Physics - Inelastic Collisions

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Inelastic Collisions

Table of Contents

Introduction

Problems

Answers

Introduction

Inelastic Collision	A collision in which Kinetic Energy is not conserved. Some of the Initial Kinetic Energy may be converted into other forms of Energy commonly Potential Energy or Thermal Energy. It is also possible for Potential Energy to be changed into Kinetic Energy making the Total Kinetic Energy greater than the Initial Kinetic Energy. This occurs when Chemical or Nuclear Energy is released.
Completely Inelastic	A collision in which the two objects stick together. The Kinetic Energy is not necessarily all transferred into other forms of energy because there can be some kinetic energy left. For example, when a football player is tackled, they do not instantly stop. Although the Kinetic Energy is not conserved, the total energy is conserved. Total momentum is always conserved.

The ballistic pendulum to the left is used to measure the speed of a bullet with a mass of m. The target is a large block of wood or other material of mass M, which is suspended by a length, l, to its center of mass. After the collision in which the projectile is imbedded in the target, the center of mass is raised to a maximum height of h. Determine the relationship between the initial speed of the projectile, v₁, and the height, h.

Assuming the collision time is extremely short and the projectile comes to rest before mass M has moved significantly, there is not net external force and momentum is conserved.

mv₁ = (M + m)v'

v' is the speed of the block and embedded projectile immediately after the collision before significant movement.

After the block begins to move gravity exerts a net force tending to pull it back to the original position. We cannot use Conservation of Momentum because of this but can use Conservation of Mechanical Energy. Initially we have only Kinetic Energy and at its maximum height we have only Potential Energy.

1/2 ( M + m )v'² = ( M + m )gh

Solve this equation for v'

v' = ( 2gh )^1/2

Substitute into the first equation

mv₁ = ( M + m ) ( 2gh )^1/2

Solve for v₁

( M + m ) ( 2gh )^1/2 ( M + m )v₁ = ---------------------------- = --------------- ( 2gh )^1/2
m m

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Problems

1.	A 20.0 g bullet is shot at a 2.00 kg block of wood in a ballistic pendulum. The maximum height reached is 150.0 cm, what was the initial velocity of the bullet?
2.	A 15.0 bullet is shot at 3.50 kg block of wood in a ballistic pendulum. If the bullet is traveling 250. m/s and the length to the center of mass is 3.00 m long, what is the horizontal component of the pendulum's displacement.
3.	An chemical explosion breaks a small asteroid into two pieces. The larger piece has twice the mass of the smaller piece. If the explosion released 5.60 x 10⁶ kJ, how much energy did each piece acquire.
4.	A 120. kg tackle moving 5.00 m/s collides and holds on to the quarterback, 80.0 kg, moving at right angle to the tackle at 8.50 m/s. What is their velocity after the collision?
5.	Uranium-232 ---> Thorium-228 + alpha The alpha particle emitted has a Kinetic Energy of 5.3 MeV (1 MeV = 1.6 x 10^-13 J). Calculate the momentum and the Kinetic Energy of the Thorium-228. Calculate the total energy released in the decay. Assume the mass numbers are the masses of the particle (Valid because we can only justify 2 significant figures).

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Answers

1.	A 20.0 g bullet is shot at a 2.00 kg block of wood in a ballistic pendulum. The maximum height reached is 150.0 cm, what was the initial velocity of the bullet?
	( M + m ) ( 2gh )^1/2 ( M + m ) v₁ = ---------------------------- = --------------- ( 2gh )^1/2 m m ( 0.0200 kg + 2.00 kg )v₁ = -------------------------------- ( 2 x 9.80 m s^-2 x 0.150 m )^1/2 = 173 m/s 0.0200 kg

2.	A 15.0 bullet is shot at 3.50 kg block of wood in a ballistic pendulum. If the bullet is traveling 250. m/s and the length to the center of mass is 3.00 m long, what is the horizontal component of the pendulum's displacement.
	mv₁ = (M + m)v' Conservation of Momentum before and after collision. 1/2 ( M + m )v'² = ( M + m )gh Conservation of Kinetic Energy before and after movement. Solve Conservation of Kinetic Energy for the common term. v' = ( 2gh )^1/2 Substitute into Conservation of Momentum. mv₁ = ( M + m ) ( 2gh )^1/2 Solve for h m²v₁² = ( M + m )² 2gh m²v₁² ( 0.0150 kg )² x ( 250. m/s )² h = -------------------- = ------------------------------------------------------- = 0.0581 m ( M + m )² 2g ( 3.50 kg + 0.0150 kg )² x 2 x 9.80 m s^-2 The diagram to the left shows the arc created in the ballistic pendulum. *Observe the right triangle formed by the light blue line which is the horizontal displacement, the segment of red line, l - h, and the hypotenuse complete red line of r.* *horizontal displacement = ( l² - ( l - h* )² )^1/2 horizontal displacement = ( ( 3.00 m )² - ( 3.00 m - 0.0581 m )² )^1/2 horizontal displacement = 0.588 m**

