Physics

1. An object has the moment of inertia of 1 kg m² rotates at a constant angular speed of 2 rad/s. What is the rotational kinetic energy of the object?

Known :

The moment of inertia (I) = 1 kg m²

The angular velocity (ω) = 2 rad/s

Wanted: The rotational kinetic energy (KE)

Solution :

The formula of the rotational kinetic energy :

KE = 1/2 I ω²

KE = the rotational kinetic energy (kg m²/s²), I = the moment of inertia (kg m²), ω = the angular velocity (rad/s)

The rotational kinetic energy :

KE = 1/2 I ω² = 1/2 (1)(2)² = 1/2 (1)(4) = 2 Joule

2. A 20-kg cylinder pulley with a radius of 0.2 m rotates at a constant angular speed of 4 rad/s. What is the rotational kinetic energy of the pulley?

Known :

Mass of cylinder pulley (m) = 20 kg

The radius of cylinder (r) = 0.2 m

The angular speed (ω) = 4 rad/s

Wanted : What is the rotational kinetic energy

Solution ;

Formula of the moment inertia of cylinder :

I = 1/2 m r²

I = the moment of inertia (kg m²), m = mass (kg), r = radius (meter)

The moment of inertia of cylinder pulley :

I = 1/2 (20)(0.2)² = (10)(0.04) = 0.4 kg m²

The rotational kinetic energy of the pulley :

KE = 1/2 I ω² = 1/2 (0.4)(4)² = (0.2)(16) = 3.2 Joule

3. A-10 kg ball with radius of 0.1 m rotates at a constant of 10 rad/s. What is the kinetic energy of the ball.

Known :

Mass of ball (m) = 10 kg

Radius of ball (r) = 0.1 m

Angular velocity (ω) = 10 rad/s

Wanted : The rotational kinetic energy

Solution :

Formula of the moment of inertia :

I = (2/5) m r²

I = moment of inertia (kg m²), m = mass (kg), r = radius (m)

Moment of inertia of the ball :

I = (2/5)(10)(0.1)² = (4)(0.01) = 0.04 kg m²

The rotational kinetic energy of the ball :

KE = 1/2 I ω² = 1/2 (0.04)(10)² = (0.02)(100) = 2 Joule

4. A 0.5-kg particle rotates at a constant angular speed of 2 rad/s. What is the rotational kinetic energy of the particle if the radius of circle is 10 cm.

Known :

Mass of particle (m) = 0.5 kg

The radius of ball (r) = 10 cm = 10/100 = 0.1 m

The angular speed (ω) = 2 rad/s

Wanted : The rotational kinetic energy

Solution :

Moment of inertia for particle :

I = m r² = (0.5)(0.1)² = (0.5)(0.01) = 0.005 kg m²

The rotational kinetic energy :

KE = 1/2 I ω² = 1/2 (0.005)(2)² = 1/2 (0.005)(4) = (0.005)(2) = 0.01 Joule

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1. A force F applied to a cord wrapped around a cylinder pulley. The torque is 2 N m and the moment of inertia is 1 kg m², what is the angular acceleration of the cylinder.

Known :

Torque (τ) = 2 N m

The moment of inertia (I) = 1 kg m²

Wanted: The angular acceleration of the cylinder

Solution :

Στ = I α

Στ = net torque, I = moment of inertia, α = angular acceleration

Angular acceleration of cylinder :

α = Στ / I = 2 / 1 = 2 rad/s²

2. A force F applied to a cord wrapped around a cylinder pulley. The magnitude of the force is 10 N, the radius of the cylinder is 0.2 m and the moment of inertia is 1 kg m2, What is the angular acceleration of the cylinder?

Known :

Force (F) = 10 N

Radius of cylinder (R) = 0.2 m

The moment of inertia (I) = 1 kg m²

Wanted: The angular acceleration of the cylinder.

Solution :

τ = F R

τ = torque, F = force, R = radius of cylinder

Torque :

τ = F R = (10 N)(0.2 m) = 2 N m

Στ = I α

Στ = net torque, I = moment of inertia, α = angular acceleration

Angular acceleration of cylinder :

α = Στ / I = 2 / 1 = 2 rad/s²

3. A force F applied to a cord wrapped around a cylinder pulley. The magnitude of force is 10 N, the radius of cylinder is 0.2 m and the mass of cylinder is 20 kg m^2,. What is the angular acceleration of the cylinder.

Known :

Force (F) = 10 N

Radius of cylinder (R) = 0.2 m

Mass of cylinder (M) = 20 kg

Wanted : Angular acceleration of cylinder

Solution :

τ = F R = (10 N)(0.2 m) = 2 N m

Moment of inertia :

I = 1⁄2 M R² = 1⁄2 (20)(0.2)² = 1⁄2 (20)(0.04) = 0.4 kg m²

Angular acceleration of cylinder :

α = Στ / I = 2 / 0.4 = 5 rad/s²

4. A 1-kg block hanging from a cord wrapped around a cylinder pulley. The moment of inertia of pulley is 1 kg m² and the radius of pulley is 0.2 m. What is the angular acceleration of the pulley. Acceleration due to gravity is 10 m/s².

