# Units 4-5 Review

Momentum and Energy

 Navigate to Answers for: [ Questions #1-#14 | Questions #15-#28 | Questions #29-#42 | Questions #43-#56 | Questions #57-#66 ] [ #43 | #44 | #45 | #46 | #47 | #48 | #49 | #50 | #51 | #52 | #53 | #54 | #55 | #56 ]

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

### Answer and Explanation:

Impulse is defined as a force acting upon and object for a given amount of time. Impulse can be computed by multiplying force*time. But in this problem, the time is not known. Never fear - the impulse equals the momentum change. The momentum change in this problem is -18 kg*m/s (see question #42). Thus, the impulse is -18 N*s.

Momentum and Impulse Connection (14 seconds)

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44. A 5-N force is applied to a 3-kg object to change its velocity from +9 m/s to +3 m/s. The impulse acts for a time period of
 a. 1.8 s b. 2.5 s c. 3.6 s d. 10 s e. none of these

### Answer and Explanation:

Use the impulse momentum change theorem with F=5 N, m=3 kg and Delta v=-6 m/s. Solving for time involves the following steps.

t = m*(delta v)/F = (3 kg)*(-6 m/s)/(5 N) = 3.6 s

Momentum and Impulse Connection (14 seconds)

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45. When a mass M experiences a velocity change of v in a time of t, it experiences a force of F. Assuming the same velocity change of v, the force experienced by a mass of 2M in a time of (1/2)t is
 a. 2F b. 4F c. (1/2)*F d. (1/4)*F e. none of these

### Answer and Explanation:

The impulse-momentum change theorem states that F*t = m*(Delta vel.). This equation can be re-arranged to locate the F by itself on one side of the equation; rearranging yields

F = m*(Delta vel.)/t

The equation shows that force is directly related to the mass, directly related to the change in velocity, and inversely related to the time. So any change in mass will result in the same change in force; and any change in time will result in the inverse effect upon the force. In this case, doubling the mass (from M to 2M) will double the force and halving the time (from t to 1/2-t) will double the force. The combined effect of these two changes will make the new force four times bigger than the old force. This is a case of where equations can be a guide to thinking about how a change in one variable (or two variables) impacts other dependent variables.

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

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46. When a mass M experiences a velocity change of v in a time of t, it experiences a force of F. Assuming the same velocity change of v, the force experienced by a mass of 2M in a time of (1/4)t is
 a. 2F b. 8F c. (1/2)*F d. (1/8)*F e. none of these

### Answer and Explanation:

The impulse-momentum change theorem states that F*t = m*(Delta vel.). This equation can be re-arranged to locate the F by itself on one side of the equation; rearranging yields

F = m*(Delta vel.)/t

The equation shows that force is directly related to the mass, directly related to the change in velocity, and inversely related to the time. So any change in mass will result in the same change in force; and any change in time will result in the inverse effect upon the force. In this case, doubling the mass (from M to 2M) will double the force and quartering the time (from t to 1/4-t) will quadruple the force. The combined effect of these two changes will make the new force eight times bigger than the old force. This is a case of where equations can be a guide to thinking about how a change in one variable (or two variables) impacts other dependent variables.

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

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47. When a mass M experiences a velocity change of v in a time of t, it experiences a force of F. Assuming the same velocity change of v, the force experienced by a mass of (1/2)M in a time of (1/2)t is
 a. 2F b. 4F c. (1/2)*F d. (1/4)*F e. none of these

### Answer and Explanation:

The impulse-momentum change theorem states that F*t = m*(Delta vel.). This equation can be re-arranged to locate the F by itself on one side of the equation; rearranging yields

F = m*(Delta vel.)/t

The equation shows that force is directly related to the mass, directly related to the change in velocity, and inversely related to the time. So any change in mass will result in the same change in force; and any change in time will result in the inverse effect upon the force. In this case, halving the mass (from M to 1/2-M) will half the force and halving the time (from t to 1/2-t) will double the force. The combined effect of these two changes will make the new force the same size as the old force. This is a case of where equations can be a guide to thinking about how a change in one variable (or two variables) impacts other dependent variables.

