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# 7.2: Examples

Difficulty Level: At Grade Created by: CK-12

## Example 1

Question: Two blocks collide on a frictionless surface. Afterwards, they have a combined mass of 10kg and a speed of 2.5m/s\begin{align*}2.5\mathrm{m/s}\end{align*}. Before the collision, block A, which has a mass of 8.0kg, was at rest. What was the mass and initial speed of block B?

Solution: To find mass of block B we have a simple subtraction problem. We know that the combined mass is 10kg and the mass of block A is 8.0kg.

10kg8.0kg=2.0kg\begin{align*}10\mathrm{kg}-8.0\mathrm{kg}=2.0\mathrm{kg}\end{align*}

Now that we know the mass of both blocks we can find the speed of block B. We will use conservation of momentum. This was a completely inelastic collision. We know this because the blocks stuck together after the collision. This problem is one dimensional, because all motion happens along the same line. Thus we will use the equation

(mA+mB)vf=mA×vA+mB×vB\begin{align*}(m_A+m_B)v_f=m_A\times v_A+m_B\times v_B\end{align*}

and solve for the velocity of block B.

(mA+mB)vf=mA×vA+mBvB(mA+mB)(vf)(mA)(vA)mB=vB\begin{align*}(m_A+m_B)v_f=m_A\times v_A+m_Bv_B \Rightarrow \frac{(m_A+m_B)(v_f)-(m_A)(v_A)}{m_B}=v_B\end{align*}

Now we simply plug in what we know to solve for the velocity.

(2.0kg+8.0kg)(2.5m/s)(8.0kg)(0m/s)2.0kg=12.5m/s\begin{align*}\frac{(2.0\mathrm{kg}+8.0\mathrm{kg})(2.5\mathrm{m/s})-(8.0\mathrm{kg})(0\mathrm{m/s})}{2.0\mathrm{kg}}=12.5\mathrm{m/s}\end{align*}

## Example 2

Question: Chris and Ashley are playing pool on a frictionless table. Ashley hits the cue ball into the 8\begin{align*}8\end{align*} ball with a velocity of 1.2m/s\begin{align*}1.2\mathrm{m/s}\end{align*}. The cue ball (c)\begin{align*}(c)\end{align*} and the 8\begin{align*}8\end{align*} ball (e\begin{align*}e\end{align*}) react as shown in the diagram. The 8\begin{align*}8\end{align*} ball and the cue ball both have a mass of .17kg\begin{align*}.17\mathrm{kg}\end{align*}. What is the velocity of the cue ball? What is the direction (the angle) of the cue ball?

Answer: We know the equation for conservation of momentum, along with the masses of the objects in question as well two of the three velocities. Therefore all we need to do is manipulate the conservation of momentum equation so that it is solved for the velocity of the cue ball after the collision and then plug in the known values to get the velocity of the cue ball.

mcvic+mevie=mcvfc+mevfe\begin{align*}m_{c}v_{ic}+m_{e}v_{ie}=m_{c}v_{fc}+m_{e}v_{fe}\end{align*}

vfc=mcvic+meviemevfemc=.17kg×2.0m/s+.17kg×0m/s.17kg×1.2m/s.17kg=.80m/s\begin{align*}v_{fc}=\frac{m_{c}v_{ic}+m_{e}v_{ie}-m_{e}v_{fe}}{m_{c}}=\frac{.17\mathrm{kg}\times 2.0\mathrm{m/s}+.17\mathrm{kg}\times 0\mathrm{m/s}-.17\mathrm{kg}\times 1.2\mathrm{m/s}}{.17\mathrm{kg}}=.80\mathrm{m/s}\end{align*}

Now we want to find the direction of the cue ball. To do this we will use the diagram below.

