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Inelastic collision

An inelastic collision is one in which the internal kinetic energy changes (it is not conserved).

[link] shows an example of an inelastic collision. Two objects that have equal masses head toward one another at equal speeds and then stick together. Their total internal kinetic energy is initially 1 2 m v 2 + 1 2 m v 2 = m v 2 . The two objects come to rest after sticking together, conserving momentum. But the internal kinetic energy is zero after the collision. A collision in which the objects stick together is sometimes called a perfectly inelastic collision    because it reduces internal kinetic energy more than does any other type of inelastic collision. In fact, such a collision reduces internal kinetic energy to the minimum it can have while still conserving momentum.

Perfectly inelastic collision

A collision in which the objects stick together is sometimes called “perfectly inelastic.”

The system of interest contains two equal masses with mass m. One moves to the right and the other moves to the left with the same magnitude of velocity represented by V. Due to this their total momentum and net force remains zero. The internal kinetic energy is mv power 2. After collision the system of interest has no net velocity, no total momentum and no internal kinetic energy. This is true for all inelastic collisions.
An inelastic one-dimensional two-object collision. Momentum is conserved, but internal kinetic energy is not conserved. (a) Two objects of equal mass initially head directly toward one another at the same speed. (b) The objects stick together (a perfectly inelastic collision), and so their final velocity is zero. The internal kinetic energy of the system changes in any inelastic collision and is reduced to zero in this example.

Calculating velocity and change in kinetic energy: inelastic collision of a puck and a goalie

(a) Find the recoil velocity of a 70.0-kg ice hockey goalie, originally at rest, who catches a 0.150-kg hockey puck slapped at him at a velocity of 35.0 m/s. (b) How much kinetic energy is lost during the collision? Assume friction between the ice and the puck-goalie system is negligible. (See [link] )

The first picture shows an ice hockey goal keeper of mass m 2 bent on his knees, turning to the left on a frictionless ice surface with zero velocity and a hockey puck of mass m 1 and velocity V 1 moving toward the right. The total momentum of the system is p 1 which is the momentum of the puck and the net force is zero. The second picture shows the goalie to catch the puck. The puck moves with velocity V 1prime and the goalie with velocity V 2 prime and their magnitudes are equal. The momentum of the puck is p 1 prime and the goalie is p 2 prime. The total momentum remains same as before collision. But the kinetic energy after collision is lesser than the kinetic energy before collision. This is true for inelastic collisions.
An ice hockey goalie catches a hockey puck and recoils backward. The initial kinetic energy of the puck is almost entirely converted to thermal energy and sound in this inelastic collision.

Strategy

Momentum is conserved because the net external force on the puck-goalie system is zero. We can thus use conservation of momentum to find the final velocity of the puck and goalie system. Note that the initial velocity of the goalie is zero and that the final velocity of the puck and goalie are the same. Once the final velocity is found, the kinetic energies can be calculated before and after the collision and compared as requested.

Solution for (a)

Momentum is conserved because the net external force on the puck-goalie system is zero.

Conservation of momentum is

p 1 + p 2 = p 1 + p 2 size 12{p rSub { size 8{1} } +p rSub { size 8{2} } = { {p}} sup { ' } rSub { size 8{1} } + { {p}} sup { ' } rSub { size 8{2} } } {}

or

m 1 v 1 + m 2 v 2 = m 1 v 1 + m 2 v 2 . size 12{m rSub { size 8{1} } v rSub { size 8{1} } +m rSub { size 8{2} } v rSub { size 8{2} } =m rSub { size 8{1} } { {v}} sup { ' } rSub { size 8{1} } +m rSub { size 8{2} } { {v}} sup { ' } rSub { size 8{2} } } {}

Because the goalie is initially at rest, we know v 2 = 0 size 12{v rSub { size 8{2} } =0} {} . Because the goalie catches the puck, the final velocities are equal, or v 1 = v 2 = v size 12{ { {v}} sup { ' } rSub { size 8{1} } = { {v}} sup { ' } rSub { size 8{2} } =v'} {} . Thus, the conservation of momentum equation simplifies to

m 1 v 1 = m 1 + m 2 v . size 12{m rSub { size 8{1} } v rSub { size 8{1} } = left (m rSub { size 8{1} } +m rSub { size 8{2} } right )v'} {}

Solving for v size 12{v'} {} yields

v = m 1 m 1 + m 2 v 1 . size 12{v'= { {m rSub { size 8{1} } } over {m rSub { size 8{1} } +m rSub { size 8{2} } } } v rSub { size 8{1} } } {}

Entering known values in this equation, we get

v = 0.150 kg 70.0 kg + 0.150 kg 35 .0 m/s = 7 . 48 × 10 2 m/s . size 12{v'= left ( { {0 "." "150"`"kg"} over {"70" "." 0`"kg"+0 "." "150"`"kg"} } right ) left ("35" "." 0`"m/s" right )=7 "." "48" times "10" rSup { size 8{ - 2} } `"m/s" "." } {}

Discussion for (a)

This recoil velocity is small and in the same direction as the puck’s original velocity, as we might expect.

Solution for (b)

Before the collision, the internal kinetic energy KE int size 12{"KE" rSub { size 8{"int"} } } {} of the system is that of the hockey puck, because the goalie is initially at rest. Therefore, KE int size 12{"KE" rSub { size 8{"int"} } } {} is initially

Practice Key Terms 2

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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