Known :

Moment of inertia of pulley (I) = 1 kg m²

Mass of block (m) = 1 kg

Acceleration due to gravity (g) = 10 m/s²

Weight (w) = m g = (1 kg)(10 m/s²) = 10 kg m/s² = 10 N

Radius of pulley (R) = 0.2 m

Wanted : Angular acceleration

Solution :

Torque :

τ = F R = w R = (10 N)(0.2 m) = 2 N m

Moment of inertia :

I = 1 kg m²

Angular acceleration :

α = Στ / I = 2 / 1 = 2 rad/s²

5. A 1-kg block hanging from a cord wrapped around a cylinder pulley. The mass of pulley is 20 kg and the radius of pulley is 0,2 m. What is the angular acceleration of the pulley and the free fall acceleration of the block. Acceleration due to gravity is 10 m/s².

Known :

Mass of pulley (M) = 20 kg

Radius of pulley (R) = 0,2 m

Mass of block (m) = 1 kg

Acceleration due to gravity (g) = 10 m/s²

Weight (w) = m g = (1 kg)(10 m/s²) = 10 kg m/s² = 10 N

Wanted : the angular acceleration of the pulley and the free fall acceleration of the block.

Solution :

The torque :

τ = F R = w R = (10 N)(0.2 m) = 2 N m

The moment of inertia of cylinder pulley :

I = 1⁄2 M R² = 1⁄2 (20)(0.2)² = (10)(0.04) = 0.4 kg m²

The angular acceleration of the pulley :

α = Στ / I = 2 / 0.4 = 5 rad/s²

The free fall acceleration of the block :

a = R α = (0.2)(5) = 1 m/s²

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The moment of inertia of the particle

1. A 100-gram ball connected to one end of a cord with a length of 30 cm. What is the moment of inertia of ball about the axis of rotation AB? Ignore cord’s mass.

Known :

The axis of rotation at AB

Mass ball (m) = 100 gram = 100/1000 = 0.1 kg

The distance between ball and the axis rotation (r) = 30 cm = 0.3 m

Wanted: Moment of inertia of ball (I)

Solution :

I = m r² = (0.1 kg)(0.3 m)²

I = (0.1 kg)(0.09 m²)

I = 0.009 kg m²

2. A 100-gram ball, m₁, and a 200-gram ball, m₂, connected by a rod with a length of 60 cm. The mass of the rod is ignored. The axis of rotation located at the center of the rod. What is the moment of inertia of the balls about the axis of rotation?

Known :

Mass of ball 1 (m₁) = 100 gram = 100/1000 = 0.1 kg

The distance of ball 1 and the axis of rotation (r₁) = 30 cm = 30/100 = 0.3 m

Mass of ball (m₂) = 200 gram = 200/1000 = 0.2 kg

The distance of ball 2 and the axis of rotation (r₂) = 30 cm = 30/100 = 0.3 m

Wanted : moment of inertia of the balls

Jawab :

I = m₁ r₁² + m₂ r₂²

I = (0.1 kg)(0.3 m)² + (0.2 kg)(0.3 m)²

I = (0.1 kg)(0.09 m²) + (0.2 kg)(0.09 m²)

I = 0.009 kg m² + 0.018 kg m²

I = 0.027 kg m²

3. A 200-gram ball, m₁ and a 100-gram ball, m_2, connected by a rod with length of 60 cm. Ignore rod’s mass. The axis of rotation located at ball m₂. What is the moment of inertia of the balls. Ignore rod’s mass.

Known :

Mass of ball 1 (m₁) = 200 gram = 200/1000 = 0.2 kg

The distance between ball 1 and the axis of rotation (r₁) = 60 cm = 60/100 = 0.6 m

Mass of ball 2 (m₂) = 100 gram = 100/1000 = 0.1 kg

The distance between ball 2 and the axis of rotation (r₂) = 0 m

Wanted : Moment of inertia of the balls

Solution :

I = m₁ r₁² + m₂ r₂²

I = (0.2 kg)(0,6 m)² + (0.2 kg)(0)²

I = (0.2 kg)(0.36 m²) + 0

I = 0.072 kg m²

4. The mass of each ball is 100 gram, connected by cord. The length of cord is 60 cm and the width of cord is 30 cm. What is the moment of inertia of the balls about the axis of rotation. Ignore cord’s mass.