We know that the momentum in the y\begin{align*}y\end{align*} direction of the two balls is equal. Therefore we can say that the velocity in the y\begin{align*}y\end{align*} direction is also equal because the masses of the two balls are equal.

mcvcy=meveyvcy=vey\begin{align*}m_cv_cy=m_ev_ey \rightarrow v_cy=v_ey\end{align*}

Given this and the diagram, we can find the direction of the cue ball. After 1 second, the 8\begin{align*}8\end{align*} ball will have traveled 1.2m\begin{align*}1.2\mathrm{m}\end{align*}. Therefore we can find the distance it has traveled in the y\begin{align*}y\end{align*} direction.

sin25o=oppositehypotenuse=x1.2mx=sin25×1.2m=.51m\begin{align*}\sin 25^o=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{x}{1.2\mathrm{m}} \rightarrow x=\sin 25\times 1.2\mathrm{m}=.51\mathrm{m}\end{align*}

Therefore, in one second the cue ball will have traveled .51m\begin{align*}.51\mathrm{m}\end{align*} in the y\begin{align*}y\end{align*} direction as well. We also know how far in total the cue ball travels in one second (.80m)\begin{align*}(.80\mathrm{m})\end{align*}. Thus we can find the direction of the cue ball.

sin1oppositehypotenuse=sin1.51m.80m=40o\begin{align*}sin^{-1}\frac{\text{opposite}}{\text{hypotenuse}}=sin^{-1}\frac{.51\mathrm{m}}{.80\mathrm{m}}=40^o\end{align*}

## Momentum Conservation Problem Set

1. You find yourself in the middle of a frozen lake. There is no friction between your feet and the ice of the lake. You need to get home for dinner. Which strategy will work best?
1. Press down harder with your shoes as you walk to shore.
2. Take off your jacket. Then, throw it in the direction opposite to the shore.
3. Wiggle your butt until you start to move in the direction of the shore.
4. Call for help from the great Greek god Poseidon.
2. You jump off of the top of your house and hope to land on a wooden deck below. Consider the following possible outcomes:
1. You hit the deck, but it isn’t wood! A camouflaged trampoline slows you down over a time period of 0.2\begin{align*}0.2\end{align*} seconds and sends you flying back up into the air.
2. You hit the deck with your knees locked in a straight-legged position. The collision time is 0.01\begin{align*}0.01\end{align*} seconds.
3. You hit the deck and bend your legs, lengthening the collision time to 0.2\begin{align*}0.2\end{align*} seconds.
4. You hit the deck, but it isn’t wood! It is simply a piece of paper painted to look like a deck. Below is an infinite void and you continue to fall, forever.
1. Which method will involve the greatest force acting on you?
2. Which method will involve the least force acting on you?
3. Which method will land you on the deck in the least pain?
4. Which method involves the least impulse delivered to you?
5. Which method involves the greatest impulse delivered to you?

3. You and your sister are riding skateboards side by side at the same speed. You are holding one end of a rope and she is holding the other. Assume there is no friction between the wheels and the ground. If your sister lets go of the rope, how does your speed change?
1. It stays the same.
2. It doubles.
3. It reduces by half.

