# Units 4-5 Review

Momentum and Energy

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29. A 4 kg ball has a momentum of 12 kg*m/s. The ball's speed is ___ m/s.
 a. 3 b. 4 c. 12 d. 48 e. none of these.

This is a relatively simple plug-and-chug into the equation p=m*v with m=4 kg and p=12 kg*m/s.

Momentum (7 seconds)

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30. A piece of putty moving with 1 unit of momentum strikes and sticks to a heavy bowling ball that is initially at rest. After the putty sticks to the ball, both are set in motion with a combined momentum that is ___.
 a. less than 1 unit b. more than 1 unit c. 1 unit d. not enough information

Since momentum must be conserved, the total momentum of the ball and putty must be 1 unit. (Before the collision, the total system momentum is also 1 unit - all due to the motion of the putty.)

Momentum Conservation Principle (16 seconds)

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31. A 2 kg mass has a velocity of 4 m/s. The kinetic energy of the mass is ___ Joules.
 a. 4 b. 8 c. 16 d. 32 e. none of these

This is a relatively simple plug-and-chug into the equation KE=0.5*m*v2 with m=2 kg and v=4 m/s.

Kinetic Energy (4 seconds)

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32. A car moving at 50 km/hr skids 20 meters with locked brakes. How far will the car skid with locked brakes if it is traveling at 150 km/hr?
 a. 20 m b. 60 m c. 90 m d. 120 m e. 180 m

Recall the reasoning from the stopping distance-speed lab (with the Hot Wheels car and the computer box) that the stopping distance is related to the square of the initial speed. So any modification in the initial speed of a skidding car will lead to a square of that same modification in the stopping distance. A change in speed from 50 km/hr to 150 km/hr is a 3-fold increase. The distance in turn must change by nine-fold (3^2). So take the original stopping distance of 20 m and multiply it by 9.

Analysis of Situations Involving External Forces (21 seconds)

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33. A 50 kg diver hits the water below (at a zero height) with a kinetic energy of 5000 Joules. The height from which the diver dove was ____ meters.
 a. 5 b. 10 c. 50 d. 100

The kinetic energy of the diver must be equal to the original potential energy. Thus,

m*g*hi = KEf

(50 kg)*(10 m/s/s)*h = 5000 J

So, h = 10 m

Mechanical Energy Conservation (13 seconds)

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34. A large force acting for a long amount of time on a small mass will produce a ______.
 a. small velocity change b. large velocity change c. small momentum change d. small acceleration e. two of the above

A large force on a small mass will result in a large acceleration (a=F/m) and subsequently a large velocity change (Delta v = a*t). This rules out choices A and D. A large force and for a long time will result in a large impulse and therefore a large momentum change. This rules out choice C.

Momentum and Impulse Connection (14 seconds) | Real-World Applications (15 seconds)

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35. Force and time pertains to momentum change in the same manner as force and distance pertains to ___________.
 a. impulse b. work c. energy change d. velocity e. none of these.

A force multiplied by a time gives an impulse which will cause (and be equal to) a momentum change. In the same manner, a force multiplied by a distance gives work which will cause (and be equal to) an energy change. Re-read those two sentences because it relates two big concepts.

Momentum and Impulse Connection (14 seconds) | Analysis of Situations Involving External Forces (21 seconds)

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36. A job is done slowly, and an identical job is done quickly. Both jobs require the same amount of work, but different amounts of ___________.
 a. energy b. power c. both of these d. none of these

Power refers to the rate at which work is done. Thus, doing two jobs - one slowly and one quickly - involves doing the same job (i.e., the same work) at different rates or with different power.

Power (13 seconds)

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37. Which requires more work: lifting a 50 kg sack vertically 2 meters or lifting a 25 kg sack vertically 4 meters?
 a. lifting the 50 kg sack b. lifting the 25 kg sack c. both require the same amount of work

Work involves a force acting upon an object to cause a displacement. The amount of work done is found by multiplying F*d*cos(Theta). The equation can be used for these two motions to find the work.

 Lifting a 50 kg sack vertically 2 meters W = (500 N)*(2 m)*cos(0) W = 1000 N (Note: The weight of a 50-kg object is approx. 500 N; it takes 500 N to lift the object up.) Lifting a 25 kg sack vertically 4 meters W = (250 N)*(4 m)*cos(0) W = 1000 N (Note: The weight of a 50-kg object is approx. 250 N; it takes 250 N to lift the object up.)

Definition and Mathematics of Work (10 seconds) | Calculating the Amount of Work Done by Forces (10 seconds)

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38. A 50 kg sack is lifted 2 meters in the same time as a 25 kg sack is lifted 4 meters. The power expended in raising the 50 kg sack compared to the power used to lift the 25 kg sack is _________.
 a. twice as much b. half as much c. the same

The power is the rate at which work is done. Power is found by dividing work by time. It requires the same amount of work to do these two jobs (see question #37) and the same amount of time. Thus, the power is the same for both tasks.

Power (13 seconds)

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39. A TV set is pushed a distance of 2 m with a force of 20 N that is in the same direction as the set moves. How much work is done on the set?
 a. 2 J b. 10 J c. 20 J d. 40 J e. 80 J

This is a relatively simple plug-and-chug into the equation W=F*d*cos(Theta) with F=20 N and d=2 m and Theta = 0 degrees.

Calculating the Amount of Work Done by Forces (10 seconds)

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40. It takes 40 J to push a large box 4 m across a floor. Assuming the push is in the same direction as the move, what is the magnitude of the force on the box?
 a. 4 N b. 10 N c. 40 N d. 160 N e. none of these

This is a relatively simple plug-and-chug into the equation W=F*d*cos(Theta) with W=40 J and d=4 m and Theta = 0 degrees.

Calculating the Amount of Work Done by Forces (10 seconds)

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41. Using 1000 J of work, a toy elevator is raised from the ground floor to the second floor in 20 seconds. How much power does the elevator use?
 a. 20 W b. 50 W c. 100 W d. 1000 W e. 20000 W

This is a relatively simple plug-and-chug into the equation P=W/t with W=1000 J and t=20 s.

Calculating the Amount of Work Done by Forces (10 seconds) | Power (13 seconds)

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42. A 5-N force is applied to a 3-kg object to change its velocity from +9 m/s to +3 m/s. The momentum change of the object is:
 a. -2.5 kg*m/s b. -10 kg*m/s c. -18 kg*m/s d. -45 kg*m/s e. none of these

Don't make this harder than it is; the momentum change of an object can be found if the mass and the velocity change are known. In this equation, m=3 kg and the velocity change is -6 m/s. When finding the velocity change, always subtract the initial velocity from the final velocity (vf - vi). The momentum change can also be found if the force and the time are known. Multiplying force*time yields the impulse and the impulse equals the momentum change.

Momentum and Impulse Connection (14 seconds)

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