Two blocks of masses \(m_1\) and \(m_2\) separated by a horizontal distance \(D\) are dropped from the same height \(h\) at the same time. Find the vertical position of the center of mass at a time \(t \) before the two blocks strike the ground. Assume no air resistance.
Hint.
Free fall of Center of Mass.
Answer.
\(Y_\text{cm} = h -\dfrac{1}{2}\, g\, t^2\text{.}\)
Solution.
We use positive \(y \) coordinate pointed up with origin at the ground. Since only external force on the two blocks is the gravity, we have a free-fall motion of the center of mass. This will give the following equation for the \(Y_\text{cm}\) at instant \(t \text{.}\)
Here \(a_y = - g \text{,}\)\(V_\text{cm}(0) = 0 \text{,}\) and
\begin{equation*}
Y_\text{cm}(0) = \dfrac{m_1 h + m_2 h}{m_1 + m_2} = h.
\end{equation*}
Therefore,
\begin{equation*}
Y_\text{cm} = h -\dfrac{1}{2}\, g\, t^2.
\end{equation*}
54.Center of Mass - Two Particles Moving in Circles.
Two particles of masses \(m_1\) and \(m_2=\dfrac{m_1}{2}\) move uniformly in different circles of radii \(R_1\) and \(R_2\) about origin in the \(xy\) plane. The \(x\) and \(y\) coordinates of the center of mass and that of particle 1 are given as follows where length is in meters and \(t\) in seconds.
From \(x_2\) and \(y_2\text{,}\) we get the radius of the circle in which 2 moves.
\begin{equation*}
r_2 = 1.0.
\end{equation*}
55.Center of Mass of Iron and Aluminum Rods Fused Together.
Find the center of mass of a one-meter long rod, made of \(50\text{-cm}\) long iron of density \(8\text{ g/cc}\) and \(50\text{ cm}\) long aluminum of density \(2.7\text{ g/cc}\) rods.
Hint.
First replace each rod by a point mass at the center.
Answer.
\(12.4\text{ cm}\) from center towards the iron rod.
Solution.
Let subscript 1 denote iron and 2 the aluminum. Let \(A \) be the area of cross-sections of the rods, \(L_1\) and \(L_2\) their lengths, and \(\rho_1\) and \(\rho_2\) their densities. The masses of each rod are
\begin{align*}
\amp m_1 = \rho_1 A L_1 = 400\, A, \\
\amp m_2 = \rho_2 A L_2 = 135\, A,
\end{align*}
in units of \(\text{grams}\) if \(A\) is in \(\text{cm}^2\text{.}\) We leave the units alone since they will cancel out in the end giving us the answer in the same unit as the unit of \(L\text{.}\)
Replacing each rod at the center by its mass by point masses we get a system of two point masses. Let us use \(x \) axis that has origin at the center of the iron rod and the positive \(x \) axis towards the aluminum rod.
This gives the point masses at \(x_1=0\) and \(x_2 = 50\text{ cm}\text{.}\) Hence \(X_\text{cm}\) is
We can write this also as \(25-12.6 = 12.4\text{ cm}\) from the center towards the iron rod.
56.Collision of a Bullet with a Block and Bullet Getting Embedded in the Block.
A bullet of mass \(200\text{ g}\) traveling horizontally towards East with speed \(400\text{ m/s}\) strikes a block of mass \(1.5\text{ kg}\) at rest on a frictionless table. After striking the block the bullet is imbedded in the block and the block and the bullet move together as one unit.
(a) What is the magnitude and direction of the velocity of the block/bullet combination immediately after the impact?
(b) What is the magnitude and direction of the impulse by the block on the bullet.
(c) What is the magnitude and direction of the impulse from the bullet on the block?
(d) If it took \(100\text{ msec}\) for the bullet to change the speed from \(400\text{ m/s}\) to the final speed after impact, what is the average force between the block and the bullet during this time?
Hint.
(a) Use conservation of momentum. (b) \(J_\text{on bullet} = \Delta p_\text{bullet}\text{.}\) (c) \(\vec J_\text{on block}^{by bullet} = -\vec J_\text{on bullet}^{by block} \text{.}\) (d) Use \(F = J/\Delta t\text{.}\)
Answer.