Known :

Mass of ball = m₁ = m₂ = m₃ = m₄ = 100 gram = 100/1000 = 0.1 kg

The distance between ball and the axis of rotation (r₁) = 30 cm = 30/100 = 0.3 m

The distance between ball 2 and the axis of rotation (r₂) = 30 cm = 30/100 = 0.3 m

The distance between ball 3 and the axis of rotation (r₃) = 30 cm = 30/100 = 0.3 m

The distance between ball 4 and the axis of rotation (r₄) = 30 cm = 30/100 = 0.3 m

Known : Moment of inertia

Solution :

I = m₁ r₁² + m₂ r₂² + m₃ r₃² + m₄ r₄²

I = (0.1 kg)(0.3 m)² + (0.1 kg)(0.3 m)² + (0.1 kg)(0.3 m)² + (0.1 kg)(0.3 m)²

I = (0.1 kg)(0.09 m²) + (0.1 kg)(0.09 m²) + (0.1 kg)(0.09 m²) + (0.1 kg)(0.09 m²)

I = 0.036 kg m²

The moment of inertia of rigid object

5. What is the moment of inertia of a 2-kg long uniform rod with length of 2 m. The axis of rotation located at the center of the rod.

Known :

Mass of rod (M) = 2 kg

The length of rod (L) = 2 m

Wanted: Moment of inertia

Solution :

The formula of the moment of inertia when the axis of rotation located at the center of long uniform rod :

I = (1/12) M L²

I = (1/12) (2 kg)(2 m)²

I = (1/12) (2 kg)(4 m²)

I = (1/12)(8 kg m²)

I = 8/12 kg m²

I = 2/3 kg m²

6. What is the moment of inertia of a 2-kg long uniform rod with a length of 2 m? The axis of rotation located at one end of the rod.

Known :

Mass of rod (M) = 2 kg

The length of rigid rod (L) = 2 m

Wanted: Moment of inertia

Solution :

The formula of the moment of inertia when the axis of rotation located at one end of the rod :

I = (1/3) M L²

I = (1/3) (2 kg)(2 m)²

I = (1/3) (2 kg)(4 m²)

I = (1/3)(8 kg m²)

I = 8/3 kg m²

7. A 10-kg solid cylinder with a radius of 0.1 m. The axis of rotational located at the center of the solid cylinder, shown in the figure below. What is the moment of inertia of the cylinder?

Known :

Mass of solid cylinder (M) = 10 kg

Radius of cylinder (L) = 0.1 m

Wanted: The moment of inertia

Wanted: The moment of inertia

Solution :

The formula of moment inertia when the axis of rotation located at the center of cylinder :

I = (1/2) M R²

I = (1/2) (10 kg)(0.1 m)²

I = (1/2) (10 kg)(0.01 m²)

I = (1/2)(0.1 kg m²)

I = 0.05 kg m²

8. A 20-kg uniform sphere with the length of 0.1 m. The axis of rotation located at the center of the sphere shown in the figure below.

Known :

Mass of sphere (M) = 20 kg

The radius of sphere (L) = 0.1 m

Wanted: a moment of inertia

Solution :

The formula of the moment of inertia when the axis of rotation located at the center of the sphere :

I = (2/5) M R²

I = (2/5)(20 kg)(0.1 m)²

I = (2/5)(20 kg)(0.01 m²)

I = (2/5)(0.2 kg m²)

I = 0.4/5 kg m²

I = 0.08 kg m²

9. A 2-kg rectangular thin plate with a length of 0.5 m and width of 0.2 m. The axis of rotation located at the center of the rectangular plat shown in the figure below. What is the moment of inertia of the rectangular?

Known :

Mass of rectangular plat (M) = 2 kg

The length of plat (a) = 0.5 m

The width of plat (b) = 0.2 m

Wanted : Moment of inertia

Solution :

Formula of moment of inertia when the axis of rotation located at the center of plat :

I = (1/12) M (a² + b²)

I = (1/12)(2)(0.5² + 0.2²)

I = (2/12)(0.25 + 0.04)

I = (1/6)(0.29)

I = 0.29/6 kg m²

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1. A force P is applied to one end of a beam with a length of 2 m. What is the magnitude of the torque? The axis of rotation at point A.

Known :

Force (F) = 10 N

Length of AB (r_AB) = 2 m

Force F is perpendicular to the beam.

The lever arm (l) = r_AB sin 90^o = (2 m)(1) = 2 m

Wanted: The torque about the axis of rotation

Solution :

The torque :

τ = F l = (10 N)(2 m) = 20 N m

The plus sign because the beam rotates counterclockwise rotation.

2. The length of a beam AB is 2 m and the magnitude of force F is 10 N. What is the magnitude of the torque? The axis of rotation at point A.

Known :

Force (F) = 10 N

Length of AB (r_AB) = 2 m

The lever arm (l) = r_AB sin 60^o = (2 m)(0.5√3) = √3 m

Wanted: The torque about the axis of rotation

Solution :

The torque :

τ = F l = (10 N)(√3 m) = 10√3 N m

The plus sign because the force F causes the beam rotates counterclockwise rotation.

3. The length of a beam is 2 m. The magnitude of F₁ is 10 N and the magnitude of F₂ is 15 N. Determine the net torque about the center of the beam.

The axis of rotation at the center of the beam.

Known :

Force 1 (F₁) = 10 N

The distance between F₁ and the center of beam (r₁) = 1 m

The lever arm 1 (l₁) = r₁ sin 90^o = (1 m)(1) = 1 m

Force F₁ is perpendicular to the beam.

