July 9, 2020

Solution Set 10

Physics 207-02 – Introductory Physics I
Solution Set 10
Nazareth College
Department of Chemistry & Biochemistry
Robert F Szalapski, PhD – Adjunct Lecturer
Fall 2011

§ Chapter 7, Problem 3

A 0.145\;\mathrm{kg} baseball pitched 39.0\;\mathrm{m}/\mathrm{s} is hit on a horizontal line drive back toward the pitcher at 52.0\;\mathrm{m}/\mathrm{s}. If the contact time between the bat and ball is 3.00\times 10^{{-3}}\;\mathrm{s}, calculate the average force between the ball and bat during the contact.


\displaystyle F \displaystyle= \displaystyle\frac{\Delta p}{\Delta t}

For our coordinate system choose the positive x axis pointing from the batter towards the pitcher. Watch the signs!

\displaystyle F \displaystyle= \displaystyle\frac{\Delta p}{\Delta t}
\displaystyle= \displaystyle\frac{p_{f}-p_{i}}{\Delta t}
\displaystyle= \displaystyle\frac{(0.145\;\mathrm{kg})(52.0\;\frac{\mathrm{m}}{\mathrm{s}})-(0.145\;\mathrm{kg})(-39.0\;\frac{\mathrm{m}}{\mathrm{s}})}{3.00\times 10^{{-3}}\;\mathrm{s}}
\displaystyle= \displaystyle 4400\;\mathrm{N}
\displaystyle= \displaystyle 4.4\;\mathrm{kN}

§ Chapter 7, Problem 4

A child in a boat throws a 6.40\;\mathrm{kg} package out horizontally with a speed of 10.0\;\mathrm{m}/\mathrm{s}. Calculate the velocity of the boat immediately after assuming it was initially at rest. The mass of the child is 26.0\;\mathrm{kg}, and that of the boat is 45.0\;\mathrm{kg}. Ignore water resistance.

Use conservation of linear momentum. We cant treat the child and the boat as one object of mass 26.0\;\mathrm{kg}+45.0\;\mathrm{kg}=71.0\;\mathrm{kg} since they will be moving together at the same speed. Since initially everything is at rest, the initial linear momentum is zero. Hence

\displaystyle 0 \displaystyle= \displaystyle(71.0\;\mathrm{kg})v+(6.40\;\mathrm{kg})\left(10.0\frac{\mathrm{m}}{\mathrm{s}}\right)
\displaystyle v \displaystyle= \displaystyle-\frac{6.40\;\mathrm{kg}}{71.0\;\mathrm{kg}}\left(10.0\frac{\mathrm{m}}{\mathrm{s}}\right)
\displaystyle= \displaystyle-0.901\frac{\mathrm{m}}{\mathrm{s}}

The negative sign indicates that the motion of the boat is in the opposite direction as the velocity of the package.

§ Chapter 7, Problem 5

Calculate the force exerted on a rocket given that the propelling gases are expelled at a rate of 1500\;\mathrm{kg}/\mathrm{s} with a speed of 4.0\times 10^{4}\;\mathrm{m}/\mathrm{s} at the moment of takeoff.

We assume that the rocket is initially at rest, and so the initial linear momentum is zero. Whatever momentum is imparted to the propelling gases, the equal bot opposite momentum will be imparted to the rocket. For problems of this type I like to consider the momentum in an arbitrary time period \Delta t. During this time the mass of propelling gas that is expelled is

\displaystyle\Delta m \displaystyle= \displaystyle\left(1500\frac{\mathrm{kg}}{\mathrm{s}}\right)\Delta t

which will have a momentum

\displaystyle\Delta p \displaystyle= \displaystyle\Delta m\; v
\displaystyle= \displaystyle\left(1500\frac{\mathrm{kg}}{\mathrm{s}}\right)\Delta t\; v

and the force is

\displaystyle F \displaystyle= \displaystyle\frac{\Delta p}{\Delta t}
\displaystyle= \displaystyle\frac{\left(1500\frac{\mathrm{kg}}{\mathrm{s}}\right)\Delta t\; v}{\Delta t}
\displaystyle= \displaystyle\left(1500\frac{\mathrm{kg}}{\mathrm{s}}\right)\; v
\displaystyle= \displaystyle\left(1500\frac{\mathrm{kg}}{\mathrm{s}}\right)\left(4.0\times 10^{4}\frac{\mathrm{m}}{\mathrm{s}}\right)
\displaystyle= \displaystyle 6.0\times 10^{7}\;\mathrm{N}
\displaystyle= \displaystyle 60.\;\mathrm{MN}

§ Chapter 7, Problem 7

A 12,600\;\mathrm{kg} railroad car travels alone on a level frictionless track with a constant speed of 18.0\;\mathrm{m}/\mathrm{s}. A 5350\;\mathrm{kg} load, initially at rest, is dropped on the car. What will be the car’s new speed?