(a) \(47\text{ m/s}\) in the original direction of the bullet; (b) \(70.6\text{ N.s}\text{,}\) towards bullet; (c) \(70.6\text{ N.s}\text{,}\) towards block; (d) magnitude: \(706\text{ N}\text{.}\)
Solution1.a
Let \(x\) be the direction of the motion of the bullet before collision as shown in Figure 7.99. Since block is at rest, the resulting motion will also be towards the positive \(x \) axis. The conservation of \(x\) component of the momentum gives us the following equaiton for the speed \(V\) of the bullket/block combination after the collision.
\begin{equation*}
0.2\times 400 = \left( 0.2 + 1.5\right)\, V.
\end{equation*}
This gives
\begin{equation*}
V = 47.1\text{ m/s}.
\end{equation*}
Solution2.b
We now look at the motion of the bullet by itself to determine the change in momentum of the bullet since the impulse on the bullet will equal the change in the momentum of the bullet. The simplified situation is shown in Figure 7.100.
Therefore, we find that the impulse on the bullet has magnitude \(70.6 \text{ N.s}\) and has the direction opposite to its original velocity.
Solution3.c
We do not need to do any work for this part since by the third law of motion, the answer of this part will be negative of the answer for part (b). That is, the impulse on the block has magnitude \(70.6 \text{ N.s}\) and has the direction same as the direction of the bullet’s original velocity.
Solution4.d
From the impuse and duration, we get the magnitude of the average force between the block and the bullet to be
57.Collision of Two Equal Mass Billiard Balls in Two Dimensions.
A billiard ball, labeled 1, moving horizontally strikes another ball, labeled 2, at rest at a glancing angle. Before the impact, ball 1 was moving at a speed of \(3.0\text{ m/s}\) and after the impact it is moving at \(1.0\text{ m/s}\) at \(150^{\circ}\) from the original direction. If the two balls have equal masses, what is the velocity of ball 2 after the impact.
Hint.
Apply conservation of momentum along the two axes. Mass will cancel out.
Answer.
\(3.93\text{ m/s}\text{,}\)\(7.4^{\circ}\) clockwise from the original direction.
Solution.
To find the magnitude and direction of the velocity of the second ball after impact we will work out its \(x\) and \(y\) components. Refer to Figure 7.101 for the axes.
The conservation of momentum along the two axes gives us the following two equations, where we are omitting the mass since they will cancel out. We will write the initial values in each direction on the left side of the equation.
Since \((3.9, -0.5) \) is in the fourth quadrant, the direction is \(7.3^{\circ} \) clockwise from the positive \(x\) axis.
58.Collision of Two Projectiles at Their Highest Point.
Two projectiles of mass \(m_1\) and \(m_2\) with \(m_2 \lt m_1\) are fired in the opposite directions from two launch sites as shown in Figure 7.102. The two launch sites are separated by a distance \(D\text{.}\) The projectiles have the same initial speed and same angle with the horizon so that both reach the same spot in their highest point and strike there. As a result of the impact they stick together and move as a single body afterwards. Find the place they will land.
This problem has two parts: (i) the one-dimensional collision at the highest point, and (ii) a projectile motion afterwards.
Answer.
\(\left(\dfrac{m_1 D}{m_1 + m_2} \right) \) from the launch sire of \(m_1\text{.}\)
Solution.
Strategy. This problem has two parts: (i) the one-dimensional collision at the highest point, and (ii) a projectile motion afterwards. We will use the axes shown in the figure below.
(i)The momentum conservation part. Let \(v_0\) be the initial speed and \(\theta\) the initial direction. Let the projetile 1 be initially located at the origin and the positive \(x \) axis be towards projectile 2. his gives us the following for the constant \(x\) velocities of the two projectiles.
The projectile motion part. If we know the height at which this projectile starts, we can figure out the place it will land. To find the height we note that during the time 1 moved a distance \(D/2\text{,}\) the max height \(H\) was reached, where \(v_{1y}=0\text{.}\) That is,
\begin{equation*}
H = \dfrac{v_0^2\sin^2\, \theta}{2g}.
\end{equation*}
Now, we can use \(V_x\) and \(H\) to get the final point \((x, 0)\) from starting place, which is at \(x_i=D/2\) and \(y_i=H\) of the combined projectile. We get \(t\) from the \(y\) equation,
\begin{equation*}
H = \dfrac{1}{2}gt^2,
\end{equation*}
which gives
\begin{equation*}
t = \sqrt{\dfrac{2H}{g}}.