Force 2 (F₂) = 15 N

The distance between F₂ and the center of the beam (r₂) = 1 m

Force F₂ is perpendicular to the beam.

The lever arm 2 (l₂) = r₁ sin 90^o = (1 m)(1) = 1 m

Wanted : The net torque about the axis of rotation

Solution :

The torque 1 :

τ₁ = F₁ l₁ = (10 N)(1 m) = 10 N m

The plus sign because the force of F₁ causes the beam rotates counterclockwise rotation.

The torque 2 :

τ₂ = F₂ l₂ = (15 N)(1 m) = -15 N

The minus sign because the force F₂ causes the beam to rotates clockwise.

The net torque :

Στ = τ₁ – τ₂ = 10 – 15 = – 5 N m

The minus sign because the beam to rotates clockwise.

4. The length beam AB is 2 m, The magnitude of F₁ is 10 N and the magnitude of F₂ is 10 N. Determine the net torque about the center of the beam.

The axis of rotation at the center of the beam.

Known :

Force 1 (F₁) = 10 Nn

The distance between F₁ and the center of the beam (r₁) = 1 m

The lever arm 1 (l₁) = r₁ sin 60^o = (1 m)(0.5√3) = 0.5√3 m

Force 2 (F₂) = 10 N

The distance between F₂ and the center of the beam (r₂) = 1 m

Force F₂ is perpendicular to the beam.

The lever arm 2 (l₂) = r₂ sin 90^o = (1 m)(1) = 1 m

Wanted : The net torque about the axis of rotation

Solution :

The torque 1 :

τ₁ = F₁ l₁ = (10 N)(0.5√3 m) = 5√3 = 8.7 N.m

The plus sign because the force F₁ causes the beam to rotates clockwise.

The torque 2 :

τ₂ = F₂ l₂ = (10 N)(1 m) = -10 N m

The minus sign because the force F₂ causes the beam to rotates clockwise.

The net torque :

Στ = τ₁ – τ₂ = 8.7 – 10 = – 1.3 N m

The minus sign because the net force causes the beam to rotates clockwise.

5. The length of a beam is 10 m, the magnitude of F₁ is 10 N, the magnitude of F₂ is 10 N and the magnitude of F₃ is 15 N. The distance between point A and point C is 7.5 m. The Force F₂ located at the center of the beam. Determine the net torque about the point C located at 2.5 m from the point B.

The axis of rotation located at point C

Known :

Force 1 (F₁) = 10 N

The distance between F₁ and point C (r₁) = 2.5 m

Force F₁ is perpendicular to the beam.

The lever arm 1 (l₁) = r₁ sin 90^o = (2.5 m)(1) = 2.5 m

Force 2 (F₂) = 10 N

The distance between F₂ and point C (r₂) = 2.5 m

The force F₂ is perpendicular to the beam.

The lever arm 2 (l₂) = r₂ sin 90^o = (2.5 m)(1) = 2.5 m

Force 3 (F₃) = 15 N

The distance between F₃ and point C (r₃) = 7.5 m

The force F₃ is perpendicular to the beam.

The lever arm 3 (l₃) = r₃ sin 90^o = (7.5 m)(1) = 7.5 m

Wanted : The net torque about the axis of rotation

Solution :

The torque 1 :

τ₁ = F₁ l₁ = (10 N)(2.5 m) = 25 N m

The plus sign because the force F₁ causes the beam to rotates clockwise.

The torque 2 :

τ₂ = F₂ l₂ = (10 N)(2.5 m) = 25 N m

The plus sign because the force F₂ causes the beam to rotates clockwise.

The torque 3 :

τ₃ = F₃ l₃ = (15 N)(7.5 m) = -112..5 N m

The minus sign because the force F₃ causes the beam to rotates clockwise.

The net torque :

Στ = τ₁ + τ₂ – τ₃ = 25 + 25 – 112.5 = – 62.5 N m

The minus sign because the net force causes the beam to rotates clockwise.

6. The length of a beam is 10 m, the magnitude of F₁ is 10 N, the magnitude of F₂ is 10 N and the magnitude of F₃ is 10 N. Determine the net torque about point A, located 5 m from the point of application of force F₁.

The axis of rotation at point A.

Known :

Force 1 (F₁) = 10 N

The distance between F₁ and point A (r₁) = 5 m

The lever arm 1 (l₁) = r₁ sin 60^o = (5 m)(0.5√3) = 2.5√3 m

Force 2 (F₂) = 10 N

The distance between F₂ and point A (r₂) = 0

The force F₂ is perpendicular to the beam.

The lever arm 2 (l₂) = r₂ sin 90^o = (0)(1) = 0

The force 3 (F₃) = 10 N

The distance between F₃ and point A (r₃) = 10 m

The lever arm 3 (l₃) = r₃ sin 30^o = (10 m)(0.5) = 5 m

Wanted : the net torque about the axis of rotation

Solution :

The torque 1 :

τ₁ = F₁ l₁ = (10 N)(2.5√3 m) = 25√3 = 43.3 N m

The plus sign because the force F₁ causes the beam to rotates clockwise.