Use conservation of momentum realizing that both masses will be moving together after the load is dropped onto the car.

\displaystyle(12,600\;\mathrm{kg})\left(18.0\frac{\mathrm{m}}{\mathrm{s}}\right)+(5350\;\mathrm{kg})0 \displaystyle= \displaystyle\left(12,600\;\mathrm{kg}+5350\;\mathrm{kg}\right)v
\displaystyle v \displaystyle= \displaystyle 12.6\frac{\mathrm{m}}{\mathrm{s}}

§ Chapter 7, Problem 8

A 9300\;\mathrm{kg} boxcar traveling at 15.0\;\mathrm{m}/\mathrm{s} strikes a second boxcar at rest. The two stick together and move off with a speed of 6.0\;\mathrm{m}/\mathrm{s}. What is the mass of the second car?

Using conservation of momentum.

\displaystyle(9300\;\mathrm{kg})\left(15.0\frac{\mathrm{m}}{\mathrm{s}}\right)+m\times 0 \displaystyle= \displaystyle(9300\;\mathrm{kg}+m)\left(6.0\frac{\mathrm{m}}{\mathrm{s}}\right)
\displaystyle m \displaystyle= \displaystyle 14000\;\mathrm{kg}

§ Chapter 7, Problem 16

A 12\;\mathrm{kg} hammer strikes a nail at a velocity of 8.5\;\mathrm{m}/\mathrm{s} and comes to rest in a time of 8.0\;\mathrm{ms}.

§ (a)

What is the impulse given to the nail?

\displaystyle\mathrm{impulse} \displaystyle= \displaystyle\Delta p
\displaystyle= \displaystyle 0-(12\;\mathrm{kg})\left(-8.5\frac{\mathrm{m}}{\mathrm{s}}\right)
\displaystyle= \displaystyle 1\underline{0}2\frac{\mathrm{kg\; m}}{\mathrm{s}}
\displaystyle= \displaystyle 100\frac{\mathrm{kg\; m}}{\mathrm{s}}

§ (b)

What is the average force acting on the nail?

\displaystyle F \displaystyle= \displaystyle\frac{\Delta p}{\Delta t}
\displaystyle= \displaystyle\frac{1\underline{0}2\frac{\mathrm{kg\; m}}{\mathrm{s}}}{8.5\times 10^{{-3}}\;\mathrm{s}}
\displaystyle= \displaystyle 12000\;\mathrm{N}
\displaystyle= \displaystyle 12\;\mathrm{kN}

§ Chapter 7, Problem 17

Figure 1.

Chapter 7, Problem 17.

A tennis ball of mass m=0.060\;\mathrm{kg} and speed v=25\;\mathrm{m}/\mathrm{s} strikes a wall at a 45^{0} angle and rebounds with the same speed at 45^{0}. What is the impulse, magnitude and direction, given to the ball?

Remember that momentum is a vector quantity.

\displaystyle\vec{p} \displaystyle= \displaystyle m\vec{v}
\displaystyle\vec{F} \displaystyle= \displaystyle\frac{\Delta\vec{p}}{\Delta t}

Recall also

\displaystyle\mathrm{impulse} \displaystyle= \displaystyle\Delta\vec{p}
\displaystyle= \displaystyle\vec{F}\Delta t

The problem is straightforward if we work with components as indicated in the figure. Assume the usual coordinate system, and note that the y component of the velocity does not change while the x component flips direction.

\displaystyle\Delta p_{x} \displaystyle= \displaystyle-\frac{1}{\sqrt{2}}\left(0.060\;\mathrm{kg}\right)\left(25\frac{\mathrm{m}}{\mathrm{s}}\right)-\frac{1}{\sqrt{2}}\left(0.060\;\mathrm{kg}\right)\left(25\frac{\mathrm{m}}{\mathrm{s}}\right)
\displaystyle= \displaystyle-2.1\frac{\mathrm{kg\; m}}{\mathrm{s}}
\displaystyle\Delta p_{y} \displaystyle= \displaystyle\frac{1}{\sqrt{2}}\left(0.060\;\mathrm{kg}\right)\left(25\frac{\mathrm{m}}{\mathrm{s}}\right)-\frac{1}{\sqrt{2}}\left(0.060\;\mathrm{kg}\right)\left(25\frac{\mathrm{m}}{\mathrm{s}}\right)
\displaystyle= \displaystyle 0.0\frac{\mathrm{kg\; m}}{\mathrm{s}}

So the magnitude of the impulse is 2.1\;\mathrm{kg\; m}/\mathrm{s} directed outward from the wall.