\end{equation*}
Using this in the \(x \) equation gives the final \(x\) coordinate.
\begin{equation*}
x - \dfrac{D}{2} = V_x t = \dfrac{m_1 - m_2}{m_1 + m_2}\, v_{1i,x}\, \times \sqrt{\dfrac{2H}{g}}.
\end{equation*}
Now we use \(v_{1i,x}\) and \(H\) and simplify the resulting expression. We get
This expression has unknown \(v_0\) and \(\theta\text{.}\) Actually, their combination \(v_0^2\sin\,\theta\cos\,\theta\) is related to the range \(D\) of the original projectiles, since from the symmetry we can tell that if we had launched only projectile 1, it would land at a distance \(D\) where the second site is located. Working out the range gives us (this is another projectile motion problem that you would need to work out.)
\begin{equation*}
D = \dfrac{2v_0^2\sin\,\theta\cos\,\theta}{g}.
\end{equation*}
Using this information, we get our final answer in the given quantities.
59.Force Required to Maintain Constant Velocity of a Freight Car with Varying Mass Cargo.
Grains from a grain hopper falls at a rate of \(10\text{ kg/sec}\) vertically onto a freight car that is moving horizontally at a constant speed \(2\text{ m/s}\) on a straight track. What force is needed to keep the freight car moving at a constant velocity.
Hint.
Use Newton’s law in change in momentum/impulse form along horizontal direction.
Answer.
\(20\text{ N}\text{,}\) horizontal in the direction of the velocity.
Solution.
We will use Newton’s law in change in momentum/impulse form along the \(x\) axis. Let positive \(x\) axis be pointed in the direction of the velocity of the freight car as shown in the figure.
Let us look at the change in \(x \) component of momentum during an interval \(t \) to \(t+\Delta t\text{.}\) Let \(\Delta m\) denote the amount of additional grain into the car during this time.
\begin{equation*}
F\Delta t = (m + \Delta m) v - m v.
\end{equation*}
Therefore, we get
\begin{equation*}
F \Delta t = v\, \Delta m.
\end{equation*}
Now, if grain is pouring in at the rate \(\alpha \text{ kg/s}\text{,}\) then we will have
\begin{equation*}
\Delta m = \alpha\Delta t.
\end{equation*}
Hence, we get the following for the magnitude of the horizontal force needed to maintain the constant velocity of the car.
\begin{equation*}
F = v\, \alpha.
\end{equation*}
Using the numerical values given we get \(F = 20\text{ N}\text{.}\) The direction of this force is, of course, in the direction of the velocity of the car.
60.Recoil Velocity of a Tennis Ball Machine.
A tennis ball machine has a number of tennis balls in it and rests on a fricionless table. The machine together with balls has mass \(M\text{,}\) with each ball’s mass \(m\text{.}\) The machine shoots a ball in the forward direction at speed \(u \) with respect to the machine at angle \(\theta\) with horizon.
Figure7.104.
As a result, the machine recoils with a speed \(V_1\) with respect to the table in the backward direction. While the machine is moving, it shoots the second ball which comes out at the same speed with respect to the machine, but now machine is moving at speed \(V_2\) with respect to the table. (a) Find (a) \(V_1\) and (b) \(V_2\text{.}\) (c) Write a formula for the speed of the ball machine after \(N\) balls have been shot.
Hint.
Work with the \(x\) axis of a coordinate system fixed to the ground and express all velocities with respect to this frame.
Answer.
(a) \(\dfrac{m}{M}\, u\,\cos\,\theta\text{,}\) (b) \(\dfrac{m}{M-m}\, u\,\cos\,\theta\text{,}\) (c) \(\dfrac{M u \cos\theta}{M-(N-1)m}\text{.}\)
Solution1.a
We will use a coordinate system that is fixed to the ground and express all velocities with respect to this system. We will work with \(x\) coordinate only with the positive \(x\) axis in the direction in which the ball comes out. We will also denote quantities for machine with a subscript \(1 \) and that of a ball by subscript \(2\text{.}\)
The \(x\)-components of the velocities before and after the firing of the first ball with respect to the table are:
The results of (b) shows us how the process will generalize to
\begin{equation*}
V_N = \frac{M u \cos\theta}{M-(N-1)m}.