The torque 2 :

τ₂ = F₂ l₂ = (10 N)(0) = 0

The torque 3 :

τ₃ = F₃ l₃ = (10 N)(5 m) = -50 N m

The minus sign because the force F₃ causes the beam to rotates clockwise.

The net torque :

Στ = τ₁ + τ₂ – τ₃ = 43.3 + 0 – 50 = – 6.7 N m

The minus sign because the net force causes the beam to rotates clockwise.

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1. A 500-gram object, A, moving at 10 m/s and 200-gram object, B, moving at 12 m/s. The objects approach each other and collide. If the speed of object A after the collision is 6 m/s, what is the speed of object B after the collision?

Known :

Mass of object 1 (m₁) = 500 gram = 0.5 kg

Mass of object 2 (m₂) = 200 gram = 0.2 kg

Initial velocity of object 1 (v₁) = -10 m/s

Initial velocity of object 2 (v₂) = 12 m/s

The final velocity of object 1 (v₁’) = 6 m/s

The plus and minus sign indicates that the objects moves in opposite direction.

Wanted : the final velocity of object 2 (v₂’)

Solution :

m₁ v₁ + m₂ v₂ = m₁ v₁’ + m₂ v₂’

(0.5)(-10) + (0.2)(12) = (0.5)(6) + (0.2)(v₂’)

-5 + 2.4 = 3 + 0.2 v₂’

-2.6 = 3 + 0.2 v₂’

-2.6 – 3 = 0.2 v₂’

-5.6 = 0.2 v₂’

v₂’ = -5.6 / 0.2

v₂’ = -28 m/s

The speed of object 2 after collision is 28 m/s.

The plus and minus sign indicates that the objects move in the opposite direction.

2. Two equal-mass objects approach each other and collide. The speed of object 1 is 6 m/s and the speed of object 2 is 8 m/s. After the collision, object 2 moves leftward with speed of 5 m/s. What is the magnitude and direction of the velocity of object 1 after the collision?

Know :

Mass of object 1 (m₁) = m

Mass of object 2 (m₂) = m

Initial speed of object 1 (v₁) = -6 m/s

initial speed of object 2 (v₂) = 8 m/s

The final speed of object 2 (v₂’) = -5 m/s

Wanted : the magnitude and direction of object 1 after collision

Solution :

Formula of conservation of linear momentum :

m₁ v₁ + m₂ v₂ = m₁ v₁’ + m₂ v₂’

m v₁ + m v₂ = m v₁’ + m v₂’

m (v₁ + v₂) = m (v₁’ + v₂’)

v₁ + v₂ = v₁’ + v₂’

-6 + 8 = v₁’ – 5

2 = v₁’ – 5

2 + 5 = v₁’

v₁’ = 7 m/s

The speed of object 1 after collision (v₁’) is 3 m/s.

The plus and minus sign indicates that the objects move in the opposite direction.

3. A 1-kg ball 1 and 2-kg ball 2 have the same direction and collide inelastically. Before the collision, ball 1 moves with speed of 10 m/s and ball 2 moves with speed of 5 m/s. The speed of ball 2 after the collision is 4 m/s. Determine the speed of ball 1 after the collision.

Known :

Mass ball 1 (m₁) = 1 kg

Mass ball 2 (m₂) = 2 kg

The speed of ball 1 (v₁) = 10 m/s

The speed of ball 2 (v₂) = 5 m/s

The final speed of ball 2 (v₂’) = 4 m/s

The plus sign of the velocity indicates that the balls have the same direction.

Wanted : the final velocity of ball 1 (v₁’)

Solution :

m₁ v₁ + m₂ v₂ = m₁ v₁’ + m₂ v₂’

(1)(10) + (2)(5) = (1)(v₁’) + (2)(4)

10 + 10 = v₁’ + 8

20 – 8 = v₁’

v₁’ = 12 m/s

The speed of ball 1 after collision is 12 m/s.

[wpdm_package id=’1190′]

1. Linear momentum problems and solutions
2. Momentum and impulse problems and solutions
3. Perfectly elastic collisions in one dimension problems and solutions
4. Perfectly inelastic collisions in one dimension problems and solutions
5. Inelastic collisions in one dimension problems and solutions

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1. A 30-gram bullet moving at 30 m/s collide a 1-kg block at rest. Determine the speed of the block if the bullet and the block lock together as a result of the collision.

Known :

Mass of bullet (m₁) = 30 gram = 0.03 kg

Mass of block (m₂) = 1 kg

Initial speed of bullet (v₁) = 30 m/s

initial speed of block (v₂) = 0 (block at rest)

Wanted : the speed of bullet and block after collision (v’)

Solution :

m₁ v₁ + m₂ v₂ = (m₁ + m₂) v’

(0.03)(30) + (1)(0) = (0.03 + 1) v’

0.9 + 0 = 1.03 v’

0.9 = 1,03 v’

v’ = 0.9 / 1,03

v’ = 0.87 m/s

2. Billiard ball A of mass 2 kg moving with speed v_A = 6 m.s⁻¹ collides head-on with ball B of equal mass. If the collision is perfectly inelastic, what are the speeds of the two balls after the collision.