4. You and your sister are riding skateboards (see Problem 3), but now she is riding behind you. You are holding one end of a meter stick and she is holding the other. At an agreed time, you push back on the stick hard enough to get her to stop. What happens to your speed? Choose one. (For the purposes of this problem pretend you and your sister weigh the same amount.)
1. It stays the same.
2. It doubles.
3. It reduces by half.
5. You punch the wall with your fist. Clearly your fist has momentum before it hits the wall. It is equally clear that after hitting the wall, your fist has no momentum. But momentum is always conserved! Explain.
6. An astronaut is using a drill to fix the gyroscopes on the Hubble telescope. Suddenly, she loses her footing and floats away from the telescope. What should she do to save herself?
7. You look up one morning and see that a 30kg\begin{align*}30 \;\mathrm{kg}\end{align*} chunk of asbestos from your ceiling is falling on you! Would you be better off if the chunk hit you and stuck to your forehead, or if it hit you and bounced upward? Explain your answer.
8. A 5.00kg\begin{align*}5.00\;\mathrm{kg}\end{align*} firecracker explodes into two parts: one part has a mass of 3.00kg\begin{align*}3.00 \;\mathrm{kg}\end{align*} and moves at a velocity of 25.0m/s\begin{align*}25.0 \;\mathrm{m/s}\end{align*} towards the west. The other part has a mass of 2.00kg\begin{align*}2.00 \;\mathrm{kg}\end{align*}. What is the velocity of the second piece as a result of the explosion?
9. A firecracker lying on the ground explodes, breaking into two pieces. One piece has twice the mass of the other. What is the ratio of their speeds?
10. You throw your 6.0kg\begin{align*}6.0\;\mathrm{kg}\end{align*} skateboard down the street, giving it a speed of 4.0m/s\begin{align*}4.0 \;\mathrm{m/s}\end{align*}. Your friend, the Frog, jumps on your skateboard from rest as it passes by. Frog has a mass of 60kg\begin{align*}60 \;\mathrm{kg}\end{align*}.
1. What is the momentum of the skateboard before Frog jumps on it?
2. Find Frog’s speed after he jumps on the skateboard.
3. What impulse did Frog deliver to the skateboard?
4. If the impulse was delivered over 0.2\begin{align*}0.2\end{align*} seconds, what was the average force imparted to the skateboard?
5. What was the average force imparted to the Frog? Explain.

11. Two blocks collide on a frictionless surface, as shown. Afterwards, they have a combined mass of 10kg\begin{align*}10 \;\mathrm{kg}\end{align*} and a speed of 2.5m/s\begin{align*}2.5 \;\mathrm{m/s}\end{align*}. Before the collision, one of the blocks was at rest. This block had a mass of 8.0kg\begin{align*}8.0 \;\mathrm{kg}\end{align*}. What was the mass and initial speed of the second block?
12. While driving in your pickup truck down Highway 280\begin{align*}280\end{align*} between San Francisco and Palo Alto, an asteroid lands in your truck bed! Despite its 220kg\begin{align*}220 \;\mathrm{kg}\end{align*} mass, the asteroid does not destroy your 1200kg\begin{align*}1200\;\mathrm{kg}\end{align*} truck. In fact, it landed perfectly vertically. Before the asteroid hit, you were going 25m/s\begin{align*}25 \;\mathrm{m/s}\end{align*}. After it hit, how fast were you going?
13. A baseball player faces a 80.0m/s\begin{align*}80.0 \;\mathrm{m/s}\end{align*} pitch. In a matter of .020\begin{align*}.020\end{align*} seconds he swings the bat, hitting a 50.0m/s\begin{align*}50.0\;\mathrm{m/s}\end{align*} line drive back at the pitcher. Calculate the force on the bat while in contact with the ball.
14. An astronaut is 100m\begin{align*}100\;\mathrm{m}\end{align*} away from her spaceship doing repairs with a 10.0kg\begin{align*}10.0 \;\mathrm{kg}\end{align*} wrench. The astronaut’s total mass is 90.0kg\begin{align*}90.0 \;\mathrm{kg}\end{align*} and the ship has a mass of \begin{align*}1.00 \times10^4\;\mathrm{kg}\end{align*}. If she throws the wrench in the opposite direction of the spaceship at \begin{align*}10.0 \;\mathrm{m/s}\end{align*} how long would it take for her to reach the ship?
15. A place kicker applies an average force of \begin{align*} 2400 \;\mathrm{N}\end{align*} to a football of \begin{align*}.040 \;\mathrm{kg}\end{align*}. The force is applied at an angle of \begin{align*}20.0\end{align*} degrees from the horizontal. Contact time is .\begin{align*}010\end{align*} sec.
1. Find the velocity of the ball upon leaving the foot.
2. Assuming no air resistance find the time to reach the goal posts \begin{align*}40.0 \;\mathrm{m}\end{align*} away.
3. The posts are \begin{align*}4.00 \;\mathrm{m}\end{align*} high. Is the kick good? By how much?

