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Two billiard balls are at rest and touching each other on a pool table. The cue ball travels at 3.8 m/s along the line of symmetry between these balls and strikes them simultaneously. If the collision is elastic, what is the velocity of the three balls after the collision?
final velocity of cue ball is $\text{\u2212}\left(0.76\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\widehat{i}$ , final velocities of the other two balls are 2.6 m/s at ±30° with respect to the initial velocity of the cue ball
A billiard ball traveling at $\left(2.2\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\widehat{i}-\left(0.4\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\widehat{j}$ collides with a wall that is aligned in the $\widehat{j}$ direction. Assuming the collision is elastic, what is the final velocity of the ball?
Two identical billiard balls collide. The first one is initially traveling at $\left(2.2\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\widehat{i}-\left(0.4\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\widehat{j}$ and the second one at $\text{\u2212}\left(1.4\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\widehat{i}+\left(2.4\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\widehat{j}$ . Suppose they collide when the center of ball 1 is at the origin and the center of ball 2 is at the point $\left(2R,0\right)$ where R is the radius of the balls. What is the final velocity of each ball?
ball 1: $\text{\u2212}\left(1.4\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\widehat{i}-\left(0.4\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\widehat{j}$ , ball 2: $\left(2.2\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\widehat{i}+\left(2.4\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\widehat{j}$
Repeat the preceding problem if the balls collide when the center of ball 1 is at the origin and the center of ball 2 is at the point $\left(0,2R\right)$ .
Repeat the preceding problem if the balls collide when the center of ball 1 is at the origin and the center of ball 2 is at the point $\left(\sqrt{3}R\text{/}2,R\text{/}2\right)$
ball 1: $\left(1.4\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\widehat{i}-\left(1.7\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\widehat{j}$ , ball 2: $\text{\u2212}\left(2.8\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\widehat{i}+\left(0.012\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\widehat{j}$
Where is the center of mass of a semicircular wire of radius R that is centered on the origin, begins and ends on the x axis, and lies in the x , y plane?
Where is the center of mass of a slice of pizza that was cut into eight equal slices? Assume the origin is at the apex of the slice and measure angles with respect to an edge of the slice. The radius of the pizza is R .
$\left(r,\theta \right)=\left(2R\text{/}3,\pi \text{/}8\right)$
If the entire population of Earth were transferred to the Moon, how far would the center of mass of the Earth-Moon-population system move? Assume the population is 7 billion, the average human has a mass of 65 kg, and that the population is evenly distributed over both the Earth and the Moon. The mass of the Earth is $5.97\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{24}\text{kg}$ and that of the Moon is $7.34\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{22}\text{kg}$ . The radius of the Moon’s orbit is about $3.84\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{5}\text{m}$ .
You friend wonders how a rocket continues to climb into the sky once it is sufficiently high above the surface of Earth so that its expelled gasses no longer push on the surface. How do you respond?
Answers may vary. The rocket is propelled forward not by the gasses pushing against the surface of Earth, but by conservation of momentum. The momentum of the gas being expelled out the back of the rocket must be compensated by an increase in the forward momentum of the rocket.
To increase the acceleration of a rocket, should you throw rocks out of the front window of the rocket or out of the back window?
A 65-kg person jumps from the first floor window of a burning building and lands almost vertically on the ground with a horizontal velocity of 3 m/s and vertical velocity of $\mathrm{-9}\phantom{\rule{0.2em}{0ex}}\text{m/s}$ . Upon impact with the ground he is brought to rest in a short time. The force experienced by his feet depends on whether he keeps his knees stiff or bends them. Find the force on his feet in each case.
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.
a. $617\phantom{\rule{0.2em}{0ex}}\text{N}\xb7\text{s}$ , 108°; b. ${F}_{x}=2.91\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{N}$ , ${F}_{y}=2.6\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{N}$ ; c. ${F}_{x}=5265\phantom{\rule{0.2em}{0ex}}\text{N}$ , ${F}_{y}=5850\phantom{\rule{0.2em}{0ex}}\text{N}$
Two projectiles of mass ${m}_{1}$ and ${m}_{2}$ are ﬁred at the same speed but in opposite directions from two launch sites separated by a distance D . They 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.
Two identical objects (such as billiard balls) have a one-dimensional collision in which one is initially motionless. After the collision, the moving object is stationary and the other moves with the same speed as the other originally had. Show that both momentum and kinetic energy are conserved.
Conservation of momentum demands ${m}_{1}{v}_{\text{1,i}}+{m}_{2}{v}_{\text{2,i}}={m}_{1}{v}_{\text{1,f}}+{m}_{2}{v}_{\text{2,f}}$ . We are given that ${m}_{1}={m}_{2}$ , ${v}_{\text{1,i}}={v}_{\text{2,f}}$ , and ${v}_{\text{2,i}}={v}_{\text{1,f}}=0$ . Combining these equations with the equation given by conservation of momentum gives ${v}_{\text{1,i}}={v}_{\text{1,i}}$ , which is true, so conservation of momentum is satisfied. Conservation of energy demands ${\scriptscriptstyle \frac{1}{2}}{m}_{1}{v}_{\text{1,i}}^{2}+{\scriptscriptstyle \frac{1}{2}}{m}_{2}{v}_{\text{2,i}}^{2}={\scriptscriptstyle \frac{1}{2}}{m}_{1}{v}_{\text{1,f}}^{2}+{\scriptscriptstyle \frac{1}{2}}{m}_{2}{v}_{\text{2,f}}^{2}$ . Again combining this equation with the conditions given above give ${v}_{\text{1,i}}={v}_{\text{1,i}}$ , so conservation of energy is satisfied.
A ramp of mass M is at rest on a horizontal surface. A small cart of mass m is placed at the top of the ramp and released.
What are the velocities of the ramp and the cart relative to the ground at the instant the cart leaves the ramp?
Find the center of mass of the structure given in the figure below. Assume a uniform thickness of 20 cm, and a uniform density of $1{\phantom{\rule{0.2em}{0ex}}\text{g/cm}}^{3}.$
Assume origin on centerline and at floor, then $\left({x}_{\text{CM}},{y}_{\text{CM}}\right)=\left(\mathrm{0,86}\phantom{\rule{0.2em}{0ex}}\text{cm}\right)$
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