider conservation of momentum. Let $v$ and $u$ be the velocities of the object and the box, respectively. The subscripts, $i$ and $f$, represent initial and final. Thus, we have
\begin{equation}
mv_i = mv_f + mu_f
\end{equation}
The coefficient of restitution is defined as the relative velocity after collision divided by the relative velocity before collision. Namely,
\begin{equation}
e = \frac{v_f - u_f}{0 - v_i}
\end{equation}
From (1) and (2), we have
\begin{eqnarray}
v_i &=& v_f + u_f \\
-v_i e &=& v_f - u_f
\end{eqnarray}
Let's find $v_f$ in terms of $v_i$ by eliminating $u_f$.
\begin{equation}
v_f = \frac{1-e}{2}v_i
\end{equation}

(2) First, we find the final velocity of the box by eliminating $v_f$ from (3) and (4).
\begin{equation}
u_f = \frac{1+e}{2}v_i
\end{equation}
The time from the first to the second collision is the distance, $L$, divided by the relative velocities.
\begin{equation}
t = \frac{L}{u_f - v_f} = \frac{L}{ev_i}
\end{equation}
Therefore, the distance that the box moved is
\[
d = t \times u_f = \frac{L}{ev_i} \times \frac{1+e}{2}v_i = \frac{1+e}{2e}L
\]

Powered by Hirophysics.com

A sliding block on a large cart: Conservation of energy and momentum

Question:
A block B is placed on the cart A that has the frictionless surface and going to be sliding the height of $h$ until the frictional bottom (the shaded part). The cart has frictionless wheels and it moves due to block's falling on the slope. The masses of the block and the cart are $m$ and $M$, respectively. Find each final velocity.

Answer:
The momentum in the horizontal direction should be conserved. Let velocities be $V_A$ and $V_B$. The initial total momentum is zero since they are initially at rest. We assume that the final total momentum is the sum of these momenta. From conservation of momentum, we have
\begin{equation}
0 = MV_A+mV_B
\end{equation}
The total energies in initial and final states must be conserved. The LHS and RHS of the following equation are initial total and final total energies, respectively.
\begin{eqnarray}
 (0+mgh) + (0 + 0) &=& \left(\frac{1}{2}mV_B^2+0\right) + \left(\frac{1}{2}MV_A^2 + 0 \right) \nonumber  \\
\rightarrow \ \frac{1}{2}MV_A^2 &+& \frac{1}{2}mV_B^2 = mgh
\end{eqnarray}
Now, derive $V_B$ and $V_A$ from these equations. From (1), we have
\begin{equation}
V_A = -\frac{m}{M}V_B
\end{equation}
Plug it in (2).
\begin{eqnarray}
M \left(\frac{m}{M}\right)^2 V_B^2 + mV_B^2 &=& 2mgh  \\
\left(\frac{m}{M}+1\right)V_B^2 &=& 2gh  \\
V_B = \sqrt{\frac{2Mgh}{m+M}}
\end{eqnarray}
Then, plug (6) in (3).
\begin{equation}
V_A = -m\sqrt{\frac{2gh}{M(m+M)}}
\end{equation}

Powered by Hirophysics.com

Ballistic pendulum

Question:
A 0.010-kg bullet is fired toward a 2.0-kg pendulum. The bullet is sucked in the pendulum block and both are raised up to 0.20-m height together. What is the initial velocity of the bullet?

Answer:
The masses of the bullet and the block can be denoted as $m$ and $M$, respectively. The initial velocity of the bullet and the final velocity of both the bullet and block are $v$ and $V$. Consider the states (a) through (b) in the figure. According to the conservation of linear momentum, the total initial momentum must be equal to the total final momentum.
\begin{equation}
mv = (m+M)V
\end{equation}
Now, consider the states (b) to (c). For those states, we use the conservation of mechanical energy.
\begin{equation}
\frac{1}{2}(m+M)V^2 = (m+M)gh
\end{equation}
$v$ and $V$ are unknown. In order to obtain the initial velocity, solve for $V$ first from the equation (1).
\begin{equation}
V = \frac{mv}{m+M}
\end{equation}
From equation (2), we have
\begin{equation}
V = \sqrt{2gh}
\end{equation}
Therefore,
\begin{equation}
\frac{mv}{m+M} = \sqrt{2gh}
\end{equation}
The initial velocity is
\begin{equation}
v = \frac{m+M}{m}\sqrt{2gh} = \frac{0.010+2.00}{0.010}\sqrt{2\cdot 9.8\cdot 0.20} = 398\quad\mathrm{m/s}
\end{equation}

Powered by Hirophysics.com