§ Collisions in one Dimension

For the next few problems we may use the equations discussed in class and also presented in Problem 30. The scenario with one object initially at rest and the other in motion as depicted in the figure. We assume a perfectly elastic collision.

Figure 2.

An elastic collision in one dimension with one object initially at rest.

The figure shows m_{A} reversing direction, but depending upon the masses it could be moving to the right after the collision. In particular,

\displaystyle v^{\prime}_{A} \displaystyle= \displaystyle\left(\frac{m_{A}-m_{B}}{m_{A}+m_{B}}\right)v_{A}
\displaystyle v^{\prime}_{B} \displaystyle= \displaystyle\left(\frac{2m_{A}}{m_{A}+m_{B}}\right)v_{A}

§ Chapter 7, Problem 22

A ball of mass m_{A}=0.440\;\mathrm{kg} moving East (+x direction) with a speed of v_{A}=3.30\;\mathrm{m}/\mathrm{s} collides head-on with an m_{B}=0.220\;\mathrm{kg} at rest. If the collision is perfectly elastic, what will be the speed and direction of each ball after the collision?

\displaystyle v^{\prime}_{A} \displaystyle= \displaystyle\left(\frac{m_{A}-m_{B}}{m_{A}+m_{B}}\right)v_{A}
\displaystyle= \displaystyle\left(\frac{0.440\;\mathrm{kg}-0.220\;\mathrm{kg}}{0.440\;\mathrm{kg}+0.220\;\mathrm{kg}}\right)\left(3.30\frac{\mathrm{m}}{\mathrm{s}}\right)
\displaystyle= \displaystyle 1.10\frac{\mathrm{m}}{\mathrm{s}}
\displaystyle v^{\prime}_{B} \displaystyle= \displaystyle\left(\frac{2m_{A}}{m_{A}+m_{B}}\right)v_{A}
\displaystyle= \displaystyle\left(\frac{2\left(0.440\;\mathrm{kg}\right)}{0.440\;\mathrm{kg}+0.220\;\mathrm{kg}}\right)\left(3.30\frac{\mathrm{m}}{\mathrm{s}}\right)
\displaystyle= \displaystyle 4.40\frac{\mathrm{m}}{\mathrm{s}}

Since both velocities have a positive sign, both masses are moving East (+x direction).

§ Chapter 7, Problem 23

An m_{A}=0.450\;\mathrm{kg} ice puck moving East (+x direction) with a speed of v_{A}=3.00\;\mathrm{m}/\mathrm{s} collides head-on with an m_{B}=0.900\;\mathrm{kg} puck initially at rest. If the collision is perfectly elastic, what will be the speed and direction of each puck after the collision?

\displaystyle v^{\prime}_{A} \displaystyle= \displaystyle\left(\frac{m_{A}-m_{B}}{m_{A}+m_{B}}\right)v_{A}
\displaystyle= \displaystyle\left(\frac{0.450\;\mathrm{kg}-0.900\;\mathrm{kg}}{0.450\;\mathrm{kg}+0.900\;\mathrm{kg}}\right)\left(3.00\frac{\mathrm{m}}{\mathrm{s}}\right)
\displaystyle= \displaystyle-1.00\frac{\mathrm{m}}{\mathrm{s}}
\displaystyle v^{\prime}_{B} \displaystyle= \displaystyle\left(\frac{2m_{A}}{m_{A}+m_{B}}\right)v_{A}
\displaystyle= \displaystyle\left(\frac{2\left(0.450\;\mathrm{kg}\right)}{0.450\;\mathrm{kg}+0.900\;\mathrm{kg}}\right)\left(3.00\frac{\mathrm{m}}{\mathrm{s}}\right)
\displaystyle= \displaystyle 2.00\frac{\mathrm{m}}{\mathrm{s}}

§ Chapter 7, Problem 26

An m_{A}=0.220\;\mathrm{kg} softball moving with a speed of v_{A}=8.5\;\mathrm{m}/\mathrm{s} collides head-on and elastically with another ball initially at rest. Afterward the incoming softball bounces backward with a speed of
v^{\prime}_{A}=-3.7\;\mathrm{m}/\mathrm{s}, the negative sign to indicate backwards.

§ (a)

Calculate the velocity v^{\prime}_{B} of the target ball.