Known :

Mass A (m_A) = 2 kg

Mass B (m_B) = 2 kg

Initial speed of ball A (v_A) = -6 m/s

Initial speed of ball B (v_B) = 4 m/s

The plus and minus sign indicates that the objects moves in opposite direction.

Wanted: the speeds of the two balls after the collision. (v’)

Solution :

m_A v_A + m_B v_B = (m_A’ + m_B) v’

(2)(-6) + (2)(4) = (2 + 2) v’

-12 + 8 = 4 v’

-4 = 4 v’

v’ = -4 / 4

v’ = -1 m/s

3. m₁ = 3 kg and m₂ = 4 kg approach each other in opposite direction with speed of v₁ = 10 m/s and v₂ = 12 m/s. If the collision is perfectly inelastic, what are the speeds of the two balls after the collision.

Known :

Mass object 1 (m₁) = 3 kg

Mass object 2 (m₂) = 4 kg

The speed of object 1 (v₁) = -10 m/s

The speed of object 2 (v₂) = 12 m/s

Wanted : the speed after collision (v’)

Solution :

m₁ v₁ + m₂ v₂ = (m₁’ + m₂) v’

(3)(-10) + (4)(12) = (3 + 4) v’

-30 + 48 = 7 v’

18 = 7 v’

v’ = 18 / 7

v’ = 2.6 m/s

[wpdm_package id=’1157′]

1. Linear momentum problems and solutions
2. Momentum and impulse problems and solutions
3. Perfectly elastic collisions in one dimension problems and solutions
4. Perfectly inelastic collisions in one dimension problems and solutions
5. Inelastic collisions in one dimension problems and solutions

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1. A 200-gram ball, A, moving at a speed of 10 m/s strikes a 200-gram ball, B, at rest. What is the speed of ball A and ball B after the collision?

Known :

Mass of ball A (m_A) = 200 gram = 0.2 kg

Mass of ball B (m_B) = 200 gram = 0.2 kg

Speed of ball A before collision (v_A) = 10 m/s

Speed of ball B before collision (v_B) = 0

Wanted: speed of ball A (v_A’) and speed of ball B (v_B’) after a collision

Solution :

v_A’ = v_B = 0

v_B’ = v_A = 10 m/s

2. Two equal-mass objects approach each other, collide and then bounce off. The speed of mass A before the collision is 8 m/s and speed of mass B before the collision is 12 m/s. What is the magnitude and direction of the velocity of mass A and mass B after the collision!

Known :

Mass A (m_A) = m

Mass B (m_B) = m

Speed of mass A before collision (v_A) = -8 m/s

Speed of mass B before collision (v_B) = 12 m/s

The plus and minus sign indicates that the objects move in opposite direction.

Wanted: the magnitude and direction of the velocity of mass A (v_A’) and mass B (v_B’) after a collision

Solution :

In a perfectly elastic collision, if the objects of equal mass then the speed of A after the collision (v_A’) = the speed of B before collision (v_B) and the speed of B after the collision (v_B’) = the speed of A before collision (v_A).

If before collision, A moving rightward and B moving leftward, then, after collision, A moving leftward and B moving rightward.

The objects have equal mass and the collision is perfectly elastic so the both objects exchanged speed.

v_A’ = -v_B = -12 m/s

v_B’ = v_A = 8 m/s

The plus and minus sign indicates that the objects move in opposite direction.

3. A 2-kg object, A, moving at a speed of 10 m/s strikes a 4-kg ball, B, moving at a speed of 20 m/s. If the collision is perfectly elastic, what is the speed of object A and object B after the collision?

Known :

Mass A (m_A) = 2 kg

Mass B (m_B) = 4 kg

Speed of object A before collision (v_A) = 10 m/s

Speed of object B before collision (v_B) = 20 m/s

Wanted : the speed of object A (v_A’) and the speed of object B (v_B’) after collision

Solution :

In perfectly elastic collision, if the objects have equal mass and approach each other, the speed of the object after collision calculated using this formula :

Speed of object A after collision :

4. A 100-gram moving at 20 m/s strikes a wall perfectly elastic collision. What is the magnitude and direction of object’s velocity after collision.

Known :

Mass (m_A) = 100 gram = 0.1 kg

Mass of wall (m_B) = infinity

The speed of mass before collision (v_A) = 20 m/s

The speed of mass after collision (v_B) = 0

Wanted : the magnitude and direction of velocity after collision

Solution :

If v_B = 0, m_A very small and m₂ very big then v_A’ = -v_A and v₂’ = 0.

The plus and minus sign indicates that the objects move in the opposite direction.