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

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48. When a mass M experiences a velocity change of v in a time of t, it experiences a force of F. Assuming the same velocity change of v, the force experienced by a mass of (1/2)M in a time of 4t is
 a. 2F b. 8F c. (1/2)*F d. (1/8)*F e. none of these

### Answer and Explanation:

The impulse-momentum change theorem states that F*t = m*(Delta vel.). This equation can be re-arranged to locate the F by itself on one side of the equation; rearranging yields

F = m*(Delta vel.)/t

The equation shows that force is directly related to the mass, directly related to the change in velocity, and inversely related to the time. So any change in mass will result in the same change in force; and any change in time will result in the inverse effect upon the force. In this case, halving the mass (from M to 1/2-M) will halve the force and quadrupling the time (from t to 4t) will quarter the force. The combined effect of these two changes will make the new force eight times smaller (i.e., one-eighth the size) than the old force. This is a case of where equations can be a guide to thinking about how a change in one variable (or two variables) impacts other dependent variables.

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

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49. A 0.5-kg ball moving at 5 m/s strikes a wall and rebounds in the opposite direction with a speed of 2 m/s. If the impusle occurs for a time duration of 0.01 s, then the average force (magnitude only) acting upon the ball is
 a. 0.14 N b. 150 N c. 350 N d. 500 N e. none of these

### Answer and Explanation:

This is a relatively simple plug-and-chug into the equation

F*t = m*(Delta vel.)

with m=0.5 kg, t=0.01 s and Delta vel.=-7 m/s. (The change in velocity is -7 m/s since the ball must first slow down from 5 m/s to 0 m/s and then be thrown back in the opposite direction at 2 m/s.) Using these numbers and solving for force yields -350 N. The magnitude of the force is 350 N and the "-" sign indicates the direction of the force.

Momentum and Impulse Connection (14 seconds)

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50. If mass and collision time are equal, then impulses are greater on objects which rebound (or bounce).
 a. TRUE b. FALSE

### Answer and Explanation:

The impulse is equal to the momentum change. And when there is a rebound, the momentum change is larger since there is a larger velocity change. For instance, a ball thrown at a wall at 5 m/s may rebound at -3 m/s yielding a velocity change of -8 m/s. An egg thrown at the same wall at the same speed of 5 m/s hits and stops, thus yielding a velocity change of -5 m/s. More velocity change means more momentum change and thus more impulse.

Effect of Rbounding (15 seconds)

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51. An unfortunate bug strikes the windshield of a bus in a head-on collision. Which of the following statements are true?

1. The magnitude of the force encountered by the bug is greater than that of the bus.
2. The magnitude of the impulse encountered by the bug is greater than that of the bus.
3. The magnitude of the momentum change encountered by the bug is greater than that of the bus.
4. The magnitude of the velocity change encountered by the bug is greater than that of the bus.
5. The magnitude of the acceleration encountered by the bug is greater than that of the bus.

### Answer and Explanation:

Answer: The true statements are D and E.

In any collision between two objects, the force, impulse, and momentum change are the same for each object. (This makes statements A, B, and C false.) However, the smaller mass object encounters a greater acceleration and velocity change. (This makes statements D and E true).

The Law of Action-Reaction (Revisited) (8 seconds)

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### Problem-Solving:

52. A 0.80-kg ball strikes a wall moving at 5.0 m/s and rebounds in the opposite direction at 3.5 m/s. If the collision with the wall endures for a total time of 0.0080 s, then determine the average force acting upon the ball. PSYW

### Answer and Explanation:

 Given: m=0.80 kg; vi = 5.0 m/s; vf = -3.5 m/s; and t=0.0080 s Find: F Useful Equation: F*t = m*(Delta v) Rearranging the equation to solve for F yields F = m*(Delta v)/t Substituting yields F = (0.80 kg)*(-8.5 m/s)/(0.0080 s) F = -850 N (NOTE: The velocity change is always found by from vf-vi.

