Momentum Problem Set PDF

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Problem set on momentum with comprehensive calculations and final answers....

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Guiyab, DJ B. STEM 12 Darwin A

8.3 Given: A = 55 kg x m/s South B = 60 kg x m/s and is 60O North of East or B= 30 kg x m/s + 51.96152423 kg x m/s as (x + y) component C = 30 kg x m/s West Solution: (a) Let: A = momentum A Ax = x-component of momentum A Ay = y-component of momentum A C = momentum C Cx = x-component of momentum C Cy = y-component of momentum C Solutions and formula: • Equation for x and y components of momentum A + C = ( 0 − 30 kg ∙ m/s) + (−55 kg ∙ m/s + 0) A + C = (−30 kg ∙ m/s) + (−55 kg ∙ m/s)

Final answer: The x and y component of momentum composing A and C are -30 kg x m/s and -55 kg x m/s. (b) Let: B = momentum B

Bx = x-component of momentum B By = y-component of momentum B C = momentum C Cx = x-component of momentum C Cy = y-component of momentum C

Solutions and formula: • Equation for x and y components of momentum B + C = ( 30 kg ∙ m/s − 30 kg ∙ m/s) + (51.96152423 kg ∙ m/s + 0) B + C = (0 kg ∙ m/s) + (51.96152423 kg ∙ m/s)

Final answer: The x and y component of momentum composing B and C are 0 kg x m/s and 51.96 kg x m/s. (c) Let: A = momentum A Ax = x-component of momentum A Ay = y-component of momentum A B = momentum B Bx = x-component of momentum B By = y-component of momentum B C = momentum C Cx = x-component of momentum C Cy = y-component of momentum C

Solutions and formula: • Equation for x and y components of momentum

A + B + C = (0 kg ∙ m/s + 30 kg ∙ m/s − 30 kg ∙ m/s) + (−55 kg ∙ m/s + 51.96152423 kg ∙ m/s + 0) A + B + C = (0 kg ∙ m/s) + (−3.03847577 kg ∙ m/s) Final answer: The x and y component of momentum composing A, B and C are 0 kg x m/s and -3.03847577 kg x m/s.

8.19 Given: Mass of the squid in the water in cavity = 7.50 kg Mass of the water in cavity = 1.65 kg Velocity of the squid after releasing the water = 2.70 m/s Solution: (a) Let: M1= mass of the squid with water in cavity V1= velocity of the squid with water in cavity M2= mass of the squid after releasing the water in cavity V2= velocity of the squid after releasing the water in cavity Mw= mass of the water in cavity Vw= velocity of the water expelling the speed g = acceleration due to gravity M = mass VO = initial velocity or speed Solutions and formula: • Formula for calculating the velocity of the water expelling from squid. 7.50 kg x 0 m/s = (5.85 kg x 2.70 m/s) − (1.65 kg x Vw) 0 kg ∙ m/s = 15.795 m/s − 1.65 kg x Vw Vw = 9.572727273 Final answer: The needed speed of the squid needed to expel the water to achieve a speed of 2.70 m/s to escape is 9. 57 m/s. (b) Let:

KEnet = kinetic energy of the squid created by this maneuver M1= mass of the squid without water V1= velocity of the squid after expelling the water M2= mass of the water in cavity V2= velocity of the squid needed to expel the water

Solutions and formula: • Formula for the kinetic energy. 𝐾𝐾 = 1/2 (5.85 𝐾𝐾 )(2.70 𝐾 /𝐾) 2 + 1/2 (1.65 𝐾𝐾)(9.572727273 𝐾/𝐾) 2 VF = 3.533049674 m/s Final answer: The kinetic energy of the squid created by this maneuver is 96.89J.

8.41 Given: Mass of yellow car travelling east is 950 kg Maa of red car travelling north is 1900 kg Speed of the slide is 16.0 m/s Direction of slide is 24.0° east of north

Solution: Let: M1=mass of the car travelling east M2=mass of the car travelling north V1i= initial velocity of the car travelling east V2i= initial velocity of the car travelling north VF= final velocity after the collision or the slide’s speed

Solutions and formula: • Formula for the collision. (950 kg)V1i + (1900 kg)( 0 m/s) = (950 kg + 1900 kg)16cos 24° V1i = 19.52335887 m/s

(950 kg)(0 m/s) + (1900 kg)V2i = (950 kg + 1900 kg)16sin24° V1i = 21.92509098 m/s Final answer: The initial velocity of the car travelling east before the collision is 19.52 m/s, whereas the initial velocity of the car travelling north befor the collision is 21.93 m/s

8.47 Given: Mass of block A = 6.00 kg Mass of block B = 10.50 kg Initial velocity of block A = 7.00 m/s Solution: (a) Let: MA = mass of block A MB = mass of block C VA = initial velocity of block A VB = initial velocity of block B VF = final velocity of the blocks since they are moving together U = maximum energy

Solution and formula: • Formula for final velocity. • Formula for the maximum energy MAVA + MBVB = (MA + MB)VF (6. 00 kg)(7.0 m/s) + ((10.50 kg)(0 m/s) = (6.00 kg + 10.50 kg)VF (42 kg ∙ m/s) + 0 kg ∙ m/s = (16.50 kg)VF VF =2.5454545 m/s U =12(6.00 kg)(7 m/s)2 −12(6.00 kg + 10.50 kg)(2.5454545 m/s)2 U = 147 J − 53.45454355 U = 93.54545645 J

Final answer: The maximum energy stored in the spring bumpers is 93.55 J the speed of each block are both 2.55 m/s. (b) Let: MA = mass of block A MB = mass of block C VAi = initial velocity of block A VBi = initial velocity of block B VAF = final velocity of block A VBF = final velocity of block B Solutions and formula: • Formula for velocity. • Formula for coefficient of restitution. ����� + ����� = ����� + ����F (6.0 kg)(7 m/s) + (10.50 kg )(0 m/s) = (6.0 kg)VAF + (10.50 kg)VBF (6.0 kg)VAF + (10.50 kg)VBF = 42 kg ∙ m/s VAF + (1.75 kg)VBF = 7 m/s

Let: VAF + (1.75kg)VBF = 7 m/s be the relation 1. � = ��� − ��� / ��� – ��� ��� − ��� = ��� − ��f (7m/s) − (0 m/s) = VBF − VAF VBF − VAF = 7 m/s

Let: VBF − VAF = 7 m/s be the relation 2. ��� + (�. �� ��)��� = � �/� −��� + ��� = � �/s

Answer: VAF= -1.9090909 m/s VBF= 5.090909

Final answer: The final velocity of block A and block B after moving apart are -1.91 m/s and 5.09 m/s.

8.51 Let: m = the mass of object A and m = the mass of object B a

b

(a) & (b) 2m v = 2m v + mBv a

a

a

(2m - m ) v = (-2m ) v 2 a

b

2

2

a

2

a

The law of energy can be transformed into following equations. 4m v = 4m v + m v 2

A

A

A

2

B

2

(4m - m ) v = 4m v 2 A

(2m

2

B

A− m

A

A

)2

B

(4mA− mB)=

4m

2

A

4mA

4m - 4mA m + mB2= 4mA2- mA mB 2 A

B

m = 3mA B

(A) Final answer: Object B (B) Final answer: Object B is three times greater than Object A

(c) m v = - m v + 3m v/2 A

A

A

A

2v = - 2v + 3v A

v = 1/2 v = 3 A

Final answer: The velocity of object A after the collision is 3 m/s...