\end{equation*}
61.Bend or Not to Bend Your Knee When Landing after a Jump.
A \(65\)-kg person jumps from first floor window of a burning building and lands almost vertically on the ground. Just before impact the horizontal component of the velocity is \(3.0\text{ m/s}\) and the vertical component is \(-9.0\text{ m/s}\text{.}\)
Upon impact with the ground the jumper comes to rest in a short time, which is not zero. The force experienced by his feet depends on whether he keeps his knees stiff or bends them. In this problem you will work out forces in the two cases as instructed below.
(a) First find the impulse on the person from the impact on the ground. Calculate both its magnitude and direction.
(b) Find the average force on the feet if the person keeps his leg stiff and straight and his CM drops by only \(1.0\text{ cm}\) vertically and moves \(1.0\text{ cm}\) horizontally during the impact. You will need to find the time the impact lasts by making reasonable assumptions about the deceleration. Although the force is not constant during the impact, working with constant average force for this problem is acceptable.
(c) Find the average force on the feet if the person bends his legs throughout the impact so that his CM drops by \(50\) cm vertically and moves \(5\) cm horizontally during the impact.
(d) Compare the results of part (b) and (c), and draw conclusions about which way is better.
Hint.
(a) Use change of momentum along \(x\) and \(y\) axes. (b) Assume constant acceleration. This would give \(\Delta x = v_{av,x} \Delta t\text{.}\) (c) Same as (b) but with different values. (d) You should get (c) to be better.
Answer.
(a) \(617\text{ N.s}, 72^{\circ}\) clockwise from \(-x\) axis, (b) \(F_x=-29,100\text{ N}\text{,}\)\(F_y = 26,590 \text{ N}\text{,}\)\(F_\text{av} = 39,000\text{ N}\text{,}\) (c) \(F_x=- 5,909\text{ N}\text{,}\)\(F_y = 5,318 \text{ N}\text{,}\)\(F_\text{av} = 8,000\text{ N}\text{,}\) (d) Bending your knee is better since you reduce the force of impact on you by spreading the impulse over longer interval.
Solution1.a
The impulse on the person is equal to change in his mometum during the impact, i.e., during the time he hits the ground and comes to rest. We have following for \(x \) and \(y \) components of his velocity instant before impact and when he comes to rest. Here \(x\) is horizontal and positive \(y\) is vertically up.
Since \((-195, 585) \) is in the second quadrant, the negative angle gives the dfirection as the clockwise angle with respect to the negative \(x\)axis.
Solution2.b
We assume constant accelerations in the \(x \) and \(y\) directions as a result of coming to rest within assumed distances. The formulas that will give \(\Delta t\) will be
\begin{align*}
\Delta x \amp = v_{\text{av},x}\,\Delta t = \dfrac{v_{xi}}{2} \Delta t. \\
\Delta y \amp = v_{\text{av},y}\,\Delta t = \dfrac{v_{yi}}{2} \Delta t.
\end{align*}
We assume constant accelerations in the \(x \) and \(y\) directions as a result of coming to rest within assumed distances as above. The formulas that will give \(\Delta t\) will again be
\begin{align*}
\Delta x \amp = \dfrac{v_{xi}}{2} \Delta t. \\
\Delta y \amp = \dfrac{v_{yi}}{2} \Delta t.
\end{align*}
Bending your knee is better since you reduce the force of impact on you by spreading the impulse over longer interval.
62.Falling Chain Weighs More Than Just Its Weight.
A chain of mass \(M\) and length \(L\) is suspended vertically with its bottom touching a pan of mass \(m\) on a scale. The chain is then let go from rest. As chain falls, the scale reads the force on the pan needed to balance the weight of the chain that is on the pan and the chain link that is impacting the pan at that instant. Assume each chain link is infinitesimally small and comes to rest instantaneously upon impact so that at each instant the entire momentum of an infinitesimal part of the chain link falling on the pan is transferred to the pan.
(a) Find an expression of the reading of the scale when a length \(x\) has fallen.
(b) What is the reading on the scale at the instant the whole chain has fallen to the fan.
(c) What is the reading on the scale after the whole chain has fallen and everything has come to rest?