We have

\displaystyle v^{\prime}_{A} \displaystyle= \displaystyle\left(\frac{m_{A}-m_{B}}{m_{A}+m_{B}}\right)v_{A}
\displaystyle-3.7\frac{\mathrm{m}}{\mathrm{s}} \displaystyle= \displaystyle\left(\frac{0.220\;\mathrm{kg}-m_{B}}{0.220\;\mathrm{kg}+m_{B}}\right)\left(8.5\frac{\mathrm{m}}{\mathrm{s}}\right)


\displaystyle v^{\prime}_{B} \displaystyle= \displaystyle\left(\frac{2m_{A}}{m_{A}+m_{B}}\right)v_{A}
\displaystyle v^{\prime}_{B} \displaystyle= \displaystyle\left(\frac{2(0.220\;\mathrm{kg})}{0.220\;\mathrm{kg}+m_{B}}\right)\left(8.5\frac{\mathrm{m}}{\mathrm{s}}\right)

We have two equations in two unknowns. Notice that the first equation (for v^{\prime}_{A}) has only one unknown, m_{B}, so I will solve for that first.

\displaystyle-3.7\frac{\mathrm{m}}{\mathrm{s}} \displaystyle= \displaystyle\left(\frac{0.220\;\mathrm{kg}-m_{B}}{0.220\;\mathrm{kg}+m_{B}}\right)\left(8.5\frac{\mathrm{m}}{\mathrm{s}}\right)
\displaystyle-3.7\frac{\mathrm{m}}{\mathrm{s}}\left(0.220\;\mathrm{kg}+m_{B}\right) \displaystyle= \displaystyle\left(0.220\;\mathrm{kg}-m_{B}\right)\left(8.5\frac{\mathrm{m}}{\mathrm{s}}\right)
\displaystyle\left(8.5\frac{\mathrm{m}}{\mathrm{s}}-3.7\frac{\mathrm{m}}{\mathrm{s}}\right)m_{B} \displaystyle= \displaystyle\left(0.220\;\mathrm{kg}\right)\left(8.5\frac{\mathrm{m}}{\mathrm{s}}+3.7\frac{\mathrm{m}}{\mathrm{s}}\right)
\displaystyle m_{B} \displaystyle= \displaystyle 0.5\underline{5}9\;\mathrm{kg}

We may substitute this into the second equation.

\displaystyle v^{\prime}_{B} \displaystyle= \displaystyle\left(\frac{2(0.220\;\mathrm{kg})}{0.220\;\mathrm{kg}+0.5\underline{5}9\;\mathrm{kg}}\right)\left(8.5\frac{\mathrm{m}}{\mathrm{s}}\right)
\displaystyle= \displaystyle 4.8\frac{\mathrm{m}}{\mathrm{s}}

§ (b)

Calculate the mass m_{B} of the target ball.

From part (a):

\displaystyle m_{B} \displaystyle= \displaystyle 0.5\underline{5}9\;\mathrm{kg}
\displaystyle= \displaystyle 0.56\;\mathrm{kg}

§ Chapter 7, Problem 46

Figure 3.

Chapter 7, Problem 46.

Find the center of mass of the three mass system. Specify relative to the left-hand 1.00\;\mathrm{kg} mass.

\displaystyle x_{\mathrm{cm}} \displaystyle= \displaystyle\frac{\sum _{i}m_{i}x_{i}}{\sum _{i}m_{i}}
\displaystyle= \displaystyle\frac{(1.00\;\mathrm{kg})(0.00\;\mathrm{m})+(1.50\;\mathrm{kg})(0.50\;\mathrm{m})+(1.10\;\mathrm{kg})(0.75\;\mathrm{m})}{1.00\;\mathrm{kg}+1.50\;\mathrm{kg}+1.10\;\mathrm{kg}}
\displaystyle= \displaystyle 0.44\;\mathrm{m}

to the right of the left-hand 1.00\;\mathrm{kg} mass.

§ Chapter 7, Problem 47

Figure 4.

Chapter 7, Problem 47.

The distance between a carbon atom (m_{\mathrm{C}}=12\;\mathrm{u}) and an oxygen atom (m_{\mathrm{O}}=16\;\mathrm{u}) in the CO molecule is 1.13\times 10^{{-10}}\;\mathrm{m}. How far from the carbon atom is the center of mass of the molecule?

As in the figure we choose to place the origin coincident with the carbon atom.

\displaystyle x_{\mathrm{cm}} \displaystyle= \displaystyle\frac{\sum _{i}m_{i}x_{i}}{\sum _{i}m_{i}}
\displaystyle= \displaystyle\frac{(12\;\mathrm{u})\left(0.00\times 10^{{-10}}\;\mathrm{m}\right)+(16\;\mathrm{u})\left(1.13\times 10^{{-10}}\;\mathrm{m}\right)}{12\;\mathrm{u}+16\;\mathrm{u}}
\displaystyle= \displaystyle 0.65\times 10^{{-10}}\;\mathrm{m}

§ Chapter 7, Problem 58

A 55\;\mathrm{kg} woman and an 80\;\mathrm{kg} man stand 10.0\;\mathrm{m} apart on frictionless ice.