Momentum and Impulse Connection (14 seconds)

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53. A 16.0-kg ball is thrown with a speed of 22.0 m/s to a 55-kg clown who is at rest on ice. The clown catches the ball and glides across the ice. Determine the velocity of the clown (and ball) immediately following the catch. PSYW

### Answer and Explanation:

Momentum conservation principles must be used to solve this collision problem. Set the initial momentum of the system equal to the final momentum of the system and solve for the unknown velocity. The information can be organized inside of a momentum table as shown below. (NOTE: units have been left off quantities to avoid "messiness".)

0
p=(55 kg)*v

p= 55*v

### Ball

p=(16 kg)*(22.0 m/s)

p= 352

p=(16 kg)*v

p= 16*v

### Total

352
55*v + 16*v

352 = 55*v + 16*v

352 = 71*v

v = (352/71) = 4.96 m/s

Using Equations as a "Recipe" for Algebraic Problem-Solving (12 seconds)

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54. A 16.0-kg ball is thrown with a speed of 22.0 m/s to a 55-kg clown on ice. At the time that the clown catches the ball, she is moving with a speed of 3.0 m/s in the same direction as the ball. The clown catches the ball and continues to glide across the ice. Determine the velocity of the clown (and ball) immediately following the catch. PSYW

### Answer and Explanation:

Momentum conservation principles must be used to solve this collision problem. Set the initial momentum of the system equal to the final momentum of the system and solve for the unknown velocity. The information can be organized inside of a momentum table as shown below. (NOTE: units have been left off quantities to avoid "messiness".)

### Clown

p=(55 kg)*(5.0 m/s)

p= 165

p=(55 kg)*v

p= 55*v

### Ball

p=(16 kg)*(22.0 m/s)

p= 352

p=(16 kg)*v

p= 16*v

### Total

517
55*v + 16*v

517 = 55*v + 16*v

517 = 71*v

v = (517/71) = 7.28 m/s

Using Equations as a "Recipe" for Algebraic Problem-Solving (12 seconds)

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55. A 0.050-kg billiard ball moving at 1.2 m/s strikes a second 0.050-kg billiard ball which is moving in the same direction with a speed of 0.40 m/s. If the faster ball slows down to a speed of 0.65 m/s, then what is the speed of the second ball? PSYW

### Answer and Explanation:

Momentum conservation principles must be used to solve this collision problem. Set the initial momentum of the system equal to the final momentum of the system and solve for the unknown velocity. The information can be organized inside of a momentum table as shown below. (NOTE: units have been left off quantities to avoid "messiness".)

### 1st Ball

p=(0.05 kg)*(1.2 m/s)

p= 0.060

p=(0.05 kg)*(0.65 m/s)

p= 0.0325

### 2nd Ball

p=(0.05 kg)*(0.40 m/s)

p= 0.020

p=(0.05 kg)*v

p= 0.05*v

### Total

0.080
0.0325 + 0.05*v

0.080 = 0.0325 + 0.05*v

0.0475 = 0.05*v

v = (0.0475/0.05) = 0.95 m/s

Using Equations as a "Recipe" for Algebraic Problem-Solving (12 seconds)

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56. A 0.050-kg billiard ball moving at 1.5 m/s strikes a second 0.050-kg billiard ball which is at rest on the table. If the first ball slows down to a speed of 0.10 m/s, then what is the speed of the second ball? PSYW

### Answer and Explanation:

Momentum conservation principles must be used to solve this collision problem. Set the initial momentum of the system equal to the final momentum of the system and solve for the unknown velocity. The information can be organized inside of a momentum table as shown below. (NOTE: units have been left off quantities to avoid "messiness".)

### 1st Ball

p=(0.05 kg)*(1.5 m/s)

p= 0.075

p=(0.05 kg)*(0.1 m/s)

p= 0.005

### 2nd Ball

p=(0.05 kg)*(0.0 m/s)

p= 0

p=(0.05 kg)*v

p= 0.05*v

### Total

0.075
0.005 + 0.05*v

0.075 = 0.005 + 0.05*v

070 = 0.05*v

v = (0.070/0.05) = 1.4 m/s

Using Equations as a "Recipe" for Algebraic Problem-Solving (12 seconds)

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