3.	An chemical explosion breaks a small asteroid into two pieces. The larger piece has twice the mass of the smaller piece. If the explosion released 5.60 x 10⁶ kJ, how much energy did each piece acquire.
	0 = m₁v'₁ + m₂v'₂ Conservation of Momentum PE_explosion = 1/2 m₁v'₁² + 1/2 m₂v'₂² Conservation of Mechanical Energy m₁ = 2m₂ Large piece is twice the mass of the smaller piece. For Momentum to be conserved, the lighter mass must have twice the velocity of the heavier mass. Of course they will be traveling in opposite directions. KE_{small particle} = 1/2 ( 1/2 mass of large ) x ( twice velocity of large )² The smaller particle will have twice the Kinetic Energy of the larger particle. Thus it will have 2/3 of the total Kinetic Energy. Kinetic Energy_{small particle} = 2/3 x 5.60 x 10⁶ KJ = 3.73 x 10⁶ KJ Kinetic Energy_{large particle} = 1/3 x 5.60 x 10⁶ KJ = 1.87 x 10⁶ KJ

4.	A 120. kg tackle moving 5.00 m/s collides and holds on to the quarterback, 80.0 kg, moving at right angle to the tackle at 8.50 m/s. What is their velocity after the collision?
	Conservation of Momentum - X m_qv_q = m_qv'Sin + m_tv'Sin = ( m_q + m_t ) v'Sin Conservation of Momentum - Y m_tv_t = m_qv'Cos + m_tv'Cos = ( m_q + m_t ) v'Cos Solve Conservation of Momentum - X for Sin m_qv_q Sin = -------------------- ( m_q + m_t ) v' Solve Conservation of Momentum - Y for Cos m_tv_t Cos = -------------------- ( m_q + m_t ) v' Divide Conservation of Momentum - X by Conservation of Momentum Y m_qv_q -------------------- Sin ( m_q + m_t ) v' m_qv_q -------- = ----------------------- = Tan = --------- Cos m_tv_t m_tv_t ---------------------- ( m_q + m_t ) v' Solve for . m_qv_q 80.0 kg x 8.50 m/s = Tan^-1 ( -------- ) = Tan^-1 ( -------------------------- ) = 48.6^o m_tv_t 120. kg x 5.00 m/s Solve Conservation of Momentum - X for v' ( You can also use Y ) m_qv_q = ( m_q + m_t ) v'Sin m_qv_q 80.0 kg x 8.50 m/s v' = ------------------------- = ---------------------------------------- = 4.53 m/s ( m_q + m_t ) Sin ( 80.0 kg + 120. kg ) Sin 48.6^o They are moving 4.53 m/s at 48.6^o to the left of the path of the Tackle.

5.	Uranium-232 ---> Thorium-228 + alpha The alpha particle emitted has a Kinetic Energy of 5.3 MeV (1 MeV = 1.6 x 10^-13 J). Calculate the momentum and the Kinetic Energy of the Thorium-228. Calculate the total energy released in the decay. Assume the mass numbers are the masses of the particle (Valid because we can only justify 2 significant figures).
	0 = m_Thv'_Th + m_Hev'_He Conservation of Momentum KE_He = 1/2 m_Hev'_He² Kinetic Energy Definition KE_Th = 1/2 m_Thv'_Th² Kinetic Energy Definition KE_decay = KE_He + KE_Th Conservation of Mechanical Energy Solve Conservation of Momentum for v'_Th. - m_Hev'_He 4.0 amu v'_He v'_Th = --------------- = -------------------- = 0.018 v'_He m_Th 228 amu Kinetic Energy of Thorium-228 relative to alpha particle. KE_Th 228 amu x ( 0.018 v'_He )² -------- = ----------------------------------- = 0.018 KE_He 4.0 amu x v'_He² The Thorium particle has 0.018 the Kinetic Energy of the Alpha Particle. KE_Th = 0.018 x 5.3 MeV = 0.095 MeV Convert Units and Solve Kinetice Energy of Thorium for v'_Th. 1.6 x 10^-13 J 0.095 MeV x ------------------- = 1.5 x 10^-14 J 1 MeV 1.00 g 1 kg 228 amu x ------------------------ x ------------ = 3.8 x 10^-25 kg 6.02 x 10²³ amu 1000 g KE_Th = 1/2 m_Thv'_Th² 2 KE_Th v'_Th = ( ------------- )^1/2 = 8.9 x 10 ⁴ m/s m_Th Substitute into the Momentum of Thorium Equation. p_Th = m_Thv'_Th = 3.8 x 10^-25 kg x 8.9 x 10⁴ m s^-1 = 3.4 x 10^-20 kg m s^-1

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