5. Two balls, A with a mass of 2-kg and ball B with a mass of 5-kg, before and after the collision are shown in the figure below. If the collision is perfectly elastic, what is the speed of ball A before the collision?

Known :

Mass of ball A (m_A) = 2 kg

Mass of ball B (m_B) = 5 kg

Speed of ball B before collision (v_B) = 2 m/s

Speed of ball A after collision (v_A‘) = 5 m/s

Speed of ball B after collision (v_B‘) = 4 m/s

Both balls are move to the right so their speed signed positive.

Wanted : Speed of ball A before collision (v_A)

Solution :

m_A v_A + m_B v_B = m_A v_A‘ + m_B v_B‘

2(v_A) + (5)(2) = (2)(5) + (5)(4)

2(v_A) + 10 = 10 + 20

2(v_A) + 10 = 30

2(v_A) = 30 – 10

2(v_A) = 20

v_A = 20/2

v_A = 10 m/s

6. Object A and B travel along horizontal plane, as shown in figure below. If the collision is perfectly elastic and the speed of object B after collision is 15 m/s¹, what is the speed of object A after collision.

Known :

Mass of object A (m_A) = 5 kg

Mass of object B (m_B) = 2 kg

Speed of object A before collision (v_A) = 12 m/s

Speed of object B before collision (v_B) = 10 m/s

Speed of object B after collision (v_B‘) = 15 m/s

Both objects, before and after collision, move to right so their velocity signed positive.

Wanted : Speed of object A after collision (v_A‘)

Solution :

m_A v_A + m_B v_B = m_A v_A‘ + m_B v_B‘

(5)(12) + (2)(10) = (5)v_A‘ + (2)(15)

60 + 20 = (5)v_A‘ + 30

80 = (5)v_A‘ + 30

80 – 30 = (5)v_A‘

50 = (5)v_A‘

v_A‘ = 50/5

v_A‘ = 10 m/s

[wpdm_package id=’1156′]

1. Linear momentum problems and solutions
2. Momentum and impulse problems and solutions
3. Perfectly elastic collisions in one dimension problems and solutions
4. Perfectly inelastic collisions in one dimension problems and solutions
5. Inelastic collisions in one dimension problems and solutions

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1. A small ball is thrown horizontally with a constant speed of 10 m/s. The ball hits the wall and reflected with the same speed. What is the change in linear momentum of the ball?

Known :

Mass (m) = 0.2 kg

Initial speed (v_o) = -10 m/s

Final speed (v_t) = 10 m/s

The plus and minus sign indicates that the objects moves in opposite direction.

Wanted : the change in linear momentum (Δp)

Solution :

Formula of the change in linear momentum :

Δp = m v_t – m v_o = m (v_t – v_o)

The change in linear momentum :

Δp = 0.2 (10 – (-10)) = 0.2 (10 + 10)

Δp = 0.2 (20)

Δp = 4 kg m/s

2. A 10-gram ball falls freely from a height, hits the floor at 15 m/s, then reflected upward at 10 m/s. Determine the impulse!

Known :

Mass (m) = 10 gram = 0.01 kg

Initial velocity (v_o) = -15 m/s

Final velocity (v_t) = 10 m/s

Wanted : Impulse (I)

Solution :

The impulse (I) equals the change in momentum (Δp)

I = m v_t – m v_o = m (v_t – v_o)

Impulse :

I = 0.01 (10 – (-15)) = 0.01 (10 + 15)

I = 0.01 (25)

I = 0.25 kg m/s

3. A 200-gram ball thrown horizontally with a speed of 4 m/s, then the ball was hit in the same direction. The duration of the ball in contact with the bat is 2 milliseconds and the ball speed after leaving the bat is 12 m/s. The magnitude of force exerted by the batter on the ball is …

Known :

Mass (m) = 200 gram = 0.2 kg

Initial velocity (v_o) = 4 m/s

Final velocity (v_t) = 12 m/s

Time interval (t) = 2 milliseconds = (2/1000) seconds = 0.002 seconds

Wanted : The magnitude of the force (F)

Solution :

Formula of impulse :

I = F t

Formula of the change in momentum :

m v_t – m v_o = m (v_t – v_o)

The impulse (I) equals the change in momentum (Δp)

I = Δp

F t = m (v_t – v_o)

F (0.002) = (0.2)(12 – 4)

F (0.002) = (0.2)(8)

F (0.002) = 1.6

F = 1.6 / 0.002

F = 800 Newton

[wpdm_package id=’1155′]

1. Linear momentum problems and solutions
2. Momentum and impulse problems and solutions
3. Perfectly elastic collisions in one dimension problems and solutions
4. Perfectly inelastic collisions in one dimension problems and solutions
5. Inelastic collisions in one dimension problems and solutions

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1. An object travels at a constant 10 m/s. Calculate the linear momentum of the object.