16. In the above picture, the carts are moving on a level, frictionless track. After the collision all three carts stick together. Determine the direction and speed of the combined carts after the collision.

1. Your author’s Italian cousin crashed into a tree. He was originally going \begin{align*}36 \;\mathrm{km/hr}\end{align*}. Assume it took \begin{align*}0.40\end{align*} seconds for the tree to bring him to a stop. The mass of the cousin and the car is \begin{align*}450\;\mathrm{kg}\end{align*}.
1. What average force did he experience? Include a direction in your answer.
2. What average force did the tree experience? Include a direction in your answer.
3. Express this force in pounds.
4. How many g’s of acceleration did he experience?
2. The train engine and its four boxcars are coasting at \begin{align*}40 \;\mathrm{m/s}\end{align*}. The engine train has mass of \begin{align*}5,500 \;\mathrm{kg}\end{align*} and the boxcars have masses, from left to right, of \begin{align*}1,000 \;\mathrm{kg}\end{align*}, \begin{align*}1,500 \;\mathrm{kg}\end{align*}, \begin{align*}2,000 \;\mathrm{kg},\end{align*} and \begin{align*}3,000 \;\mathrm{kg}\end{align*}. (For this problem, you may neglect the small external forces of friction and air resistance.)
1. What happens to the speed of the train when it releases the last boxcar? (Hint: Think before you blindly calculate.)
2. If the train can shoot boxcars backwards at \begin{align*}30 \;\mathrm{m/s}\end{align*} relative to the train’s speed, how many boxcars does the train need to shoot out in order to obtain a speed of \begin{align*}58.75 \;\mathrm{m/s}\end{align*}?
3. Serena Williams volleys a tennis ball hit to her at \begin{align*}30\;\mathrm{m/s}\end{align*}. She blasts it back to the other court at \begin{align*}50 \;\mathrm{m/s}\end{align*}. A standard tennis ball has mass of \begin{align*}0.057 \;\mathrm{kg}\end{align*}. If Serena applied an average force of \begin{align*}500\;\mathrm{N}\end{align*} to the ball while it was in contact with the racket, how long did the contact last?
4. Zoran’s spacecraft, with mass \begin{align*}12,000 \;\mathrm{kg}\end{align*}, is traveling to space. The structure and capsule of the craft have a mass of \begin{align*}2,000 \;\mathrm{kg}\end{align*}; the rest is fuel. The rocket shoots out \begin{align*}0.10 \;\mathrm{kg/s}\end{align*} of fuel particles with a velocity of \begin{align*}700 \;\mathrm{m/s}\end{align*} with respect to the craft.
1. What is the acceleration of the rocket in the first second?
2. What is the average acceleration of the rocket after the first ten minutes have passed?