Hint.
Look at momentum change during interval \(t \) to \(t + \Delta t\text{,}\) where \(t\) is the instant when \(x\) length has fallen and are already at rest on the pan.
We look at the normal force on the pan, which is the reading in the scale, duration \(t \) to \(t + \Delta t\text{.}\) During this time the chain between \(x\) and \(x+\Delta x\) strikes the pan and deposits its momentum. The pan already has \(m = (M/L)x\) at rest on the pan and another mass \(\Delta m = (M/L)\Delta x\) with speed \(v = \sqrt{2gx}\text{,}\) due to free fall by distance \(x\text{,}\) deposits its momentum. Therefore, the normal force on the pan will be
\begin{equation*}
N(t) = \left( \dfrac{M}{L}x\right) g + \left( \dfrac{M}{L} \Delta x \right) \sqrt{2gx}\dfrac{1}{\Delta t},
\end{equation*}
The \(\Delta x/\Delta t\) in the second term is just the speed at this instant, which is \(v = \sqrt{2gx}\text{.}\) Therefore, we have
\begin{equation*}
N(t) = \left(\dfrac{M}{L}x\right) g + \left( \dfrac{M}{L} \right) 2gx = 3 \left( \dfrac{M}{L} \right) x g.
\end{equation*}
Solution2.b
Putting \(x=L\) in the answer in (a), we get \(N = 3Mg\) for the reading at this instant.
Solution3.c
After everything has come to rest, the reading will just be \(Mg\text{.}\)
63.(Calculus) Chain Falling on the Side of a Frictionless Table.
A chain of length \(L\) rests on a frictionless horizontal table. When a small portion of length \(\delta\) of the chain is pushed over the side of the table and let go from rest, the entire chain falls off the table.
Figure7.105.
(a) What is the length of the chain that has fallen off the table by time \(t\text{?}\) (b) What is the speed with which chain is falling?
Hint.
Set up \(\vec F \Delta t = \Delta \vec p\text{.}\) Also note that external force is weight of only the hanging part.
Figure 7.106 shows situation at instant \(t\text{.}\) At an instant \(\Delta t\) later, speed of the entire chain will be \(v + \Delta v\text{.}\) Let \(\lambda\) be mass per unit length. Noting that external force is only the weight of the hanging part, we will get the following \(y\)-equation of motion.
Figure7.106.
\begin{equation*}
(v_y+\Delta v_y) \lambda L - v_y \lambda L = \lambda y g \Delta t.
\end{equation*}
Here \(v_y = dy/dt\text{,}\) which is equal to \(v\) in the figure. Dividing by \(\Delta t\) and taking the infinitesimal time limit, we get
\begin{equation*}
\frac{dv_y}{dt} = \frac{g}{L} y.
\end{equation*}
This is same as
\begin{equation*}
\frac{d^2y}{dt^2} = \frac{g}{L} y.
\end{equation*}
The general solution of this differential equation is
\begin{equation*}
y = A e^{\sqrt{g/L}\,t} + B e^{-\sqrt{g/L}\,t},
\end{equation*}
where \(A\) and \(B\) are constants. To fix them, we will use \(y=\delta\) and \(v_y=0\) at \(t=0\text{.}\) First we get expression for \(v_y\) by taking a derivative.
\begin{equation*}
v_y = \sqrt{g/L}\, \left( A e^{\sqrt{g/L}\,t} - B e^{-\sqrt{g/L}\,t} \right).
\end{equation*}
Thus at \(t=0\text{,}\) we have
\begin{align*}
\amp A + B = \delta.\\
\amp \sqrt{g/L}\,(A - B) = 0.
\end{align*}
Therefore,
\begin{equation*}
A = B = \frac{\delta}{2}.
\end{equation*}
Hence, we have the following expressions for position and velocity of the bottom of the falling chain.
A can is placed in an inverted position on a steady stream of water coming out a hose vertically (Figure 7.107). Initially, the can is pushed up, but at some height the can becomes suspended in air. Suppose speed of water at the hose is \(v_0\) in \(\text{m/s}\text{,}\)\(V\)\(\text{kg/m}^3\) of water flow per second, \(\rho\) the density of water in \(\text{kg/m}^3\text{,}\) and \(M\) the mass of the can in \(\text{kg}\text{.}\) What will be the height if we assume that after hitting the can, water hits the can elastically, i.e., water is turned back at the same speed? (Adapted from Kleppner and Kolenkow)
Figure7.107.