Known :

Mass (m) = 1 kg

Velocity (v) = 10 m/s

Wanted : linear momentum (p)

Solution :

Formula of the linear momentum :

p = m v

p = linear momentum, m = mass, v = velocity

The linear momentum :

p = m v = (1)(10) = 10 kg m/s²

2. A 2-kg block and a 4-kg blocks moves at a constant 20 m/s. What is the linear momentum of the blocks.

Known :

Mass of block A (m_A) = 2 kg

Mass of block B (m_B) = 4 kg

Velocity of block A (v_A) = 20 m/s

Velocity of block B (v_B) = 20 m/s

Wanted : Linear momentum of block A (p_A) and linear momentum of block B (p_B)

Solution :

Linear momentum of block A :

p_A = m_A v_A = (2)(20) = 40 kg m/s

Linear momentum of block B :

p_B = m_B v_B = (4)(20) = 80 kg m/s

3. A 2-kg block moves at a constant 2 m/s and a 2-kg block moves at a constant 4 m/s. Determine the linear momentum of the blocks.

Known :

Mass of block A (m_A) = 2 kg

Mass of block B (m_B) = 2 kg

Velocity of block A (v_A) = 2 m/s

Velocity of block B (v_B) = 4 m/s

Wanted : the linear momentum of blocks

Solution :

Linear momentum of block A :

p_A = m_A v_A = (2)(2) = 4 kg m/s

Linear momentum of block B :

p_B = m_B v_B = (2)(4) = 8 kg m/s

4. A 5-kg object at rest. What is the linear momentum of the momentum.

Known :

Mass (m) = 5 kg

Object’s velocity (v) = 0

Wanted : the linear momentum (p)

Solution :

p = m v = (5)(0) = 0

[wpdm_package id=’1155′]

1. Linear momentum problems and solutions
2. Momentum and impulse problems and solutions
3. Perfectly elastic collisions in one dimension problems and solutions
4. Perfectly inelastic collisions in one dimension problems and solutions
5. Inelastic collisions in one dimension problems and solutions

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Power is the rate at which work is done in a given period. Mathematically, power is the work/time ratio.

P = W/t

Description: P = power (Joule/second = Watt), W = work (Joule), t = time interval (second)

Based on this equation, it can be concluded that the greater the work rate, the greater the power. On the other hand, the smaller the work rate, the smaller the power. Work rate refers to the rate at which work is done.

Power is a scalar. The SI unit of power is the Joule/second. The Joule/second = Watt (abbreviated as W), named so to pay homage to James Watt. The British imperial unit for power is foot-pound per second.

This unit is too small for practical purposes, so the greater unit horsepower (abbreviated as hp) is used. One horsepower = 550 foot-pound per second = 764 Watt = ¾ kilowatt.

The amount of work can also be expressed in power x time units, for instance, kilowatt-hour or kWh. One kWh refers to the work done at a constant rate of 1 kiloWatt for one hour.

1. A 50-kg person runs up the stairs 10 meters high in 2 minutes. Acceleration due to gravity (g) is 10 m/s². Determine the power.

Known :

Mass (m) = 50 kg

Height (h) = 10 meters

Acceleration due to gravity (g) = 10 m/s²

Time interval (t) = 2 minute = 2 (60) = 120 seconds

Wanted : Power (P)

Solution :

Formula of power :

P = W / t

P = power, W = work, t = time

Formula of Work :

W = F s = w h = m g h

W = work, F = force, w = weight, d = displacement, h = height, m = mass, g = acceleration due to gravity

W = m g h = (50)(10)(10) = 5000 Joule.

P = W / t = 5000 / 120 = 41.7 Joule/second.

2. Calculate the power required of a 60-kg person who climbs a tree 5 meters high in 10 seconds. Acceleration due to gravity is 10 m/s².

Known :

Mass (m) = 60 kg

Height (h) = 5 meters

Acceleration due to gravity (g) = 10 m/s²

Time interval (t) = 10 seconds

Wanted : Power

Solution :

Work :

W = m g h = (60)(10)(5) = 3000 Joule

Power :

P = W / t = 3000 / 10 = 300 Joule/second.

3. A rotary comedy with a power of 300 watts and period of 5 minutes rotates 5 rounds. The energy it uses is ….

A. 15 kJ

B. 75 kJ

C. 90 kJ

D. 450 kJ

Known :

Power (P) = 300 Watt = 300 Joule/second

Period (T) = 5 minutes = 5 (60 seconds) = 300 seconds

Number of rotation = 5

Wanted: Energy used by the rotary comedy

Solution :

The correct answer is D.

[wpdm_package id=’1190′]

1. Work was done by force problems and solutions
2. Work-kinetic energy problems and solutions
3. Work-mechanical energy principle problems and solutions
4. Gravitational potential energy problems and solutions
5. The potential energy of elastic spring problems and solutions
6. Power problems and solutions
7. Application of conservation of mechanical energy for free fall motion
8. Application of conservation of mechanical energy for up and down motion in free fall motion
9. Application of conservation of mechanical energy for motion on a curved surface
10. Application of conservation of mechanical energy for motion on an inclined plane
11. Application of conservation of mechanical energy for projectile motion

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