5. In Sacramento a \begin{align*}4000 \;\mathrm{kg}\end{align*} SUV is traveling \begin{align*}30 \;\mathrm{m/s}\end{align*} south on Truxel crashes into an empty school bus, \begin{align*}7000 \;\mathrm{kg}\end{align*} traveling east on San Juan. The collision is perfectly inelastic.
1. Find the velocity of the wreck just after collision
2. Find the direction in which the wreck initially moves
6. A \begin{align*}3 \;\mathrm{kg}\end{align*} ball is moving \begin{align*}2 \;\mathrm{m/s}\end{align*} in the positive \begin{align*}x-\end{align*}direction when it is struck dead center by a \begin{align*}2 \;\mathrm{kg}\end{align*} ball moving in the positive \begin{align*}y-\end{align*}direction. After collision the \begin{align*}3\;\mathrm{kg}\end{align*} ball moves at \begin{align*}1 \;\mathrm{m/s}\end{align*} \begin{align*}30\end{align*} degrees from the positive \begin{align*}x-\end{align*}axis. Once you have filled out the Table below, solve the following problems:
1. Find the velocity and direction of the \begin{align*}2 \;\mathrm{kg}\end{align*} ball.
2. Use the table to prove momentum is conserved.
3. Prove that kinetic energy is not conserved.
To 2-significant digit accuracy fill out the table.
\begin{align*}3 \;\mathrm{kg}\end{align*} ball \begin{align*}p_x\end{align*} \begin{align*}3 \;\mathrm{kg}\end{align*} ball \begin{align*}p_y\end{align*} \begin{align*}2 \;\mathrm{kg}\end{align*} ball \begin{align*}p_x\end{align*} \begin{align*}2\;\mathrm{kg}\end{align*} ball \begin{align*}p_y\end{align*}
Momentum before
Momentum after collision
1. Students are doing an experiment on the lab table. A steel ball is rolled down a small ramp and allowed to hit the floor. Its impact point is carefully marked. Next a second ball of the same mass is put upon a set screw and a collision takes place such that both balls go off at an angle and hit the floor. All measurements are taken with a meter stick on the floor with a co-ordinate system such that just below the impact point is the origin. The following data is collected:
1. no collision: \begin{align*}41.2 \;\mathrm{cm}\end{align*}
2. target ball: \begin{align*}37.3 \;\mathrm{cm}\end{align*} in the direction of motion and \begin{align*}14.1 \;\mathrm{cm}\end{align*} perpendicular to the direction of motion
1. From this data predict the impact position of the other ball.
2. One of the lab groups declares that the data on the floor alone demonstrate to a \begin{align*}2\end{align*}% accuracy that the collision was elastic. Show their reasoning.
3. Another lab group says they can’t make that determination without knowing the velocity the balls have on impact. They ask for a timer. The instructor says you don’t need one; use your meter stick. Explain.
4. Design an experiment to prove momentum conservation with balls of different masses, giving apparatus, procedure and design. Give some sample numbers.

1. .
2. .
3. .
4. .
5. .
6. .
7. .
8. \begin{align*}37.5 \;\mathrm{m/s}\end{align*}
9. \begin{align*}v_1 = 2v_2\end{align*}
1. \begin{align*}24 \frac{kg-m}{5}\end{align*}
2. \begin{align*}0.364 \;\mathrm{m/s}\end{align*}
3. \begin{align*}22 \frac{kg-m}{5}\end{align*}
4. \begin{align*}109 \;\mathrm{N}\end{align*}
5. \begin{align*}109 \;\mathrm{N}\end{align*} due to Newton’s third law
10. \begin{align*}2.0 \;\mathrm{kg}, 125 \;\mathrm{m/s}\end{align*}
11. \begin{align*}21 \;\mathrm{m/s}\end{align*} to the left
12. \begin{align*}3250 \;\mathrm{N}\end{align*}
1. \begin{align*}90 \;\mathrm{sec}\end{align*}
2. \begin{align*}1.7 \times 10^5 \;\mathrm{sec}\end{align*}
1. \begin{align*}60 \;\mathrm{m/s}\end{align*}
2. \begin{align*}.700 \;\mathrm{sec}\end{align*}
3. yes, \begin{align*}8.16 \;\mathrm{m}\end{align*}
13. \begin{align*}0.13 \;\mathrm{m/s}\end{align*} to the left
1. \begin{align*}11000 \;\mathrm{N}\end{align*} to the left
2. tree experienced same average force of \begin{align*}11000 \;\mathrm{N}\end{align*} but to the right
3. \begin{align*}2500 \;\mathrm{lb}\end{align*}.
4. about \begin{align*}2.5\end{align*} “g”s of acceleration
1. no change
2. the last two cars
1. \begin{align*}0.00912 \;\mathrm{s}\end{align*}
1. \begin{align*}0.0058 \;\mathrm{m/s}^2\end{align*}
2. \begin{align*}3.5 \;\mathrm{m/s}^2\end{align*}
1. \begin{align*} 15 \;\mathrm{m/s}\end{align*}
2. \begin{align*}49^\circ \;\mathrm{S}\end{align*} of \begin{align*}\;\mathrm{E}\end{align*}
14. b. \begin{align*}4.6 \;\mathrm{m/s} \ 68^\circ\end{align*}

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