Hint.
Balance weight of the can by the force of water on can. Find force of water in can by finding force of can on water in changing the momentum of water.
Strategy: By looking at the rate of change of momentum of water, we will get the force exerted by the can on water. By third law, water will be exerting the same magnitude force on the can. Then, we will balance forces on the can to find the necessary condition for the problem.
Impulse on water by the can will equal the rate at which momentum of water is changed by hitting the can. Let \(m\) denote the mass of water that strikes the can in duration \(\Delta t\text{.}\) Let \(F\) be the magnitude of force between can and water. Let positive \(y\)-axis be pointed up in Figure 7.107.
\begin{equation*}
-F \Delta t = m (-v) - m v = - 2 mv.
\end{equation*}
Here \(v\) is the speed with which water hits the can. Since water had vertical velocity \(v_0\) a height \(h\) below the can, we have
\begin{equation*}
v = \sqrt{v_0^2 - 2 g h}.
\end{equation*}
And \(m\) will be
\begin{equation*}
m = \rho V \Delta t.
\end{equation*}
Therefore, the magnitude of force by water on the can will be
\begin{equation*}
F = 2\rho V \sqrt{v_0^2 - 2 g h}.
\end{equation*}
Now, this force balances the weight of the can to keep the can suspended. Therefore,
\begin{equation*}
2\rho V \sqrt{v_0^2 - 2 g h} = Mg.
\end{equation*}
65.Cannon Balls Fired at Once Versus one After the Other.
There are \(N\) cannons on a railway car. The total mass of the car and the cannons is \(M\text{.}\) The cannons can fire cannon balls of mass \(m\) at a speed \(u\) with respect to the car regardless of the state of motion of the car. Initially the car is at rest. If cannons are fired horizontally in the direction of the back of the car, the car moves forward.
What is the speed of the car when all cannons are fired at the same time?
What is the final speed of the car if cannons are fired one after another?
Which is larger? Why?
Solution1.a
Case of all cannonballs fired at the same time. The horizontal momentum of the [(cannon + car) + cannonballs] system is conserved. We also need to express the velocities with respect to an inertial frame. Suppose the cars are moving the positive \(x\)-direction. Let the \(x\)-component of the rest of the structure is \(v_x\) and all cannon balls were fired at \(x\)-velocity equal to \(-u\) when cannon was at rest. Therefore, the equation for the conservation of momentum along \(x\)-axis with respect to the ground will be
\begin{equation*}
v_x = \left(\frac{N m}{M - N m} \right)\ u.
\end{equation*}
Note: \(M\) is mass of (cannon plus car) and all the cannon balls together. So, denominator will never be zero.
Solution2.b
Case of firing one cannonball at a time starting from rest. We again start with \(N\) cannonballs in the car with a total mass = \(M+ Nm\text{,}\) and shoot one cannonball after another. The \(x\)-component of the velocity of the (cannon + car) after the first cannonball is fired is obtained by simply setting \(N = 1\) in (a).
\begin{equation*}
v_{1x} = \frac{m}{M+(N-1)m}\ u
\end{equation*}
The \(v_{2x}\) after the second cannonball is fired will obey
To compare the two for a general \(N\) is difficult since this requires summing up the series. Comparing for \(N=2\) would suffice for our purpose here. We will try to estimate the ratio \(v\)(all) over \(v\)(one after another).
\begin{equation*}
\frac{v(\textrm{all})}{v(\textrm{one after another})} = \frac{2m}{M-2m}\times \frac{M(M+m)}{m(2M+m)}
\end{equation*}
Let \(x = m/M\) and take \(x \ll 1\) approximation.
\begin{align*}
\frac{v(\textrm{all})}{v(\textrm{one after another})} \amp = \frac{2}{1-2x}\times \frac{1+x}{2+x}\\
\amp = a(1+2x+\cdots) \ \frac{1+x}{2} \ (1-\frac{x}{2}+\cdots)\\
\amp \approx 1+ \frac{5}{2} x \ \ \textrm{(Keeping terms to only one power in x)}\\
\amp \gt 1\ \ \ \textrm{(Since )}
\end{align*}