§ (a)

Figure 5.

Chapter 7, Problem 58(a).

How far from the woman is their center of mass (CM)?

Notice that I have chosen the coordinate system so the woman is initially standing at the origin.

\displaystyle x_{\mathrm{cm}} \displaystyle= \displaystyle\frac{\sum _{i}m_{i}x_{i}}{\sum _{i}m_{i}}
\displaystyle= \displaystyle\frac{(55\;\mathrm{kg})(0.0\;\mathrm{m})+(80\;\mathrm{kg})(10.0\;\mathrm{m})}{55\;\mathrm{kg}+80\;\mathrm{kg}}
\displaystyle= \displaystyle\underline{5}.93\;\mathrm{m}

While we were given the distance between the two to three significant figures, the man’s mass is only given to one and the woman’s mass to two significant figures. Because the woman is at the origin, the CM is \underline{5}.93\;\mathrm{m} from her.

§ (b)

Figure 6.

Chapter 7, Problem 58(b).

If each holds one end of a rope, and the man pulls on the rope so that he moves 2.5\;\mathrm{m}, how far from the woman will the man be now?

There will be tension in the rope only if he pulls on it and she pulls back. The tension can only draw them towards each other. (You cannot push away with a flexible rope.) If he has moved 2.5\;\mathrm{m}, then his new coordinate is at 7.5\;\mathrm{m} as indicated in the figure. We do not know her position, so let’s call it x. We can now calculate their CM again with their new positions, but it must be the same as the CM in part (a) since their are no external forces to change their CM.

\displaystyle x_{\mathrm{cm}} \displaystyle= \displaystyle\frac{\sum _{i}m_{i}x_{i}}{\sum _{i}m_{i}}
\displaystyle\underline{5}.93\;\mathrm{m} \displaystyle= \displaystyle\frac{(55\;\mathrm{kg})x+(80\;\mathrm{kg})(7.5\;\mathrm{m})}{55\;\mathrm{kg}+80\;\mathrm{kg}}
\displaystyle(55\;\mathrm{kg})x+(80\;\mathrm{kg})(7.5\;\mathrm{m}) \displaystyle= \displaystyle(\underline{5}.93\;\mathrm{m})(55\;\mathrm{kg}+80\;\mathrm{kg})
\displaystyle= \displaystyle 800\;\mathrm{kg\; m}
\displaystyle x \displaystyle= \displaystyle\frac{800\;\mathrm{kg\; m}-(80\;\mathrm{kg})(7.5\;\mathrm{m})}{(55\;\mathrm{kg})}
\displaystyle= \displaystyle\underline{3}.64\;\mathrm{m}

To obtain their separation use the distance formula for one dimension.

\displaystyle\mathrm{distance} \displaystyle= \displaystyle\left|\underline{3}.64\;\mathrm{m}-7.5\;\mathrm{m}\right|
\displaystyle= \displaystyle\underline{3}.9\;\mathrm{m}

§ (c)

Figure 7.

Chapter 7, Problem 58(c).

How far will the man have moved when he collides with the woman?

Now they are both at the same unknown position which we shall label as x. Again computer their center of mass and equate it with the original value since no external forces have acted upon the system.

\displaystyle x_{\mathrm{cm}} \displaystyle= \displaystyle\frac{\sum _{i}m_{i}x_{i}}{\sum _{i}m_{i}}
\displaystyle\underline{5}.93\;\mathrm{m} \displaystyle= \displaystyle\frac{(55\;\mathrm{kg})x+(80\;\mathrm{kg})x}{55\;\mathrm{kg}+80\;\mathrm{kg}}
\displaystyle= \displaystyle\frac{(55\;\mathrm{kg}+80\;\mathrm{kg})x}{55\;\mathrm{kg}+80\;\mathrm{kg}}
\displaystyle= \displaystyle x

In other words, they will be standing at their mutual center of mass. With a little experience a student of physics knows intuitively that this is the only possibility. To calculate the distance the man has traveled we again use the distance formula.

\displaystyle\mathrm{distance} \displaystyle= \displaystyle\left|\underline{5}.93\;\mathrm{m}-10.0\;\mathrm{m}\right|
\displaystyle= \displaystyle\underline{4}.1\;\mathrm{m}


Call Me Dr Rob logo